Cengage MECHANICS 1

Motion in One Dime,nsion 4.17 2But v = u + at =} s = 1 [u + u + atl x I Vav = -54--- 21-0 = -2.5 -l s = ut + 21 aI' rus 3. Here X, = 5 m, X, = -5 m, I, = 4 s, I, = 7 s So the required value of average velocity is given by Third Equation of Motion Vav = -5-5 -10 -3.3 ms- 1 --- = -- = 7- 4 3 s= 1 + OB) aD = 1 +' OB) aD (i) tlWltiURID' The velocity versus time curve of a mov~ 2(OA 2(DC As OA = CD BC ing point is shown in figure. Find the retardation of the par· AC ticle, (ii) Now, acceleration a = slope of velocity time graph = c DB - DC or aD = DB - DC ,_, 20 a A0~1~0~:2:0:3-0+4,0,-5t0,6-0~7~0~~D -~:a:D:-- Using equations (i) and (ii) 1 y- 2 (DC+DB) Fig. 4.46 or DB' - DC' = 2as Sol. The slope of velocity-time graph represents acceleration or retardation of the particle during motion. If slope is positive or v2 - u2 = 2as or v2 = u2 + 2as it represents acceleration and if slope is negative it represents retardation, The section CD of the graph represents retardation lH!1tlijjllD The position versus time graph for a cer· and magnitude of retardation is tain particle moving along the x·axis is shown in Fig. 4.45. lill = Change in velocity = 60 = 2 mis' Find the average velocity'in the time intervals (i) 0 to 2 s (ii) 2 s to 4 s (iii) 4 s to 7 s. Time taken (70 - 40) x(m) IDID!RE: The velocity versus time graph ofa linear 10 motion is shown in Fig, 4.47. Find the distance and displace. 8 6 ment from the origin after 8 s. 4 2 t4 0 -2 8 -4 tin s -6 Fig. 4,47 Fig. 4.45 Sol. We are given the values of time, but the values of the Sol. The area under velocity-time graph represents distance and displacement of the particle, position are to be obtained from the graph corresponding to the 1. Distance travelled by the particle is the total length of the given time interval under consideration, We know that slope of path travelled by the particle, Hence distance travelled by the particle the displacement-time graph represents velocity, S = Total sum of arca under the graph X2 -Xl = 1 2 + 4) x 4 + 1 + 4) x 2= 12 + 6 = 18 m Formula to be used: Vav = - - - where Xl and X2 are the -( -(2 22 t2 - t] 2. The displacement of the particle is the change of position, initial and final positions\" respectively and I, and I, are initial The displacement of the particle is positive for area above and final values of time, respectively, the time axis and negative for the area below the time axis.. Hence displacement of the particlc is = x,1. Here X, 0 m, = 10 m, II =0 S, I, = 2 s Therefore the required average velocity is given by 10 - 0 -l Vav = 2 _ 0, = 5 ms 2. Here XI = 10m, X2 =5 m, I, = 2 S, 12 = 4:s L'>x = I + 4) x 4 - 1 + 4) x 2 = 12 - . So the required value of average velocity is given by 6= 6m 2(2 2(2

4.18 Physics for IIT·JEE: Mechanics I 5. At t = 0, a particle starts from rest and moves along a straight line, whose acceleration-time graph is shown in ,---+ Concept Application Exercise 4.4 f----, Fig. 4.50. 1. a. What can you say about velocity in each of the follow· 5[----, ing position-time graphs? f--,--~--,--,--l> tis (i) (ii) (iii) 2 46 8 -5 ______ ..L.._ _ _I Fig. 4.50 Convert this graph into velocity-time graph. From the velocity-time graph, find the maximum velocity attained (iv) (v) (vi) by the particle. Also find from v-I graph, the displacement xx and distance travelled by the particle from 2 to 6 s. ~'~i 6. Answer the following question giving reasons in brief: (vii) (viii) Is the time variation of position, shown in the Fig. 4.51, observed in nature\" (IIT-JEE,1979) Fig. 4.48 i b. The slope of velocity-time graph is equal to accelera- Position (x)--Jl> tion. (True/False) Fig. 4.51 c. What does the area under acceleration-time graph rep- resent? A ball is thrown upwards with an initial d. Can velocity-time graph be parallel to velocity axis? velocity of 10 ms-I. Considering bighcst point as the origin (Yes/No). Why\" and vertically downward direction as positive direction, find the signs of position, velocity and acceleration of the object c. What is the slope of v-I graph in uniform motion? under motion during it\" upward and dmvnward jOUl'ney. 2. a. A ball is thrown vertically upward. After some time it rNegative (0,0) Highest point returns to the thrower. Draw the ve/ocity-tirne graph and speed-time graph. ll'ositive b. A ball is dropped from some height. After rebounding Fig. 4.52 from the 11001' it ascends to the same height. Draw the velocity-time graph and speed-time graph. Sol. As the particle always remains below (0, 0) and downward direction is positive, so position x will remain positive dtJring 3. A body starts at t = 0 with veloeity II and travels along both upwards and downwards motion. a straight line. The body has a constant acceleration a, Draw the acceleration-time graph, velocity-time graph During upward motion, velocity is negative and during down- and displacement-time graph from t = 0 to t = 10 s for ward motion velocity is positive. the following cases: a. 1I =8 ms..!.} , a =2 ms- 2 Acceleration is positive during both upward and downward motion, It is because acceleration due to gravity (g) is downward h. u =8 ms,--l, a = -2 ms·-2 and our downward direction is positive. c. u = --Sms-l,a =2ms-2 d. u = -8 ms'-'), a = -2 ms-2 4. See I:\"ig. 4.49 and find the average acceleration in first 20 s. alms··2 20 to o \"---\"'toc----o2\"'o--e3-0\".. tis Fig. 4.49 (I-Iint: Area under a-t graph is equal to change in velocity)

Motion in One Dimension 4.19 A car takes 20 s to move around a round- Dn = u + a2: (21! - 1) about of radius 14 m. Calculate 1. average speed, and Putting the values, Dn = 5 - 22: [2 x 2 - I] = 2111, which is 2. magnitude of average velocity. Sol. the required distance travelled. 1. Average speed is the rata of total distance to total time taken. A particle starts from rest with a constant So, distance covered in one revolution = 2ITr acceleration a = 1 m/s2• 22 1. Determine tbe velocity after 2 s. 2. Calculate the distance travelled in 3 s. = 2 x -- x 14 = 88 m. 3. Find the distance travelled in the third second. 7 4. lftbe particle was initially moving with a velocity of 5 mis, Average speed = 88/20 = 4.4 mls then find the distance travelled in the third second. 2. In one complete revolution displacement of car is zero. I v;v I = I?ispI~(:erncI1t = ...~.. = 0 ms-' Time 10 Sol. 1. We need the velocity-time relation to solve this question. A train travels from city A to city B witb Formula used: v = II + al So wc have v = 0 + (I )(2) = 2 m/s a constant speed of 10 ms·I and returns back to city A with a constant speed of 20 ms· I • Find its average speed during 2. Here u = 0, I = 3 s. a = I m/s2, s = ? the entire Journey. Sol. Given VA = lOms-l, V/3 = 20 111S-- 1 We need the distance-time relation. In a question where mo- Average speed is always calculated from the ratio of total tion is under acceleration and it is not circular or rotatory, the distance to the total time taken. distance-time relation which is used is s = +I Average speed is not the arithmetic average of speeds in a ut \"2(/(2. journey. So. we have s = 0 + 2I: x I X (3)2 = 4.5 m Consider, the distance between the two cities A and B == x lTI. 3. Here n ;;;:; 3, Dn ;;;:; ? -=-Time taken by the train to travel from A to B = = fl (say) Again we need the distance-time relation. But here the dis- x 10 tance has to be determined in a part.icular second. So in such Time taken to come back from Jj to A = 20 = 12 (say) +a a case the formula used is DII = U 2(2n - 1). .'. average speed = T_o..t_al.d_is-ta-n-c-e = x +x 2x So, we have Dn = 0 + I x 3 - I) = 2.5 m. whieh is Total time t] + tz 2:(2 the required distance travelled. 2 x 20 4. Here u = 5 mls 3 So, we have Dn = 5 + 1 x 3- I) = 7.5 m which is the re- 2:(2 quired distance travelled. 1\\vo halls of different masses m 1 and 1n2 Consider a particle initially moving with arc dropped from two different heights h [ and It2, respec- tively. Find the ratio of time taken hy the two balls to drop a velocity of 5 ms- 1 which starts decelerating at a constant through these distances. rate of 2 ms·2• Sol. Analysis of situation: As the two balls have been dropped 1. Determine the time at which the particle becomes station- ary. so the initial velocity in both the cases must he zero. Also the 2. Find the distance travelled in tbe 2nd second. two heights involved are different Hence the time taken in both Sol. the cases will be different due to the same initial velocity. 1. Here u;;;:; 5 ITIs·- 1, v;;;:; 0, a;;;:; -2 ms- 2, [;;;:;? ,. = ul + I We have to find the time when the particle becomes stationary, Formula used: y 2al2 i.e, it comes to rest. It means the final velocity will be zero. Symbol '(l' carries a negative sign as the statement involves Now for ball 1: y = hI, t = fl, as the motion is under gravity, retardation or deceleration. so a = g (taking the downward direction as positive) Formula used: v = u + at (Formula that relates the initial h, = O I g l 2, I2 and tinal velocities to time.) I+2: =} h, = 2:g/, So, we have 0 = 5 - 21 =} I = 2.5 s )2h'=} I, = (i) 2. Here u ;:;; 5 mIs, a ;;:; - 2 111S2 , n = 2, DIl ;;;:; ? g Formula used: For the distance travelled in the nth second and for ball 2: y = h 2, t = [2, as the motion is under gravity, so the relation used is a = g (taking the downward direction as positive) I gl,, ~ 2: 12 = -ii--Y+112 = 012 (ij) =}

4.20 Physics for IIT-JEE: Mechanics I (h,\"From equations (i) and (ii), we have!!. = train by 1 ms-I • After 50 s, train P crosses the engine of the train Q. Find out what was the distance between the trains Vhzt2 initially, provided the length of each train is 400 m. This expression conveys that the time of fall is independent of the mass. Sol. Let the initial distance between two trains = x m A child throws a ball up with an initial First of all we need to find out the total distance the two trains velocity of 20 ms-I • Find out the maximum height that the travelled in 50 s as the difference of these two distances must be ball can achieve and how long will it take to come back to the child's hands? equal to the initial distance between the two trains plus the sum = =Sol. Initial velocity 20 mis, final velocity 0, The final velocity of length of the two trains, is zero as ball must be temporarily at rest at the maximum height. Distance travelled by train P, Sp = ut + I 2 We take the point of projection as the origin and the upward ),at direction as positive direction. =} 20 x I x I x 502 = 2250 m We need [Maximum height = s = ?, Time = ?J 50+), Formula used: v2 - u2 = 2as which gives 02 - 202 = 2( - 10) s =} s = 20 m Distanee travelled by train Q, Here we have taken a = - g = -10 ms2 because the upward direction is positive. So the maximum height attained is 20 m. SQ = ut + I at 2 = 20 x 50 + I x Ox 502 = 1000 m To find time, formula used: v = u + at. Let us use this for- z ), mula from bottom to top, It gives: 0 = 20 - lOt =} t = 2 s, It means the ball went up Now, Sp - SQ = 2250 - 1000 = x + 800 to the top in 2 s. For motion under gravity the time of ascent and the time of descent are exactly equal. (Provided air resistance is x = 450m negligible,) So the total time of travel = (2 + 2) s = 4 s, A ball is dropped from the top of a high A balloon is at a height of 40 m and is buiildiing at t = O. At a later time I = 10, a second ball is thrown downward with initial speed Vo. Obtain an expres- ascending with a velocity of 10 ms- I • A bag of 5 kg weight is sion for the time t at which the two balls meet. dropped from it. When will the bag reach the surface of the Sol. Let the distance covered by the first ball = Yl (for vertical =earth? Given g 10 ms-2• distance symbol Y is preferred), Sol. Herc we can takcpoint of dropping as origin and downward Similarly the distance covered by the second ball = Y2, direction as positive. Given, h = +40ms~l, u = -IOms-\"l, ~ I~ a = g = + 10 ms-2 y~ LI. x '\"I I I {III \"\" Fig. 4.53 Fig. 4.54 The initial velocity of the bag will be same as that of the We can use the condition that at a particular time (as given) distances covered by both the balls are same, But the cauti<,>n is velocity of balloon. As it moves in opposite direction of the that the first ball has covered Yl in to time and second ball has covered yz in (t - to) time, balloon, so it will have negative sign with it. As we have values Let us assume that the origin be at the top of the building with of h, u and g, so time can be obtained by the distance formula, downward direction positive. If the two balls meet al time t, then +z~ -)- 1-+2 . For Yl, t = to, Vo = 0, a = g =} 12 c, formula used: S = ut at (where symbols have their Yl = ), gto usual meaning), For Y2, t = t - to, initial speed = va, a = g So 40 = -lOt + I l 0 t 2 =} 5t 2 - 10 t - 40 = 0 12 Z =} yz = vo(t - to) +),g (t - to) =} 5t 2 - 20 t -I- 10 t - 40 = 0 From the given condition, 5t(t -4)+ lO(t -4)=0 =} (t -4)(5t + 10)=0 12 12 =}t =4sand-2s -2g. t 2 get - to) As timc cannot bc negative, so time taken by body to reach + -Yl = Y2 = =} vo(t - to) the ground is t = 4 s, After simplifying and solving, we get Two trains P and Q are moving on par- allel tracks with a uniform speed of 20 ms- I , The driver =} t = [vo - gto/2] to Vo - gto of train P decides to overtake train Q and accelerates the [!!!lHll!l!IiI\" A balloou starts rising upward with a constant acceleration a and after time to second, a packet

is dropped from it which reaches the ground after t second Motion in One Dimension 4.21 (Fig. 4.55). Determine the valne of t. at which the ball reaches its maximum height, (b) the max- imum height, (c) the time at which the hall returos to the height from which it was thrown, (d) the velocity of the ball at this instant, (e) the velocity and position of the ball at t = 5.00 s. B IS\"\" 2 S YB= 20 m vyB= 0 ayE'\" 10 ms·..2 Fig. 4.55 c tc\"\"4 s Sol. Analysis of situation: t = 0 is the time when the balloon L YC\"\"'Om started rising up. At t = to when the packet is dropped. the bal- I v).c\"\" --20 ms--! I loon is moving up with velocity v = 0 + ato = ato. Hence initial aye = --10 mg·..2 velocity of the packet will be Un = ato (upward). As the balloon I has started rising upwards with constant acceleration, a, so after I I tg.to seconds its height from the ground is Yo = 2I(1 1, I, 12 I D to\"\"5.00s For packet: s = vt - 'igt =} -'iato = atot - 'igt I YD = ---22.5 m VyiJ '\"\"' --30 ms-! =} gt 2 - 2atot - atJ = 0 50.0m I ayD\"\" 10ms-2 I Solving the quadrat.ic equation, we get I ato I t= - I g I A particle, moving with uniform acceler- I ation from A to along a straight line, has velocities VI and I V2 at A and B, respectively. If C is the mid-point between A I \"«+ ·If. and B then determine the velocity of the particle at C. e> Sol. Let v be the velocity of the particle at C. Assume accelera- tion of the particle to be a and distance between A and B to be x. Fig. 4.57 ••Vi V v, Sol. (a) Analysis of situation: At position (B) velocity of the ball must be zero. It means velocity has changed hy 20 mls as it AC • is momentarily at rest at (B). Whenever an object goes up in the air, it is under the negative effect ofgravity; its velocity decreases •x B by 10 mls in each second (since the retardation or deceleration Fig. 4.56 is 10 ms'). So the ball must take 2 s to move from (A) to (B). • Formula used: vYH = vYA + ayt lAs we have to find time with To find the velocity at point C, we can easily express the given initial and final velocities]. ·desired result in terms of given and assumed quantities, i.e., VVB = 0 (velocity at B), vYA = 20 m/s (velocity at A). V!, V2 and x. The formula that easily.relates these quantities is a,. = -g = -IOm/s2, t = tB =? v2 _ u2 = 2as, x v; = 2a - Now from A to C: v2 - 2 From C to B: v,i - v2 = 2a -X o= 20 - JOt =} t = 2 s. 2 (b) Analysis of situation: The ball is thrown from (A). At (B) Solve to get: v = ) v 2 2+ 7 the velocity of the ball becomes zero. The time taken to reach 1 2 from (A) to (B) is 2 s. Hence the length A B will be equal to ~ A ball, thrown from the top of a build- maximum height. ing which is 50 m high, is given an initial velocity of 20.0 Formula to be used .IS Y =1Yo + Vo 2+ -ay.t ms-I straight upward, On its backward journey the ball just misses the edge of the roof on its way down, as shown 2' in Fig, 4.57, Considering the position (of projection of the Here it will be in the following form: ball) as origin andHme at A(tA) = 0, determine (a) the time + V.vAt + 1 2 2Ymax= = ay t Yll YA

4.22 Physics for IIT-JEE: Mechanics I Substituting values, )'A = 0, Vri\\. = 20 ms~l, )If) = Yc + V.vC t + 1 y t 2 a,=-g=-lOm/s2,1=2s, . -2a .I oi1 = -20 - 5 = -25 m we haveYB = 0 + (20.0) (2) + 2( -10)(2)2 = (40 - 20) = 0 - 20(1)+ -(-10) 2 bALiHml Plot the graphs for the variation of the =20 m Alternatively: . velocity with time in each of the following cases: Because thc avcragc velocity for this first interval is 10 mls (thc average of 20 mls and 0 m/s) and the ball travels for about 2 s, 1. when the acceleration has constant value. 2. when the acceleration increases with time. we expect the ball to travel about 20 m. 3. when the acceleration decreases with time. (c) There is no reason to believe that the ball's motion from (8) to (e) is anything other than the reverse of its motion from (A) Sol. to (8). The motion from (A) to ee) is symmetric. Thus, the time needed for it to go from (A) to (e) should be twice the time 1. When the acceleration has a constant value, then velocity will change linearly with time. Hence graph is a straight line eneeded for it to go from (A) to (8). Note that A and are at starting from origin. . Yo + vot + I same level. Formula to be used IS Y = 2ayl'. Where y = )'c = 0, Yo = YA = 0, V)'A = 20 mis, y A 'ar=-g=-IOm/s',t=tc . . 12 i Yc = YA + VyAt + zayt ,-v i;Substituting these values, we have 0= 0 + 20.0 t, - 5 o~----------~x This is a quadratic equation and so there must be two Fig. 4.58 solutions for t = fe. The equation can be factored to give Ic(20.0 - 51c) = 0 2. When acceleration is increasing with time, the slope of V- T graph will also increase with time. Due to this, the graph will One solution is tc = 0, COlTcsponding to the time the ball be a concave upward curve (Fig. 4.59). starts its motion. The other solution is te = 4 s, When the ball is back at thc height from which it was thrown (position e), the y coordinate is again zero. Note: It is dOl,bIe the vallie weca!clilaiedfor fB' y (d) Analysis of Situation: Again, we expect everything at (e) A to be the same as it is at (A), except that the velocity is now in the opposite direction. The value of t found in (e) can be inserted i\" into equation vf = Vi + at to give ,-o ke::-----,---.....x Formulae: vrc = VyA + ayl = 20.0 + (-lOx4) = -20.0 mls Fig. 4.59 The velocity of the ball when it arrives back at its original 3. When acceleration is decreasing with time, the slope of height is equal in magnitude to its initial velocity but opposite V- T graph also decreases accordingly. Hence the V - T in direction (indicated by -ve sign). graph will be a concave downward curve (Fig. 4.60). (e) Analysis of Situation: To find velocity and position at t = 5,00 S, we can use any position during motion as the initial y condition provided we have required values of velocity and time at that position. We can also consider B as OUf new initial position ,_ A besides A. i Case (i) When B is the initial position, then again using Vy = u y v +ayt o '-----------.....x Time of travel =(5.00 - 2) s =2 s =3 sand VyB = 0, Fig. 4.60 a,. = -g = -10 mis' VylJ = VyB +ayt :::::> VyD = -10 x 3 = -30 m/s liiiifiidllDi An object is in nniform motion along a Case (ii) When A is initial position, straight line. What will be the position-time graph for the motion of the object if: then 1= 5.00s, V\"A = 20 mis, a y = -g = -10 mis' =1. Xo +ve, v = + ve +lIyD = VrA ayt = 20 - 10 x 5 = -30 m/s 2. xo=+ve, v:-ve Position of ball at 10 = 5.00 s w.r.t. IA = 0, let us fut1her extend the idea of choosing initial time. Here we can consider 3. Xo =- ve, v =+ve tc as our new initial time. eSo time of travcl bctween and D = 5 - 4 = I s, y, = 0 a = -g = -lOm/s2, Vy' = -20 m/s

Motion in One Dimension 4.23 4. both Xo and v are negative? The letters Xo and v represent Consider the followiug vx-t graph to be parabolic. aeceleration-time graph and aualyse the mo- =position of the object at time t 0 and uuiform velocity tion of the particle from A to E. of the object, respectively. v\" Sol. The equation of motion of an object at any time I moving with uniform velocity along a straight line is given by c x = Xo + vt. E 1. When Xo :::: +ve, v :::: +ve Fig. 4.65 In this case x will take positive values with the passage of time. so the graph has to be a straight line in first quadrant. Sol. Acceleration-time graph is as follows: We know that slope of velocity-time graph is equal to acceleration. Slope at x A is maximum, decrea'ies linearly and- becomes zero at C and then starts increasing in negative side and becomes maximum 0\\----- negative at E. a Fig. 4.61 A B 2. When Xo= +ve. v = - ve c Here with increasing value of I. the value of x will go on decreasing and will take negative values. So graph is a straight D line starting from highest value ofx in question and will move E on into fourth quadrant x Fig. 4.66 So~ at A acceleration is maximum, at C zero and at E accel- eration is negative maximum, Analysis of motion: Fig. 4.62 I I IVx < 0, moving In -x IA POSItIve slope, direction, slOWing fA = 0 + .L. ...--= _~ a -\"\"--\"'\"\"-,...x 3. When Xo :::: -ve and v:::: +ve \" II so ax > 0 down - -~----f--- In this case. value of x will go on increasing with time as (VI) I! Vx = 0, t>a Instantaneously at will go on taking higher value. So the graph is straight line IB POSItIve slope, rest, about to move f8 - ...+--~ - - - - - - - - x v= 0 te ~----~~\",0--,-\"\"-\"..~----~-x1 ----+1-----vx> 0; C zero slope, that will start from most negative value of x in question and so ax > 0 In +x directIon will cross over to first quadrant where x has +vc values, moving in +x direction, at x so ax = 0 maximum speed H---' instantaneously at I Vx = 0; 0\\-+--\" 0 negative slope, rest, about to move to ....-.------ 1 so ax < 0 in -x direction v= 0 --.~~.----.-~~--I moving in - x , -4---q---- I IVx < 0; E negative slope, direction, t£ ._-.- -- .\"\"\"- ·· ..\"-,·!:'--------x so ax < 0 speeding up . Fig. 4.63 ~~~--~~~---------~ 4. When Xo = -ve and v = -ve 1. Between points A and B Values of x will go on decreasing with time Xo and vI both Acceleration is positive and decreasing in magnitude. Also will result in negative values. So the straight line will statt magnitude of velocity is decreasing, because velocity and from a negative value of Xo and wHl keep on taking negative acceleration are in opposite direction. values in the fourth quadrant. 2. At poiut B x The velocity is instantaneously zero at this moment. So parti- cle is at rest momentarily. But the particle is still accelerating 0\\----- as the slope has non-zero value, Acceleration is positive and decreasing in'magnitude, From here the particle will start --Xo moving along positive x direction, Fig. 4.64 3. Between points Baud C Velocity is positive and increasing. So the acceleration is positive but decreasing in magnitude,

4.24 Physics for IIT-JEE: Mechanics I 4. At point C Between these points. slope of x-I graph is increasing, it Velocity at point C is maximum as graph peaks at this point. means magnitude of velocity is increasing. It means velocity At this point slope is zero, therefore the acceleration is zero. and acceleration both are in same direction. So acceleration is positive. Note that acceleration is constant between A and B. S. Between points C and D Velocity goes on decreasing between C and D. Due to neg- Also the graph is concave up, so the acceleration is ative slope acceleration is also negative. Acceleration is in- positive. creasing in magnitude. Velocity and acceleration are in op- 2, At point B : posite direction. Particle is at origin at this point At this point slope of x-I 6. At point D graph is maximum, so velocity is maximum at this point. Velocity is zero at this point. But slope is still negative, so the Before point B acceleration is positive, but after point B acceleration is negative. Particle will start moving in negative acceleration will be negative as the slope of x-I graph will direction. Acceleration is increasing in magnitude. start decreasing after this (Fig, 4,70), 7, Between points D and E x=o Particle is moving in - ve direction. The slope~ of graph is negative in this region. It means the acceleration is negative; Fig, 4,70 but magnitude of slope is increasing, it means the magnitude of acceleration is increasing. Velocity and acceleration are in We cannot define acceleration at point B. same direction. 3. Between points Band C: Consider the following x-I graph to be Between these points, x has positive value, it means particle parabolic. velocity-time graph and acceleration-time is to the right of origin. But value of x is increasing, so the particle is moving away from origin. It means velocity of graph and analyse the motion of the particle regarding its the particle is positive. velocity and acceleration. Betweenthese points, slope ofx-I graph is decreasing, it means magnitude of velocity is decreasing. It means velocity xc and acceleration are in opposite directions. So acceleration is negative. Note that acceleration is constant between Band C. E v x • x~o ..q...--a Fig. 4,71 Fig. 4.67 Sol. Velocity-time graph and acceleration-time graph are as Also the graph is concave down, so acceleration is shown: negative. v a 4. At point C: B AB Particle is maximum away from origin at this point. At this point slope of x-I graph is zero, so velocity is zero. But C acceleration is negative as the graph is concave down. At A this point, the particle will change its direction of motion. E v=o X BcD I, D x=O +-a Fig. 4,68 Fig, 4.72 Analysis of motion: So the particle will start moving towards origin. I. Between points A and B: 5, Between points C and D: Between these points, x has negative value, it means particle Between these points, x has positive value, it means particle is to the left of origin. But value of x is decreasing, so the is to the· right of origin. But value of x is decreasing, so the particle is moving towards origin. It means velocity of the particle is moving towards origin. It means velocity of the particle is positive. particle is negative. -~-'-v ->.~--->,-----x V X -->-a x=o IC Fig, 4.69 x=o +-a Fig, 4,73

Motion in One Dimension 4.25 Between these points, magnitude of slope of x-I graph Note: is increasing, it means magnitude of velocity is increasing. but in negative direction. So acceleration is negative. Note .. Velo~ity iscltangingatpoinlf B, C,D,E,F. AtB, D, that acceleration is constant between C and D. PVellldtycha!lge$ sUi14ellly fr?m ~fgative t()positi\":e alldatC; Eve[ocity challges srnoothfyfrom jJ(jsitir~ Also the graph is concave down, so acceleration is to lIegative. . .••... ..• • . . . i ...•.. .. negat.ive, .. AtB,D, F velo~ity shallgesvery.lJuic.kly$~ (Mae,. 6. At point D: celeratioll mus/be very large. Particle is to the right of origin at this point. At this point slope of x-t graph is negative maximum, so velocity is Position-time graph: The diagram given itself conveys maximum at this point in negative direction. Before point D position-time graph (Fig. 4.75). acceleration is negative, but after point D acceleration will be positive (as the graph is concave up after D). We cannot ~c t,,,,p t,,,,,,,f: :,,,,F define acceleration at point D. , , y (m) 7. Between points D and E: Between these points, x has positive value, it means particle - - - r, - - - ' - - - - T,- - - . is to the right of origin. But value of x is decreasing, so the particle is moving towards origin. It means velocity of the o'-_-'L--_-'-_-'-_....L_L-__. t (s) particle is negative. Between these points, magnitude of slope of x-I graph Fig. 4.75 is decreasing, it means magnitude of velocity is decreasing, but in negative direction. So acceleration is positive. Note Velocity-time graph: As the slope of v-T graph changes thrice that acceleration is constant between D and E. from negative to positive during the bounces. so V- T graph must observe sharp changes at these points (Fig. 4.76). Also the graph is concave up, so acceleration is positive. iitbltlUlm A rubber ball is released from a height of about 1.5 m. It is caught after three bounces. Sketch graphs of its position, velocity and acceleration as fnnctions oftime. Provided positive y~direction as upward direction. o @1.5 ~® 6Y \\ ~ v] 10 o Fig. 4.76 'GO. (........·. . . .. °05 Acceleration-time graph: Slope of V- T graph remains same 0o0. .(..;.9.•.0...· 000C•g•.t0i..0. & (-I0 mls2 ) till bounces as velocity holds, i.e., it has single value during free tall. At the time of contact with fioorit changes sub- <' stantially during a very short time interval. So the acceleration will be large and positive which has to be represented in the form 0.0 of straight upward lines (Fig. 4.77). ® @ CD ,----I7l---7-c-~'---+-_H1I-f_ I (s) Fig. 4.74 ,,, ,, ,-----~----- __ 4 __ · Sol. ,,,, ,, ----~---. Checks the following points: \" --,.-_. ,,-----r----- 1. Position and time graph will have the same shape as given , , ,,\" \"\" graph. \"\" __ ,~_-J:L-_~ ~ \" 2. Slope of position-time graph gives velocity. Hence change in velocity can be observed as per the slope of position-time ----r\" --- T graph. L-_~'_~ 3. Slope of velocity-time graph gives acceleration. 4. Acceleration in free fall r~mains constant with time. II II \\2 ______ I ______ IL _____ J I _____ IL ____ I ___ I __ . ~ ~ ~ Fig. 4.77

4.26 Physics for IIT-JEE: Mechanics I EXERCISES Solutions on page·4.46 10. A train of 150 m length is going towards north direction at a speed of 10 mh A bird flies at a speed of 5 mls towards south 1. Fig. 4.78 shows a particle starting from point A, travelling direction parallel to the railway track. Find the time taken by upto B with a speed s. then upto point C with a speed 2 s and finally upto A with a speed of 3 s. Determine its average the bird to cross the train. speed. 11. A car starts moving with constant acceleration and covers the distance between two points ISO m apart in 6 s. Its speed as ~' it passes the second point is 45 mis, Find !20\" a. its acceleration. o B b. its speed when it was at the first point c. thc distance from the first point whcn it was at rest. 12. From a lift moving upward with a uniform acceleration a, a A man throws a ball vertically upwards with a velocity v relative Fig. 4.78 to the lift The ball comes back to the man after a time t, Show that a + g = 2v/L 2. A particle moving in a straight line covers half the distance 13. A balloon is ascending vertically with an acceleration of with speed of 3 m/s. The other half of the distance is covered l m/s2. Two stones are dropped from it at an interval of 2 s. in two equal time intervals with a speeds of 4.5 mls and 7.5 mis, respectively. Find the average speed of the particle Find the distance between them 1.5 s after the second stone during this motion. is released, 14. A train starts from station A with uniform acceleration at for 3. Find the ratio of the distance moved by a freely falling body some distance and then goes with uniform retardation ([2 for from rest in 4th and 5th second of its journey. some more distance to come to rest at station B. The distance 4. Two balls of different masses (one lighter and other heavier) between stations A and B is 4 km and the train takes IllS h to are thrown vertically upwards with the same speed. Which one will pass through the point of projection in their down- complete this journey, If accelerations are in km per minute ward direction with the greater speed? I + -I ::: 2. unit then show that: - al £12 5. A car runs at a constant speed on a circular track of radius 15. A stone is let to fall from a balloon ascending with an accel- 200 111, taking 62.S s on each lap. Find the average velocity and average speed on each lap. eration f. After t time a second stone is dropped. Prove that 6. A train accelerates from the rest for time t1 at a constant rate the distance between the stones after time (I since the second ex and then it retards at the constant rate f3 for time (2 and stone is dropped is 'Ii(f + g)l(t + 21'), comes to rest. Find the ratio: (11(2. 16. A stone falling from the top of a vertical tower has descended x rn when another is let to fall from a point y m below the 7. An athlete swims the length of 50 m pool in 20 s and makes top, If they hIll from rest and reach the ground together. show the return trip to the starting position in 22 s. Determine his average velocity in +(x V)2 a. the first half of the swim. that the height of tower is ~ m. b. the second half of the swim, 4x c. the round trip. 17. Divide a plane 10 m long and 5 m high into three parts so that a body starting from rest takes equal time to slide down these parts, Also find the time taken then, 8. A train stops at two stations d distance apart and takes time lR. The driver of a car moving at 30 mls suddenly sees a truck that is moving in the same direction at 10 m/s and is 60 m t on the journey from one station to the other. Assuming ahead. The maximum deceleration of the car is 5 m/s2• a. Will the collision occur if the driver's reaction time is zero? that its motion is first of uniform acceleration ex and then immediately of uniform retardation f3, show that -I + -f3I I' If so, when\" h. If the car driver's reaction time 0[0.5 s is included, what is a =-, the minimum deceleration required to avoid the collision? 2d 9. The speed of a train increases at a constant rate a from zero 19. A steel ball is dropped from the roof of a building, A man standing in front of a 1 m high window in the building notes to v and then remains constant for an interval and finally that the ball takes 0, I s to 1'111 from the top to boltom of the window The ball continues to fall and strike the ground, On decreases to zero at a constant rate f3. If I be the total distance striking the ground the ball gets rebounded with the same described, prove that the total time taken is + ,~1)[~+ ~] speed with which it hits the grouncL If the ball reappears at 2 a the bottom ofthe window 2 s after passing the bottom of the v fJ window on the way down, find the height of the bUilding.

Motion in One Dimension 4.27 20. A particle is dropped from the top of a tower h III high and at velocity the same moment another particle is projected upward from the bottom. They meet when the upper one has descended a Fig. 4.79 distance h/n. Show that the velocities or the two when they meet arc in the ratio 2:(n - 2) and that the initial velocity of e. The slope of positionldisplacement-time graph is equal to the particle projected up is J(I/2)ngh. _ _ _(Velocity/Speed). 21. An elevator whose floor to the ceiling distance is 2.50 m) starts ascending with a constant acceleration of 1.25 m/52. One f. The slope of distance-time graph is equal to second aftcr the start, a bolt begins falling from the elevator. __ .__.._(Velocity/Speed). Calculate: a. free rall time of the bolt. g. A position-tim,e graph is shown. Is it possible practically? b. the displacement and distance covered by the bolt during (Yes/No) IIp,,,i(ioo the frcc fall in the reference frame of ground. -~; 22. In a car race, car A takes a time t less than car B at the finish and passes the finishing point with a velocity v more than Fig. 4.80 the car B. Assuming that the cars start form rest and travel with constant accelerations al and a2, respectively, show that h. Which of the following position-time graphs represent negative velocity? v = (Jaia,) t. u. h. c. d. 23. A stone is dropped from the top of a cliff of height h. n second later a second stone is projected downward from the i. A position~time graph is shown in Fig 4.81. What does same cliiT with a vertically downward velocity u. Show that the slope of line segment PQ indicate. (instantaneous ve- the two stones will reach the bottom of the cliff together, if locity/Average velocity/Nothing) 8h(\" - gn)2 = gn'(2u - gnJ2• What can you say about the limiting value of n? xrl2iHr,HHj. 24. A particle moving in a straight line is observed to be at a \"I·~l distance 'a' from a marked point initially, at a distance 'b' : : time after an interval of n seconds, to be at a distance 'c' after Fig. 4.81 2n seconds and at a distance'd' after 311 seconds. Prove that if the acceleration is uniform, d - a= 3(c - hJ and that the j. What can you say about acceleration in each of the following velocity-time graphs') c + a - 21> I' 1· acceleration is equal to - - - ,- - . ~, LL, 11 (i) (ii) (iii) 25. Two motor cars start from A simultaneously and reach B after 2 h. The first car travelled half the distance at a speed ~[ ~I~\" , I' \" of 30 kmh- i and the other half at a speed of 60 kmh--'. The (iv) (v) (vi) second car started from rest and covered the entire distance 28. You are given some graphs in helow. Look at them care- with a constant acceleration. At what instant of time were the fully and find out which of the graphs will represent the one- speeds of both the vehicles same? Will one of them overtake dimensional motion, the other? 26. A train of length I = 350 III starts moving rectilinearly with constant acceleration a::: 3.0x 10--2 ms-- 2. t::.:::: 30 s after start, the locomotive headlight is switched on (event 1), and 60 s after this event the tail signal light is switched on (even! 2). a. Find the distance between these events in the reference frame fixed to the train and to the earth. h. How and at what constant velocity v relative to the earth must a certain reference frame R move for the two events to occur.in it at the same point? 27. a. Can position-time graph be a straight line parallel to po- sition axis? (Yes/No) b. Area enclosed by velocity-time graph is equal to___ ..__._._._.__.{displacement/distanceJ. c. Area enclosed by specd~time graph is equal to _~ ___(displacement/distanceJ. d. For a one dimensional motion, a velocity-time graph is shown in Fig. 4.79. Convert this graph into a speed~til11e graph.

4.28 Physics for IIT·JEE: Mechanics I ~'~(i) (Ii) Distance -1 Time (iii) Displacement (iv) Displacement Time Time !-70.!-.--'1~02--'3-\"74--75--;6\"--- I(s) Fig. 4.83 (vl spccd (vi) Velocity x B A c Time Time (vii) Velocity (viii) Total path length Time Time Fig. 4.84 29. Figure 4.82 given below shows the displacement-time graph 32. The diagram given below represents the motion of a biker in for a particle moving along a straight line path. the form of position-time graph (Fig. 4.85). 5 i i4 tOO 200 300 400 500 Fig. 4.85 § a. The time intervals during which the biker was stationary E3 are______. \"•\"u§ E 2 h. The time intervals during which the bike is moving in the positive x direction are'____ 0-(\"' c. The time interval for which bike is moving in negative x :=: direction is _____ L-~--~2---T3---4r---5~~6~-47~X 33. A physics professor leaves her house and walks along the sidewalk towards campus. Aftcr 5 min it starts raining and time(s) --+ she returns home. Her distance from her house as a function of time is shown in Fig. 4.86. Fig. 4.82 x(m) State True or False. IV a. Time during which the particle was at rest is a s to 2s. 400 300 b. The maximum velocity of the particle is -2.5 m/s2 . 200 30. A position-time graph for a particle moving along the x axis toO is shown in Fig. 4.83. State True or False. oc-l\"'.'-cL.1....J.-L..L.L.:L t (min) =a. The average velocity in the time interval t 1.5 s to 2345678 Fig. 4.86 t = 4.0 s is -2.5 mls. b. The instantaneous velocity at t = 2.00 s by measuring the sklpe of the tangent line shown in the graph is - 7.4 m/s. The velocity is zero at t = 4 s. 31. You are given the position-time graph of three different bod· ies A, B. and C (Fig. 4.84). Find which will have greater velocity and which will have least velocity.

Motion in One Dimension 4.29 At which of the labelled points is her velocity acceleration a VS. t, indicating numerical values at significant a. zero? points of the graph. b. constant and p.ositivc. c. constant and negative. t' d. increasing in magnitude. e. decreasing in magnitude. v (m/s) 34. The graph shown in Fig. 4.87 shows the velocity of a police officers motorcycle plotted as a function of time. v., (mJs) () 2 3 4 5 50 45 r(s)~ 40 Fig. 4.89 35 30 37. Thc velocity-time graph of a particle moving in a straight line is shown in the Fig. 4.90 given below. Find the displacement 25 and the distance travelled by the particle in 6 s. 20 1-.,-';\"-<· 15 v (m/sf 10 4 E 5 F o 2 456891O 1314 ' (S) 2 ------ ------- J J Fig. 4.87 o 3 4C 5 ID 1I I (s) a. The instantaneous accelerations at t :::: 3 S, at t ;;;::: 7 s, and A 1 2B 6 att = 11 s are____. ------ b. The distances covered by the officer in the first 5 s. first 9 s and first 13 s are _~~ G 35. A cat walks in a straight line. which we shall call the Fig, 4,90 x-axis with the positive direction to the right. As an observant physicist, you make measurements of this cat's motion and 38. Fig. 4.91 shows a graph of the acceleration of a model rail- construct a graph of the cat's velocity as a function of time road locomotive moving on the x-axis. Graph its velocity and (as shown in Fig. 4.88). coordinate as functions of time if x ;:::: 0 and Vx ;:::: 0 at t == O. Vx 2 (cmJs) -+-t-LL---L.-+-----\"-~+---''----'---'-- I (5) 8 o 5 10 15 20 25 )0 35 40 7 _________ LI_---'I 6 5 2 4 3 Fig. 4.91 2 39. A woman starts from her home at 9.00 a,m., walks with a I (s) speed of 5 kmlh on a straight road up to her office 2.5 km 0 2 34 56 7 away, stays at the office up to 5.00 p.m. and returns home by an auto with a speed of25 km/h. Plot the position-time graph Fig. 4.88 of the woman taking her home as origin. a. The eat's velocities att =4.0s andatt =7.0sare _ __ 40. A runner jogs along a straight road (the +x direction) for 30 b. The cat's acceleration at t :::: 3 s is ____.. min, travelling a distance of 6 km. She then turns around and c, The distance that the cat moves during the first 4.5 sand walks back towards her starting point for 20 min, travelling 2 km during this time. State True or False: from t =0 tot =7.5 s are _ _~ a. Final displacement of the funner relative to her starting d. Sketch clear graphs of the eat's acceleration and position position is 4 km b. Her average speed for the entire trip is 0.16 km/m. as functions of time, assuming that the cat started at the c, The averagc velocity for the entire trip is 0.16 km/m. ongin. 36. Starting at x = 0, a particle moves according to the graph of v d. The runner's average velocity while jogging is 0.4 km/m. vs. t shown in Fig. 4.89. Sketch a graph of the instantaneous e. Her average velocity while walking is 0.1 km/m. 41. At the instant the traffic light turns green. a car that has been waiting at an intersection starts ahead with a constant accel- eration of 3.20 m/s2. At the same instant a truck, travelling

4.30 Physics for IIT-JEE: Mechanics I with a constant speed of 20,0 mIs, overtakes and passes the c. cannot be zero d. depends upon the particle car. a. The car ovcl1akes the truck at a distance _ _ _from its 2. If the distance covered is zero, then displacement a. 111ust be zero starting point. b. mayor may not be zero h. Speed of the car when it overtakes the truck _____, c. cannot be zero c. Sketch an x-I graph of the motion of both vehicles, Take d. depends upon the particle x = 0 at the intersection. 3. The numerical value ofthe ratio ofaverage velocity to average d. Sketch a v,-I graph of motion of both the vehicles, speed is 42. A particle moves along the x axis. Its x coordinate varies with n. always less than one b. always equal to one time according to the expression x = -4t + 2t 2, where x is c. always more than one d. equal to or less than one in ill and t is in s. The position-time graph for this motion is shown in Fig. 4,92. Note that the particle moves in the 4. The numerical value of the ratio of instantaneous velocity to negative x direction for the first second of mot.ion, is at rest at instantaneous speed is moment t ;:::; 1 s and then heads back in the positive x direction a. always less than one for 1 > I s, b. always equal to one c. always more than one x(m) d. equal to or less than one 10 8 5. A body moves 4 111 towards east and then 3 m north, The - --\" t (s) displacement and distance covered by the body arc o 2 .1 4 a. 7m, 6 m b . 6 m, 5 m Fig, 4,92 c. 5 111, 7 01 d. 4 01, 3 01 °a. Displacement of the particle in the time intervals 1 = to 6. The location of a particle is changed, What can we say about I = J sand t = I s to t = 3 s arc ______, the displacement and distance covered by the particle? °b. The average velocity in the time intervals 1 = to t = I s and t = I s to I = 3 s are _______, a. Both cannot be zero e. Instantaneous velocity of the particle at 1 = 2.5 s is b, One of the two may be zero 43. The acceleration of a particle varies with time as shown in c. Both must be zero Fig, 4,93, d. Both must be equal -2 t (s) 7. The angle between velocity and acceleration during the re- Fig. 4.93 tarded motion is a. Find an expression for velocity in terms of t. Assume that v = 0 atl = 0, a. 180\" b. 400 c. 45° d. 0\" b. Calculate the displacement of the particle in the time in- 8. A particle moves with uniform velocity, Which of the fol- terval from I = 2 s to t = 4 s, lowing statements about the motion of the particle is true? a. Its speed is zero S()it.itions' on page 4.52 b. Its acceleration is zero c. Its acceleration is opposite to the velocity 1. If displacement of a particle is zero, the distance covered d. Its speed may be variable, a. must be zero b. mayor may not be zero 9. The magnitude of the displacement is equal to the distance covered in a given interval of time if the particle a. moves with constant acceleration along any path h. moves with constant speed c. moves in same direction with constant velocity or with variable velocity d. Ploves with constant velocity 10. A moving body is covering the displacement directly propor- tional to the square of the time, The acceleration of the body is a. increasing b. decreasing c. zero d, constant 11. The magnitude of average velocity it; equal to the average speed when a particle moves a. on a curved path h. in the same direction c. with constant acceleration d. with constant retardation

Motion in One Dimen'sion 4.31 12. If a particle moves with a constant velocity be equal to zero, where t is equal to a. its acceleration is positive b. its acceleration is negative 2a b. Iai a d, zero c. its acceleration is zero a, 3b c. 3b d. its speed is zero 21. A person travels along a straight road for the first half time with a velocity v, and the second half time with a velocity _ v, v, vV21. Then the mean velocity is given by 13. The displacement s of a particle is proportional to the first + 2I I power of time t, i.e., s ex t; then the acceleration of the particle a.v=--- b.-=-+- is a. infinite c. v = ,\"\",v,2 v,d, : = Vhi- V2 b. zero V2 c. a small finite value V d. a large finite value 22, A person travels along the straight road for half the distance with velocity v, and the remaining half distance with velocity 14. The ratio of the average velocity of a train during a journey V2. Then average velocity is given by to the maximum velocity between two stations is a, V,V2 b• vi c. (v, +V2) d, 2v,v, 2 a.=l b.>l c.<l d.>or<l - (v, + V2) vf v\"23, A body covers one-third of the distance with a velocity 15. The distance travelled by a particle in a straight line motion is directly proportional to t'/2, where t = time elapsed. What the second one-third of the distance with a velocity V2 and the is the nature of motion? a. Increasing acceleration remaining distance with a velocity V3. The average velocity b. Decreasing acceleration c. Increasing retardation is + +VI V2 V3 d. Decreasing retardation a. 3 16. An athlete completes half a round of a circular track of radius + +b, 3Vl V2 V3 R, then the displacement and distance covered by the athlete VI V2 V2V3 V3Vj are + +VIV2 V2V3 V3Vj a. 2R andrr R C. b. rrR and 2R c. Rand 2nR 3 d. 2rrR and R d, VI V2V3 3 24. Between the two stations, a train accelerates from rest uni- formly at first, then moves with constant velocity and finally 17. If two balls of same density but different masses are dropped retards uniformly to come to rest. If the ratio of the time taken from a height of 100 m, then (neglect air resistance) a. both will come together on the earth be I :8: I and the maximum speed attained be 60 kmlh, then b. both will come late on the earth c. first will come first and second after that what is the average speed over the whole journey? d. second will come first and first after that a, 48 km/h b, 52 kmlh c. 54 kmlh d. 56 kmlh 25. A particle moving in a straight line covers half the distance 18. If a body starts from rest, the time in which it covers a par-' with speed of 3 m/s. The other half of the distance is covered tieuIar displacement with unifonn acceleration is a. inversely proportional to the square root of the displace- in two equal time intervals with speeds of 4.5 m/s and 7.5 ment b. inversely proportional to the displacement mis, respectively. The average speed of the particle during c. directly proportional to the displacement d. directly prop0l1ionai to the square root ofthe displacement this motion is a. 4.0 m/s b. 5.0 mls c, 5.5 mls d. 4.8 m/s. 26. Two balls of different masses m, and tnb are dropped from two different heights, viz., a and b. The ratio of times taken 19. Check up only the correct statement in the following. va :by the two to drop through these distances is d, a2 :b2 a, a:b b. boa c. y'j) a, A body has a constant velocity and still it can have a vary- 27. If two balls of same density but different masses are dropped ing speed. . from a height of 100 m, then a. both will come together on the earth b. A body has a constant speed but it can have a varying b. both will come late on the earth velocity. c. first will come first and second after that d, second will come first and first after that c. A body having constant speed cannot have any accelera- tion. 28. A body starting from rest moving with unifonn acceleration d. None of these. 20. The position x of a particle varies with time (I) as has a displacement of 16 m in first 4 sand 9 m in first 3 s. x = at' - bt3 . The acceleration at time t of the particle will The acceleration of the body is a. I ms-2 b.2 ms-' c. 3 ms-2 d. 4 ms-2

4.32 Physics for IIT-JEE: Mechanics I 29. The velocity acquired by a body moving with uniform ac- s l! celeration is 30 ms- 1 in 2 sand 60 ms-! in 4 s, The initial A velocity is a. zero h. 2 ms- i c. 3 ms-2 d. 10 ms-2 30. A particle moves along x-axis in such a way that its position coordinate (x) varies with time (1) according to the expression •I x = 2 - 51 + 61 2 Its initial velocity is a. -3 mls b. -5 mls c. 2 mls d.3 mls 31. A partide starts from the origin with a velocity of 10 m/s Fig. 4.94 and moves with a constant acceleration till the velocity in- creases to 50 m/s. At that instant, the acceleration is suddenly 40. A body is released from the top of a tower of height H m. reversed. What will be the velocity of the particle when it re- After 2 s it is stopped and then instantaneously released. What turns' to the starting point? will be its height after next 2 s? a. zero h. 10 mls c. 50 mls d. 70 mls a.(H-5)11l b.UI-IO)m 32. A particle is moving along x-axis whose instantaneous speed c. (H - 20) m d. (H - 40) m is v' = lOS - 9x'. The acceleration of particle is 41. A stone is dropped from the top of a tower of height h. After a. -9x mis' h. -18x m/s2 I s another stone is dropped from the balcony 20 In below -9.:r the top. Both reach the bottom simultaneously. What is the c. - - m/s2 value of h') Take g = 10 ms 2 2 a. 3125 In b. 312.5 m e. 31.25 m d. 25.31 In d. None of these 42. A train 100 m long travelling at 40 ms-- 1 starts overtaking 33. A hall is released from the top of a tower of height h. It takes another train 200 m long travelling at 30 ms-!. The time time T to reach the ground. What is the position of the ball taken by the first train to pass the second train completely is (from ground) after time T 13? a. 30 s h. 40 s c. 50 s d. 60 s a. hl9 m b. 7hl9 m c. 8hl9 m d. 17h/18 In 43. A person is throwing two balls into the air one after the other. 34. If x denotes displacement in time t and x = a cos t, then the He throws the second ball when first ball is at the highest acceleration is point. If he is throwing the balls every second. how high do a. a cos t h. -(J cos t c. a sin t d. -a sin t they rise? 35. Taxies leave the station X for station Y every 10 min. Simul- a. 5 In h. 3.75 m c. 2.50 m d. 1.25 m taneously, a taxi also leaves the station Y for station X every 44. A st.one thrown upwards with speed u attains maximum 10 min. The taxies move at the same constant speed and go height h. Another stone thrown upwards from the same point with speed 2u attains maximum height H. What is the rela- from X and Y or vice-versa in 2 h, How many taxies coming tion between hand H? a. 21z = H b. 311 = H c. 4h = H d. 511 = H from the other side will meet each tax-i enroute from Y and X? a.24 b.23 c.12 d.ll 45. A body dropped from the top of a tower covers a distance 36. The velocity-time relation of an electron starting from rest 7x in the last second of its journey, where x is the distance is given by v ::: kt where k :::::: 2 m/52. The distance traversed covered in first second. How much time does it take to reach in first 3 s is the ground? a.9111 b.16 m c.27 m d.36m a.3s b.4s c.5s d.6s 37. When the speed of a car is u, the rninimum distance over 46. An engine of a train moving with uniform acceleration passes which it can be stopped is s. If the speed becomes I1U, what an electric pole with velocity u and the last compartment with will be the minimum distance over which it can be stopped velocity v. The middle part of the train passes past the same during the same time? pole with a velocity of a. sin b. ns c. sln 2 d.n 2s u+v +2 v2 a. --2-- 11 38. A thief is r:unning away on a straight road in a jeep moving b · - -2- with a speed of9 ms-· 1• A policeman chases him on a motor . c. Vfu-'22+V2 d. V~----2:2:---- cycle moving at a speed of 10 ms·-I. If the instantaneous separation of the jeep from the motor cycle is 100 m, how 47. The relation between time t and distance x is long will it take for the policeman to catch the thief? a.Is b.19s c.90s d.l00s I = ax 2 + f!x 39. The displacement-time graph oftwo bodies A and B is shown where a and f3 are constants. The retardation is in Fig. 4.94. The ratio of velocity of A(VA) to velocity of a. 20<v' b.2f!v3 c.2af!v1 d.2f!2 v3 B(v/i) is b. .J3 c.l/3 d.3 48. The displacement x of a particle moving in one dimension a. 1/13 aunder the action of constant force is related to time t by the equation t = Jx -1- 3, where x is in m and f is in s, Find the

Motion in One Dimension 4.33 displacement of the parlicle when its velocity is zero. when they are at the same height, then a. bomb from BI reaches ground first a. zcro b:12m c.6m d.18m b. bomb from B2 reaches ground first c. bomb from B3 reaches ground first 49. The velocity of light emitted by a source S, observed by an d. they reach the ground simultaneously observe O. who is at rest with respect to S, is c. If the observer moves away from S with velocity v. the velocity of light as observed will be 59. A particle is dropped from rest from a large height. Assume ::J~: b. c- v g to be constant throughout the motion. The time taken by it d.c 2 to fall through successive distances of I m each will be v a. all equal. being equal to ../2/g second c2 50. The x and y coordinates of a particle at any time t are given b. in the ratio of the square roots of the integers 1,2,3•... +by x = 71 412 and y = 51. where x and yare in m and I in c. in the ratio of the difference in the square roots of the in- s. The acceleration of the pmticle at 5 s is tegers, i.e.,../T. (../2 - ../T), (~- ../2), (../4 - ~), ... a. zero b. 8 m/s2 c. 20 mls2 d. 40 m/s2 d. in the ratio of the reciprocals of the square roots of the . . III 51. The distances moved by a freely falling body (starting from rcst) during 15t. 2nd , 3rt! '\" .. , nth second of its motion are pro- Integers, I.e.. yrl,' y!2O' yI3'i\"\" portionell to a. even numbers 60. A person moves 30 Il1 north and then 20 m towards east and b. odd numbers finally 30../2 m in south-west direction. The displacement of c. all integral numbers d. squares of integral numbers the person from the origin will be a. 10m along north b. 10 Il1 along south c. 10m along west b. zero 52. A wooden block is dropped from thc top of a cliff 100 m 61. An object accelerates from rest to a velocity 27.5 mls in 10 sec. Find the distance covered by the object during next lOs. high and simultaneously a bullct of mass 10 gm is fired from a. 412.5 m b. 137.5 m c. 550 m d. 275 m thc foot of the cliff upwards with a velocity of 100 m/s. The bullet and woodcn block will meet after a time 62. A body falls freely from rest. Hcovers as much distance in the a. lO s b. 0.5 s c. 1 s d. 7 s last second of its motion as covered in the first three seconds. 53. A drunkard is walking along a straight road. He takes 5 steps The body has fallen for a time of i1lrward and 3 steps backward and so on. Each step is I m a.3s b.5s c.7s d.9s long and takcs I s. There is a pit on Ihe road II m away fron; 63. A stone is dropped from a rising balloon at a height of76 m the starting point. The drunkard will fall into the pit after above the ground and reaches the ground in6 s. What was the a.29s b.2ls c.37s d.3ls velocity of the balloon when the stone was dropped? Take g 54. A slonc is dropped from a certain height which can reach the = 10 m/s2 . ground in 5 s. It is dropped after 3 s of its fall and then it a. (52/3) m/s upward b. (52/3) mls downward again released. The total time taken by the stone to reach the c. 3 mls d. 9.8 mls ground will be 64. A body is dropped from a height of 39.2 m. After it crosses a.6s b.6.5s c.7s d.7.5s halfdistance, the acceleration due to gravity ceases to act. Thc 55. A body travels 200 cm in the first 2 sand 220 cm in the next body will hit the ground with velocity (Takc g = 10 m/s2) 4 s with deccleration. The velocity of the body at the end of a. 19.6 mls b. 20 mls c. 1.96 mls d. 196 mls the 7th second is 65. Two trains, one travelling at 15 ms- I and other at 20 ms~l, a, 5 cmls' b. 10 cmls c. IS cmls d. 20 cmls are heading towards one another along a straight track. Both 56. A body starts from rest and travels a distance S with uniform the drivers apply brakes simultaneously when thcy are 500 m apart. If each train has a retardation of 1 ms~2, the separation acccleraton, then moves a distance 25 unifonnly and finally comes to rest after moving further 55 under uniform retar- after they stop is dation. The ratio of average velocity to maximum velocity a. 192.5 m b. 225.5 m c. 187.5 m d. 155.5 m is 66. Two cars are moving in same 'direction with a speed of a.2/5 b.3/5 c.4n d.5n 30 kmlh. They are scparated by a distance of 5 km. What is the speed of a car moving in opposite direction if it meets 57. A body sliding on a smooth inclined plane requires 4 s to the two cars at an interval of 4 min? a. 60 kmlh b. IS kmlh c. 30 kmlh d. 45 km/h reach the bottom, starting from rest at the top. How much time does it take to cover one fourth the distance starting frQmrest at the top? 67. Two trains A and 13, 100 m and 60 m long, are moving in opposite directions on parallel tracks. The velocity of shorter a.ls 1l.2s train in 3 times that of the longer one. If the trains take 4 s to cross each other, the velocities of the trains arc c.4s d.16s a. VA = 10 mIs, VB = 30 mls b. VA = 2.5 m/s. VII =7.5 mls 58. B\" B2 and B3 are three balloons ascending with velocities v. 2\" and 3v, respectively. If a bomb is dropped from each

4.34 Physics for IIT-JEE: Mechanics I e. VA = 20 mis, VB = 60 mls of2.45 In from the floor of elevator, it reaches the floor of the elevator after a time (g ;::: 9.8 m/s2) d. VA = 5 mis, VB = IS mls 68. Two trains each travelling with a speed of 37.5 krn/h arc a.v7s b.l/v7s c.2s d.1/2s approaching each other on the same straight track. A bird 76. A particle slides from rest from the topmost pointofa vertical that can fly at 60 kmlh flies otT from one train when they circle of radius r along a smooth chord making an angle 0 are 90 km apart and heads directly for the other train. On with the vertical. The time of descent is rcaching the other train it flies back to the first and so on, a. least for e = 0 Total distance covered by the bird is b. maximum for f) ;::: 0 a. 90 km b. 54 km e. 36 km d. 72 km ee. least for = 45\" 69. Between two stations a train starting from rest first acceler- ed. independent of ates uniformly, then moves with constant velocity and finally retards uniformly to come to rest. If the ratio of the time taken 77. A body is thrown vertically upwards from fl, the top of a be I:8: I and the maximum speed attained be 60 kmlh, then what. is the average speed over the whole journey? tower. It reaches the ground in time fl. Ifi1 is thrown vertically a. 48km/h b. 52 km/h c. 54 km/h d. 56 km/h downwards from A with the same speed, it reaches the ground in time f2. If it is allowed to fall freely from A, then the time 70. A ban is thrown upwards with speed v from the top of a tower it takes to reach the ground is given by and it reaches the ground with speed 3v. What is the height a. t = I, + 12 2 2 of the tower? 4v2 8v2 c.I=0r tz, d.1 = v2 2v2 d.- C.- 78. Two cars A and B are travelling in the same direction with a. - b.- g gg g 71. A ball is dropped into a well in which fue water level is at a velocities VA and VB (VA> VB). When cal' A is ata distances depth h below the top. If the speed of sound be c, then the behind car B, the driver of car A applies the brakes producing time after which the splash is heard will be given by a uniform retardation a; there will be no collision when a. h [ y/[;h2+~c] (VA - VB)2 (VA - v/lf a.s<- b.s=----- 2a 2a ' b. h [ y([2;h, -~] e. s :0: (VA - VB )2 d. s < I!A - VB )2 c 2a - 2a [2 I]e. h - + - 79. Fourpersons are initially at the [our comers of asquare whose gc side is equal to d. Each person now moves with a uniform d. h [~ - ~] speed V' in such a way that the first moves directly towards '!? c the second, the second directly towards the third, the third 72. If a particle travels n equal distances with speeds VI, V2, ... , directly towards the fourth and the fourth directly towards lin, then the average speed V of the particle will be the first. The four persons wi 11 meet after a time equal to + + +a. a.d/V b.2d/3V e.2d/J3V d.d/J3V Vj li2 S Vn V= 80. Thc deceleration experienced by a moving motor boat, af- ---\"----\"- . dv n tCl' its engine is cut-off is given by ~ = _kv3, where k is t + + ... +b. -V = ---'--n-V=jV-2-··_· V.II VII <it Vj Vz V3 constant. If Vo is the magnitude of the velocity at cut-off, the· c.~ = .t. (~- + ~ + . + ~) magnitude of the velocity at a time tafter the cut~off is V' Il Vj V2 VII a. vo/2 b. Vo Vn d. -V. = V/V2j + 2 + ... + vn2 V2 d. / 73. A ball is thrown from the top of a tower in vertically upward V(2vf,kt + I) direction. Velocity at a point hm below the point ofprojection 81. For motion of an o~ject along x~axis, the velocity v depends is twice of the velocity at a point h m above the point of on the displacement x as v = 3x2 - 2x. What is the acceler- projection. Find the maximum height reached by the ball ation atx = 2 m. a. 48 ms,2 -above the top of the tower. b. 80 ms 2 a. 2h b.3h c. (5/3)h d. (4/3)h e. 18 ms·2 d. 10 ms- 2 74. A juggler keeps on moving four balls in the air throwing the 82. A stone is dropped from the 25'\" storey of a multistoricd balls after regular intervals. When one ball leaves his hand building and reaches the ground in 5 s. Tn the first second, (speed = 20 lnS-') the position of other balls (height in m) it passes through how many storeys of the building? (g ;::: 10 m/s2). will be (Take g = 10 ms·· 2) a. 10,20,10 b. 15,20, 15 a. I b.2 c. 3 d. None of these c. 5, 15,20 d.5, lO,20 83. A body is projected upwards with a velocity u. It passes through a ccrtain point above the ground after tj. The time 75. An elevator in which a man is standing is moving upwards .after which the body passes through the samc point during with a speed of I() m/s. If the man drops a coin from a height

Motion in One Dimension 4.35 the return journey is b. 23 ((1~/2-_I,I),) 92. A point moves in a straight line so that its displacement x In at time t s is given by x 2 ::: 1 + [2. Its acceleration in 111/s2 at a. (~-I~) rl. g2 time [ sis -I c. 3 (/2 _I,) I g b 'X-·3 a. x,. 84. A balloon is moving vertically with a velocity of4 m/s. When 12 1I c. d.--- it. is at a height of h~ a hody is gently released from it. If it X x3 X x2 reaches the ground in 4 s, the height of the balloon, when the 93. A point moves with uniform acceleration and VI, iJ2 and Vl body is released, is denote the average velocities in the three successive inter- a. 80 m b. 96 m c. 64 m rl. 78 m vals of time t), t2 and t3< Which of the following relations is 85. A parachutist drops first freely from an aeroplane for 10 s correct? and then parachute opens 6ut. Now he descends with a net +a. (v, - v,) : (V2 - Vl) = (I, - il) : (12 13) retardation of 2,5 m/52, If he bails out of the plane at a height + +b. (v, - V2) : (V2 - V3) = (I, 12) : (12 13) of 2495 III and g = 10 m/s2,- his velocity on reaching the c. (v, - V2) : (V2 - V3) = (I, - I,) : (1, - 13) ground will be d. (v, - V2) : (v, - V3) = (I, - /2) : (12 - 13) a. 5 mls b. 10 Illis c. 15 mls d. 20 mls 94. A 2 m wide truck is moving with a uniform speed Va ::: 8 mls along a straight horizontal road. A pedestrian starts to cross 86. A police party is chasing a dacoit in ajccp which is moving the road with a uniform speed V when the truck is 4 m away at a constant speed v, The dacoit is on a motor cycle. When from him. The minimum value of v so that he can cross the he is at a distance x from the jeep he accelerates from rest road safely is at a constant rate. Which of the following relations is true, if a. 2.62 mls b. 4.6 mls c. 3.57 Illis d. 1.414 m/s the police is abJe to catch the dacoit? a. v2 .:: ax b. v2 ::s 2ax 87. A train is moving at a constant speed V, Its driver observes GrapHital SolIitiQllS OJi page 4.58 Concepts another train in front of him on the same track and moving in the same direction with constant speed l), If the distance 1. Which of the following velocity-time graphs shows a realistic between the trains be x, what should be the minimum retar- situation for a body in motion? dation of the train so as to avoid collision? a. (V + vf (V - v)' -~- b.--- x x (V + v)2 c. (V - V)2 2x d. 2x 88. A moving car possesses average velocities of 5 ms-· j , 10 ms--) and 15 111S-- 1 in the first, second and third seconds, b. respectively, What is the total distance covered by the car in 'I ~ these 3 s'? ~i a. 15 m b. 30 m c. 55 m b. None of these c. d. 89. The average velocity of a body moving with uniform accel- 2. The velocity~time graph of a body moving in a straight line is shown in Fig. 4.95. The displacement of the hody in 10 s eration after travelling a distance of 3.06 m is 0.34 ms-· 1• If is the change in velocity of the body is 0.18 ms-' during this time, its uniform acceleration is a. 0.01 mis' b. 0.02 m/s2 c. 0.03 m/s2 d. 0.04 m/s2 90. Water drops fall from a tap on the fioor 5 m below at regular intervals of time, the first drop striking the floor when t.he 2- fifth drop begins to fall. The height of the third drop from the t- ground, at the instant when the first drop strikes the ground, 1\"---+--+---'\\c--j--...L--+l/s will be (g = to ms- 2) -t a. 1.25 III b. 2.15 m c. 2.75 m d. 3.75 m 2- - - - - - - - - - - - - 91. Drops of water fall at regular intervals from the roof of a Fig. 4.95 building of height H ::: 16 m, the first drop strikes the ground at the same moment when the nfth drop detaches itself from the roof. The distances between the different drops in air as a.4m b.6m c.8m d.lOm the first drop reaches the ground are 3. The velocity-time graph of a body is shown in Fig. 4.96. The displacement covered by the body in 8 s is a. 1m, Sm, 7m, 3m b. 1m, 3m, Sm, 7m c. 1111, 3m, 7m, Sm d. None of the ahove

4.36 Physics for IIT-JEE: Mechanics I v (mls) 7. The velocity-time graph of-a particle moving in a straight line is shown in Fig 4.100. The acceleration of the panicle at 6 t = 9 s is 4 --,, 2 0 t-+-+--+-T-r-t--+--+-., (s) -1 2 15 -4 ··6 10 Fig. 4.96 5 b.12m e.lOm d. 28 m c,,--t---.-ir---j--i--->-' (s) 2 4 6 8. 10 12 4. The variation of velocity of a particle moving along a straight Fig. 4.100 line is shown in Fig. 4.97. Thedistance travelled by the par- ticle in 12 s is a. zero b. 5 m/s2 c. - 5 m/s2 d. - 2 111/S2 8. The velocity-time graph of a body is shown in Fig. 4.101. It vlm:(l indicates that 5 - --,---.-.. oI'--+-+-+-'-+-',,-,-~_ tig A c Il -25 5 Fig. 4.97 a. 37.5 m b. 32.5 m e.35.0m d. None of these Fig. 4.101 5. The graph shows the variation of velocity of a rocket with a. at B force is zero time. The maximum height attained by the rocket is b. at B there is a force but towards motion c. at B there is a force which opposes motion vlms .j d. none of the above is true 1000 9. The velocity-time graph of a body is given in Fig. 4.102. Thc maximum acceleration in ms~2 is o'- - -'- - - - - - - ' j , c'- '1'2-0- > - t!s l'ims I 60 10 1]0 ! Fig. 4.98 20 . a. 1.1 km b.5km e. 55 km d. none of these () \"'--f-+--+----+--->-'is 20 30 40 70 6. From the velocity-time graph. given in Fig. 4.99 of a particle moving in a straight line, one can conclude that Fig. 4.102 vhn~,--l B a.4 b.3 c.2 d.1 4A 10. Which of the following velocity-time graphs is not possible practically? c~--r:::--+:--:-::'-. fis o 3 8 J2 Fig. 4.99 a. its average velocity during the 12 s interval is 2417 ms- l b. its velocity for the first 3 s is unifonn and is equal to 4ms- 1 c. the body has a constant acceleration between t = 3 sand t = 8\" d. the body has a uniform retardation from t = 8 s to t = 12 s

Motion in One Dimension 4.37 11. The velocity-time graph of a body is shown in Fig. 4.103. b·~u The ratio of average acceleration during the intervals 0 A and AB is o I~ v (m/s) 40 f)--------------- C c. d. o~-L------~-L~~ L-_ _ _- ' - _ . v - - - - - ' -o- - - = -- . , 8 I (s) u Fig. 4.103 15. From a high tower, at time t = 0, one stone is droApeci from rest and simultaneously another stone is projected vertically a. I h.1I2 c.l/3 d.3 up with an initial velocity. The graph of distance S between the two stones plotted against time t will be 12. On the displacement-time graph, two straight Jines make an- gles 60° and 30°, with time axis,as shown in Fig, 4.104. The ~. lL.~ ~5 c 'a. b. ratio of the velocilies represented by them is o t,O x (m) c. , d . 0 tOt A 16. An object is vertically thrown upwards. Then the displacement-time graph for the motion is c B I (s) ss Fig. 4.104 il2g ____ _ a. 1:2 h.l:3 c.2:1 d.3:1 b. 13. DisplacemenHime graph of a body is shown in Fig. 4.105. (m) o \"---7--:C~ 0 ulg 2uig ulg 2ulg s S ()' ' - - '- - - - ' - -- - - - - '- - -' c - - (s) c. d. z/l2g II 12 13 14 1/l2g t 0 • o'---:c~-=T--+ ufg 2ulg Fig. 4.105 Velocity-time graph of the motion of the body will be a. b. 17. The acceleration will be positive in which of the following graphs, 0 0 I, c. d. (I) (II) (III) (IV) ,i vi a. (I) and (III) h. (I) and (IV) c. (II) and (IV) d. None of these 0 I, 1_12_ r.\" 14 0 18. The graph (Fig. 4.106) below describes the motion of a ban rebollnding from a horizontal surface being released from a point above the surface. Assume the ball collides each time with the floor inelastically. The quantity represented on the .v-axis is the ball's(take upward direction as positive) 14. An object is thrown up vertically. The velocity-time graph a. displacement h. velocity for the motion of the particle is c. acceleration d. momentum

4.38 Physics for IIT-JEE: Mechanics I y 21. Two balls are dropped from the top of a high tower with a timc interval of to second, where to is smaller than the time taken 0L-----~I~,----1~2----IL3--~1> by the first ball to reach the fioor, which is perfectly inelastic. The distance s between the two balls, plotted against the time Fig. 4.106 lapse t from the instant of dropping the second ball is best represented by 19. The acceleration versus time graph of a particle is shown in Fig. 4.107. The respective v-t graphs of the particle are .1L ,lIn. a t-tr> 0 t----ft> a 'l~ \"1~ 0 o t~ 0 t~ I, I, v 22. The acceleration versus time graph of a particle moving in a Fig. 4.107 straight line is shown in Fig. 4.109. The velocity-time graph •• of the particle would be v b. 0 O. \" \" 4 I, I, vv -----:c0+------\"'2~--' (5) c. d. Fig. 4.109 0 0 ,,, a. a straight line b. a parabola \" \" \"\" c. a circle d. an ellipse 20. The displacement-time graph of a moving particle with con- 23. The acceleration-time graph of a particle moving along a stant acceleration is shown in Fig. 4.108. The velocity-time graph is given by straight line is as shown in Fig. 4.110. At what time the par- ticle acquires its initial velocity? x(m) • 5 02 , (5) --+----c4~----·'(S) Fig. 4.108 Fig. 4.110 v v a. 12 s b. 5 s c.8 s d. 16 s •. 24. Plot acceleration-time graph of the velocity-time graph given v b• in Fig. 4.111 C. 20 2 v 2 10 2I d. 1'--1--\\--1-+-.' (5) 0 -10 Fig. 4.111

Motion in One Dimension 4.39 (m/s') (m/s') 27. Which graph represents uniform motion: a a •.2 2 0 b• 5 t5 t (s) f---;+--,I\",O-,,:t- .....:2:0r + t (s) to 20 5 t5 , ,0 - L-J -2 -2 (m/s') d. (m/g') MuLtiple Correct a t (s) a Answers Type c. 2 2 0 10 20 °f -+ -5; -to; ;l- - t- -; !;: -I . , (s) 1. Check up the only correct statement in the following: 5 I]5 t5 20 -2 -2 3. A body having a constant velocity can have a varying speed. b. A body having a constant speed can have a varying veloc- 25. For velocity-time graph given in Fig. 4.112. ity. c. A body having constant speed can have an acceleration. d. If velocity and acceleration arc in same direction, then 2.5 distance is equal to displacement. ,,, ,,, 2. A block slides down asmooth inclined plane when released IL..._~'3'-_J',,-S_7--!>t (s) from the top, while another falls freely from the same point 7 a. sliding block will reach the ground first Fig. 4.112 b. freely falling block will reach the ground first The acceleration-time graph of the motion of the body is c. both the blocks will reach the ground with different speeds d. both the blocks will reach the ground with same speed (ms-') (ms-2) 3. A car accelerates from rest at a constant rate of 2 ms\"-2 for some time. Then it retards at a constant rate of 4 ms·-z and •. at b• at comes to rest. It remains in motion for 6 s. a. Its maximum speed is 8 ms- l 0 ($) 0 h. Its maximum speed is 6 ms--\" I c. It travelled a total distance of 24 m d. It travelled a total distance of 18 m (ms-·2) (ms\") 4. At I = 0, an arrow is fired vertically upwards with a speed of 100 ms- l A second arrow is fired vertically upwards with c. at d. at the same speed at I = 5 s. Then 0 (,) a. the two arrows will be at the same height above the ground. t3_ _5 7 at 1 = 12.54 s 26. The displacement-time graph of a moving particle is shown h. the two arrows will reach back their statting points at t = in Fig. 4.113. The instantaneous velocity of the pm1icle is 20sandl=25s negative at the point c. the ratio of the speeds of the first and second arrows at I = 20 s will be 2: I d. the maximum height attained by either arrow will be lOOOm 5. Two bodies of masses m \\ and m2 are dropped from heights hi \"~ D v,and hz, respectively. They reach the ground after time 1\\ and Q 12 and strike the ground with and li2, respectively. Choose ~u C the correct relations from the following: E F I, (h, I, (h, d.E a.t;=y~ b. t;=yh; Time V, {rh;,- V, h2 c. V2 = Vh2 d. =V2 hI a. D Fig. 4.113 6. From the top of a tower of height 200 m, a ball A is projected b. F c. C up with lOms- 1 and 2 s later another ball B is projected

4.40 Physics for IIT-JEE: Mechanics I vertically down with the same speed. Then v (111/S) a. both A and B will reach the ground simultaneously 10 b. hall A will hit the ground 2 s later than B hitting'the ground c. both the balls will hit the ground with the same velocity o I---'k--i--+-- t ($) d. hoth the balls will hit the ground with the different velocity -10 7. A body starts from rest and then moves with uniform accel- eration. Then - 20 a. its displacement is directly proportional to the square of Fig. 4.114 the time b. ils displacement is inversely proportional to the square of 15. Figure 4,115 shows the velocity (v) or a particle plotted the time against time (t). c. it may move along a circle d. it always moves in a straight line r----+ 21' 8. Which is/arc correct.? Fig. 4.115 a. If velocity of a body changes, it must have some acceler- ation a. The particle changes its direction of motion at SOme point b. Jf speed of a body changes, it must have some acceleration b. The acceleration of the particle remains constant c. If body has acceleration, its speed mllst change c. The displacement of the particle is zero d. If body has acceleration, its speed may change d. The initial and final speeds of the particle are the same 16. The displacement. of a particle as a function of time is shown 9. The body will speed up if in the Fig. 4.116. It indicates a. velocity and acceleration are in same direction h. velocity and acceleration are in opposite direction s c. velocity and acceleration arc in perpendicular directi()n d. velocity and acceleration are acting at acute angle w.ct. O~-+--+--+--4--+ each other I 245 10. Average acceleration is in the direction of a. initial velocity Fig. 4.116 b. final velocity c. change in velocHy a. the particle starts with a certain velocity, but the motion is d. final velocity if initial velocity is zero retarded and finally the particle stops 11. Which of the following statement is true? h. the velocity of the particle decreases a. A body can have varying speed without having varying c. the acceleration of the particle is in opposite direction to velocity b. A body can have constant speed but varying velocity Ihe velocity c. A body can have velocity without having acceleration d. the particle starts with a constant. velocity, the motion is d. A body can have acceleration without having velocity accelerated and finally the patiicle moves with another 12. A particle is projected vertically upward with velocity l/ from constant velocity a point A, when it returns to point of projection a. its average speed is u/2 17. A particle moves in a straight line with the velocity as shown h. its average velocity is zero in the Fig. 4.117. Att = lI. x = -16 m c. its displacement is zero d. its average speed is II i ,6 , 13. A particle moves along a straight line and its velocity depends v 2 - -j - - ~ on time as v = 4t - t 2 . Then for first 5 s I 30 40 a. Average velocity is 25/3 III 0 10 1824 r (s) h. Average speed is 10 Ill/S c. Average velocity is 5/3 Ill/S ~2 d. Acceleration is 4 m/s2 at t = 0 6 --------- , 14. The velocity-time plot for a particle moving on a straight line is shown in the Fig. 4.114. Fig. 4.117 a. The particle has a constant acceleration b. The particle has never turned around a. the m3ximmri value of the position coordinate of the par- c. The particle has zero displacement ticle is 54 m d. The average speed in the interval 0 to 10 s is the same as the average speed in the interval lOs to 20 s

Motion in One Dimension 4.41 h. the maximum value of the position coordinate of the par- 6. Statement I: Retardation is directed opposite to the velocity. ticle is 36 m Statement II: Retardation is equal to the time rate of decrease of velocity. c. the particle is at the position of 36 m at t = 18 s 7, Statement I: Magnitude of average velocity is equal to av- d. the particle is at the position of 36 matt = 30 s erage speed, if velocity is constant. Statement II: If velocity is constant, then there is no change 18. Velocity variations of an object moving along a straight line in the direction of motion. are rec'orded in thc enclosed Fig. 4.1l8. S. Statement I: The velocity of a particle may vary even when 2 its speed is constant. Statement II: Such a body may move along a circular path. -3 ------ -4 9. Statement I: Two balls of different masses are thrown verti- cally upward with same speed. They will pass through their Fig. 4.118 point or projection in the downward direction with the same speed. 3. The acceleration is maximum in first two seconds Statement II: The height and the downward velocity attained b. The acceleration magnitude is maximum during the ninth at the point of projection are independent of the mass of hall. second 10. Statement I: At any instant, acceleration of a body can c. The object is farthest from the starting point after 16 s change its direction without any change in direction of ve- d. The displacement is largest at 7 s locity. Statement II: At any instant, direction of acceleration is same as that of direction of change in velocity vector at that instant. AsserNon-Reasoning So[ut{ons on. page4.6J Comprehensive Type Type Soludonion pqge4.61 In the following questions, each question contains STATE- For Problems 1-2 MENT I (Assertion) and STATEMENT II (Reason). Each ques- tion has 4 choices (a), (h), (e), and (d) out of which only one is The displacement ofa body is given by 4s = M + 2Nt 4 , where correct. M and N are constants. (a) Statement I is True, Statement II is True; Statement II is a correct explanation for Statement 1. 1. The velocity of the body at any instant is (b) Statement I is True, Statement II is True; Statement II is NOT a. M +2Nt4 b.2N a correct explanation for Statement 1. d. 2Nt3 4 (e) Statement I is True, Statement TI is False. M+2N (d) Statement I is False, Statement II is True. c. 1. Statement I: The displacement of a body may be zero, though 4 its distance can be finite. Statement II: If a body moves such that finally it arrives at 2, The velocity of the body at the end of I s from the start is initial point, then displacement is zero while distance is finite. a.2N M+2N 2, Statement I: The distance and displacement arc different b. 4 physical quantities. S~atement II: Distance and displacement have same dimen- c. 2(M + N) 2~+N sion. d. 4 3. Statement I: Average velocity of the body may be eqnal to its instantaneous velocity. For Problems 3-5 Statement II: For a given time interval of a given motion, A body is dropped from the top of a tower and falls freely. average velocity is single valued while average speed can have many values. 3. The distance covered by it after n seconds is directly pro- portional to 4. Statement I: A body can have acceleration even ifits velocity is zero at a given instant. a. n2 b. 11 Statement II: A body is momentarily at rest when it reverses d. 2n2 - I its direction of velocity. C. 2n - I 5. Statement I: An object can possess acceleration even at a 4. The distance covered in the nth second is proportional to time when it has uniform speed Statement II: It is possible when the direction of motion a. n2 h. n keeps changing. c. 2n - I d. 2n 2 - I 5. The velocity of the body after n seconds is proportional to a. 112 b. n c.2n-l d. 2n2 - I

4.42 Physics for IIT-JEE: Mechanics I For Problems 6-7 :For Problems 15-16 A body, at rest, is acted upon by a constant force (it means acceleration of the body will be constant). A body is dropped from a balloon moving up with a velocity 6. What is the nature of the displacement-time graph? of 4 ms'] wheu the balloon is at a height of 120.5 m from the ground. a. Straight line 15. The height of the body after 5 s fhllll the ground is (g = b. Parabola 9.8 mls2) c. Asymmetric parabola d. Rectangular hyperbola a. 8 m b. 12m c. 18 m d.24m 16. The distance of separation between the body and the bal- loon after 5 s is 7. What is the nature of velocity-time graph? fuln.5m ~IOO.5m a. Straight line b. Parabola c. 132.5 III d. 112.5 m c. Elliptical d. Hyperbola :For Problems 17-18 For Problems 8-9 A bus starts moving with acceleration 2 ms-2 . A cyclist 96 Tn behind the bus starts simultaneously towards the bus at a constant A car accelerates from rest at a constant rate Q' for some time speed of 20 m/s. and then decelerates at a constant rate f3 to come to rest. The 17. After what time will he be able to overtake the bus? total time elapscd is /. (Irr-JEE. 1978) 8. The maximum velocity attained by the car is a. 4 s b. 8 s c. 12 s d. 16 s afJ afJ 18. After some time the bus will be left behind. If the bus b. - - I continues moving with the same acceleration, after what a. 2(a + fJ) / time from the beginning, the bus will overtake the cyclist? a+fJ 2afJ 4afJ a. lOs b.12s c.14s d. 16s c. - - - t d. --t a+fJ a+fJ 9. Total distance travelled by the car is :For Problems 19-21 Distance is a scalar quantity. Displacement is a vector quan- afJt' afJt' tity. The magnitude of displacement is always less than or equal to distance, For a moving body displacement can be zero but a. 4(a + fJ) .b. 2(a + fJ) distance cannot be zero. Same concept is applicable regarding velocity and speed. Acceleration is the rate of change of velocity, 2afJt' 4afJt 2 If acceleration is constant, then quantities of kinematics are ap- c. (a + fJ) d. (a + fJ) plicable. For one dimensional motion under the gravity hi which :For Problems 10-11 air resistance is considered, the value of acceleration depends on A body is moving with uniform velocity of 8 m/s. When the the density of medium, Each motion is measured with respect to the frame of reference. Relative velocity may be greater/smaller body just crossed another body, the second one starts and moves to the individual velocities. with uniform acceleration of 4 ms-2. 10. The time after which the two bodies meet, will be a. 2 s c. 6s d. 8 s 19. A particle moves from A to B. The ratio of distance to displacement is (Fig. 4.119) 11. The distance covered by the sccond body when they meet C2J is a.8m b. 16 m c.24m d.32m :For Problems 12-14 A0B A body is allowed to fall from a height of 100 m. If the time taken for the first 50 m is tl and for the remaining 50 m is f2' Fig. 4.119 12. Which is correct? a. 2 b. 2 a. tl = 12 h. I] > 12 c. d. 1: 1 C. 11 < 12 d. depends upon the mass 4 20. A person is going 40 m n011h, then 30 m east and then 30.J2 m southwest. The net displacement will be 13. The ratio of t] and t2 is nearly a. 10 m towards east a. 5:2 b. 3:1 c. 3:2 d. 5:3 h. 10m towards west c. 10 m towards south 14. The ratio of times to reach the ground and to reach first d. 10 m towards north half of the distance is 21. A particle is moving along the path y = 4x2 The distance a. V3:1 b• .J2:1 c. 5:2 d. 1:V3 and displacement from x = I to x = 2 is (nearly)

Motion in One Dimension 4.\"43 a. 150, 12 b. 160,20 a. 8111S\",-1 h. 7.6 ms-- I c. 200,30 d. 150,20 c. 6.4 ms-,J d.5.8me' For Problems 22-23 26. The resistive force suffered by a motor boat is ex V2 where A car is moving towards south with a speed of 20 m/s. A mo- V is instantaneolls velocity. The engine was shut down when the velocity of the boat was jlo. Find the average torcyclist is moving towards east with a speed of 15 m/s. At a velocity at any time I. certain instant, the motorcyclist is due south of the car and is at a. Vtl + V (-jIV-oV-+\"-jI-) a distance of 50 111 from the car. 2 22. The shortest distance between the mot.orcyclist and the b. - 2 car is c. V V\" log,(jI,,/ V) 2 V V\" 10g,(Vo/ jI) a. 20 Il1 b. 10 m c.40m d.30m (jill - V) d. (jill + V) 23. The time after which they are closest to each other 27. A particle moves from A to B such that x = f2 + t - 3. e. 115 s a. 1/3 s b. 8/3 s d. 8/5 s Its average velocity from t = 2 s to :;:;: 5 s is For Problems 24--28 a. 6 ms- I b. 8 ms- I Consider a particle moving along x-axis as shown in Fig. 4.120. Its distance from the origin 0 is described by the coordinate x, c. 8.5 ms- I d. 7 ms-' which varies with time. At a time II, the particle is at point P, where its coordinate is XI, and at time 12 it is at point Q, where 28. A boy throws.a ball to his friend 20 m away. The ball its coordinate is X2. The displacement during the time interval from 11 to 12 is the vector from P to Q: the x-component of this reaches to the friend in 4 s. The friend then throws the isvector (X2- Xl) and all other components are zero. ball back to boy and it reaches the boy in 5 s. Assume the It is convenient to represent the quantity X2 -Xl, the change ball travels with constant velocity during any throw. in x. by means of a notation using the Greek letter I'. (capi- tal delta) to designate a change in any quantity. Thus we write 3. The average velocity is 40/9 ms -! . .6.x = X2 - Xl in which .6.x is not a product but is to be inter- h. The average acceleration is zero. preted as a single symbol representing the change in the quan- c. The average velocity is zero but average acceleration lity x. Similarly, we denote the time interval from tl to t2 as is nonzero, 1'.1 = 12 - I,. d. Average acceleration of the motion cannot be defined. y For Problems 29-30 Two particles A and B arc initially 40 111 apart. A is behind B. Particle A is moving with uniform velocity of 10 mls towards B. ~P Q Particle B starts moving away from A with constant acceleration of2 m/s2 --~o --~~---4~-----*-----X 29. The time at which there is a minimum distance between the two is I + - - X2~\" XI = LlX - - - I a. 2 s b.4s c. 5 s d. 6 s X, 30. The minimum distance between the two is Fig, 4.120 a.20m h. 15 m c. 25 m d. 30m The average velocity of the particle is defined as the ratio For Problems 31-35 Figure 4.121 shows displacement versus time graph for a parti- of the displacement .6.x to the time interval 6.1. We represent cle moving along x-axis, Find the average velocity in the time interval average velocity by the letter v with a bar (V) to signify average value. Thus _ X2 - Xl .6.x x (m) v=----=-- 12 - I, 1'.1 24. A particle moves half the time of its journey with u. The rest of the half time it moves with two velocities V\\ and V2 such that half the distance it covers with V, and the other half with V2. Find the net average velocity. Assume straight line motion. a. u (V, + V,) + 2V, V2 . b. 2u (V, + V, ) 2(V, + V,) 2u + V, + V, c. U (jI, + V2) 2 jI, jl2 -- 5 2V, d. U + V, + jI, 6L-~~--~~--~----~ 25. A particle moves according to the equation x = 12+ 31+4. Fig. 4.121 The average velocity in the first 5 s is

4.44 Physics for IIT-JEE: Mechanics I 31. fromO-2s 40. the maximum velocity of the particle is a. 5 m/s b. 0 m/s a. 20 m/s b. 25 m1s d. IO m/s c. 30 m/s d. 40 m/s c. 15 m/s 32. fromO-4s 41. the distance travelled with uniform velocity is a. 3/4 m/s b. 41Sm/s a. 375 m b. 125 m c. 2/3 m/s d. 514 mls c. 300 m d. 450m 33. from 2 s - 4 s For Problems 42-43 Study the four graphs given below. Answer the following ques- a. -2.5 m/s b. 5mls tions on the basis of these graphs. c. 3 m/s d. 4 mls x 34. from 4 s-7 s a. 2/3 m/s b. -1013 mls / ,' T, d.15m/s c. om/s 35. from 0 - 8 s b. 20 m/s x d. IS m/s a. 10 m/s c. om/s For Problems 36-38 The velocity-time graph of a particle in straight line motion is shown in Fig. 4.122. The particle starts its motion from origin. v in ms' I, 4 (iii) (iv) 2 42. In which of the graphs, the particle has more magnitude of velocity at t[ than at t2. ::! --------------'---'\" a. (i), (iii) and (iv) -4 b. (i) aud (iii) c. (ii) and (iii) d. None of the above Fig. 4.122 43. Acceleration of the particle is positive 36. The distance travelled by the particle in 8 s is a. in graph (i) b. in graph (ii) a.18m b.16m c.8m d.6m c. in graph (iii) d. in graph (iv) 37. The distance of the particle from the origin after 8 s is For Problems 44-45 Study following graphs a.18m b. 16m c.8m d.6m x x 38. Find the average acceleration from 2 s to 6 s. -~ I I a. -2 m/s2 b. -1 mis' c. 2 m/s2 d. I m/s2 For Problems 39-41 (i) (ii) The velocity-time graph of a particle moving along a straight line is as shown in Fig. 4.123. The rate of acceleration and de- xx celeration is constant and it is equal to 5 m8-- 2. If the average velocity during the motion is 20 ms-- l , then r\\Vi\\, i~_ . t (l\\1 s 20 tllllC (iii) (iv) - Fig. 4.123 44. The paI1iclc is moving with constant speed a. in graph (i) and (iii) 39. the value of t is a. 5 s b. 10 s c. 20 s d. 5j2s

Motion in One Dimension 4.45 b. in graph (i) and (iv) 4. The displacement versus time curve is given (Fig. 4.124) c. in graph (i) and (ii) d. in graph (i) t c~---f) 45. The particle has a negative acceleration s u. in graph (i) b. in graph (ii) d. in graph (iv) c. in graph (iii) MatGl1ing ,~ Solutions on page 4.63 o ,-. (io,lumn T~pe , Fig. 4.124 1. A particle moves along a straight line such that its displace- ment S varies with time I as S = a + {3t + yt 2, ColuumI' . ColulUnl i. Acceleration at t ;\" 2. s. . .' a.1l +5 r -it Av'erage ve.locit.y du.ring'-3-fd-,-s-.-- -,b-.-2-y---- 5. Study the following v-I graphs in Column I carefully and iii.Velocityatt \"Is .' .. C~ ex .•. match appropriately with the statements given in Column II. .iv. Initial. displacement Assume that motion takes place from time 0 to T. '. d'll + 2y a. net displaccm'tmt is positive, 2. For a body projected vertically up with a velocity;;:, from the blit not zero ground, match the following I Column n h. net displacement is, nega- r.:ci._::\",-~_,_-+.ca=.-=Z::e:.::r:co\",for round trip tive, bu'f not zerO ii. U av V~\"+'lJ; , .. b. --2- over any time mterval c. ~ over.thetot.l time of its flight f --' -'- '-- -'- c- -c :- +- =vo- --- d. 3. A ball is thrown vertically upward from the top of a cliff. Take the starting position of motion as origin and upward direction as positive, Column I specifics the position, velocity andlor acceleration of the particle at any instant. Column II gives their signs (+) or (- ) at that moment. Match the columns . . .Column I . '. Column II i. When the ballis<\\bove the point of a.a!waysp projection, its displacement is .' ii. When the pall is allo.ve the pain! of b. always nega-, . projection,itsYei()city is .... live . ... iii. When the ballis above.thepoint of c. may be positive ...projection, its acceleratio~ is ' or maybe nega- .. . -- -t-ive- - . - - iv. When the.ball is.below theroint of d. may be z projection, its 'accelenHion js ---

4.46 Physics for llT-JEE: Mechanics I ANSWERS AND SOLUTIONS Sulljective Type l I t , 12 12 L.H.S. = ;:; + fj = Vo + Vo = Vo = 2d/1 = 2d rr 2rr Srr =R.H.S. AB 2r BC 31\" CA 6 r 1. t! = - = -,t2 = - = --,t3 = - = - - 9. V = \"\" = fit, (Fig. 4.129) S .I' 2.1' 2.1' 3.1' 3s 2s 12 I, c ,. 2n!3 3s 5 j) I, 6 nl2 Fig. 4.129 Fig. 4.125 1= (t, + 12 + 13 + V (2)2: = 2rrr '* -21 = 2t2 + -aV + -v v fi avo speed t, + 12 + t3 1.8 s S/2 S t2 = 8, = S, S, + 8, = S/2 t2 = : - ~ [-,,- + -,,-] 2. I, = - = -, - -, 36 4.5 7.5 V 2 a fi ~3-mls 45 mls 7.5 m/s [.!. .!.]t, + t2 + I, = -\"- + -\"- + IF : + ~ + S/2 1-+ 1-+ I I, afi v2afi Fig. 4.126 10. 1= -vS,t,\", = -15-0 = lOs 10+5 avo speed =-t, +-S2-1, = 4 mls 11. v2 - u' = 2a x 180 3. D4 (g/2)(2 x 4 - 1) = 7 '* 4S2 - u2 = 360a Ds (g/2)(2xS-I) 9 V = U+ al 4. Both will pass with same speed, because free fan motion is independent of mass. , us=•o X •_u 6s •-45m1s 5. Average velocity on each lap will be zero, because displace- P, 180m P, ment is zero. _a Fig. 4.130 2rr r 2rr x 200 avo speed = - - = - - - = 20 m/s I 62.8 '*6. Vo = 0 + at, = =Solve to get u IS mis, a S n:vs2 t, = vo/a, 0 = Vo - fit, =} t2 = vo/fi '*u2 = 0' + 2ax IS2 = 0' + 2 x 5x ,*x = 22.5 m. I, fi v=,o 12. Taking upward direction as positive, let us work in the frame of lift. Acceleration of ball relative to lift = (g + a) down- +ward, so areal = -(g a), initial velocity: Urel = v, final ve- locity: Vtd = -vas the ball will reach the man with same Fig. 4.127 speed w.r.t lift '*. -v = V + (-g - a)1 '* 2v 7. a. v oa-' = SO = 2.S mls Apply v,,, = ut\" +a\",1 -=g+a 20 I = =50 13. Let at any time I = 0, balloon is at position A, where its b. Vaa-2 22 2.27 mls =velocity is u (Fig. 4.131). At I 2 s it reaches at B, where its c. Vav is zero for round trip because displacement is zero. 12 + 1 2 I + I velocity becomes v, then 2: at , 2: 2:(a/, 2:(fi 8. Vo = \"\" = fit, , d = fil, = )/, I2)I, AB = S= u x 2 + I = 2v +2 2:a(2)2 = I + I = +I (2) = I also V = u +a x 2 = ~ +2, -vol, -vot2 -vo(t, -vol 2 2 2 2 + I2 a -/3 - 8, = u x 3.S 2:(-g)(3.5) , - S, = vx 1.5 + I, 2:( -g)(1.S) Fig. 4.128

Motion in One Dimension 4.47 15. SI = I + f)(I + 1')2, S, = +I 2:(g 2:(g f)1,2 v I = =S 2:(g + + 21') t\"\" 2 s,B SI - S, f)(I)(1 S a = 1mls 2 16. Time taken by stone I to fall throughh - x is equal to time taken by stone 2 to faU through h - y (Fig. 4.134) h- Y =. 2:I 2 = ro:::: gl , UI Y 2gx \" x t=O A Fig. 4.131 y x Required distance between the stones: h CD • x = SI + S - S2 Fig. 4.134 Solve to get x = 55 m Alternatively: If we work from the frame of balloon, then yfh - x = uII + 1 solve to geth = -(x\"-+~'-'- acceleration of each stone w.r.t. balloon will be g + a after _gI2, releasing from it (Fig. 4.\\32). Initial velocity of each stone 2 4x will be zero w.r.t. baUoon. 17. a = . 10 x -5 = 5 mis' o g sme = 10 s, s, 1=1.5$ 5m t \"\" 3.5s Fig. 4.135 Fig. 4.132 += 1 2 x, = 12 +I XI 2:al , Xl 2:a(21) SI = 2:(g a)(3.5)2 10 = XI + X2 + X3 = 1, S2 = 2I: Cr; + a)(1.5)2 2:a(31) x = SI - S, = 55 m 801vetogetxl = 10 =30 50· 2 +14. 11 I, = 4 min, v = alII = a,l, -9m,X2 = -9m,X3 -9m,t =-38 2:I 18. a. Let collision occurs at time I. 12 S = x 4v =} 4 = 2v =} v = 2 \"larc1t For car w.r.t. truck: Src1 = Urc1t + v 60 = (30 - 10)1 + I - 0)1 2 -(-5 2 I' - 81 + 24 = 0, from here .we will get no real value of t, it means collision does not occur. b. Relative distance covered in 0.5 s = (30 - 10)0.5 = 10m, 50 = (30 - 10)1 + I - 0)12 -(-a 2 aI' - 401 + 100 = 0 Fig. 4.133 To avoid collision, its discriminant, D :0 0 11 + I, = v [~ + ~J =} 4 = 2 [~ + ~J =} 402 - 4al00:o 0 =} -a :0': 4 mis' al a2 al a2 1 II 19. From Fig. 4.136 h = U x 1 - 2: g (I)2 . =} -+-=2 h + 1= U x l.l - I aj a2 2: g (1.1)' =} U = 20.5 mls

Motion in One Dimension 4.47 15. Sl = 1 + f)(t + t')2, S2 = I + f)1\" 2(g 2(g t \"'\" 2 s,B 1. S a=lmJs 2 S = Sl - S2 = 2(g + f)(t)(t + 2t') u 16. Time taken by stone 1 to fall throughh - x is equal to time =-taken by stone 2 to fall through h - y (Fig. 4.134) h-y~ 21g12,Ul =y2gx x y x Fig. 4.131 Required distance between the stones: Fig. 4.134 x = Sl + S - S2 h- x= Ult + 1 t 2 , solve to get h = ~(-x'+-''y-)'-2- _g Solve to get x = 55 m 2 4x Alternatively: If we work from the frame of balloon, then 17. a=gsin8=lOx :0=5mls2 acceleration of each stone w.r.t. balloon will be g + a after releasing from it (Fig. 4.132). Initial velocity of each stone will be zero w.r.t. balloon. o s, s, t =1.5s Sm t ~ 3.5s Fig. 4.135 Fig. 4.132 Sl = 1 + a)(3.5)2 = =+1 1 2(g Xl 2a12, Xl X2 2a(21)2 S2 = I + a)(1.5)2 10 = Xl + X2 + X3 = I, 2(g 2a(31) x = Sl - S, = 55 m solvetogetxl = 910 m,x, = 930 m,X3 = 950· m,t = 32 8 14. tl + t2 = 4 min, v = altl = a2t, 18. a. Let collision occurs at time I. 1, S = 21 x 4v ='> 4 = 2v ='> v = 2 zare1t For car w.r.t. truck: Srel = Urelt + 60 = (30 - 10)1 + I - 0)t 2 -(-5 2 12 - 81 + 24 = 0, from here we will get no real value of t, it means collision does not occur. b. Relative distance covered in 0.5 s = (30 - 10)0.5 = 10 m, 50 = (30 - lO)t + 1 - , -(-a O)t 2 aI2 -40t+100=0 Fig. 4.133 To avoid collision, its discriminant, D :s 0 tl + t2 = V [~ + ~] ='> 4 = 2 [~ + ~J ='> 402 - 4a 100 :s 0 ='> -a ?: 4 mls2 a1 a2 al a2 19. From Fig. 4.136 h = u x 1 - 21g(1)2' 11 2h + 1 = U x 1.1 - 1 ='> - + - = 2 g(1.1)2 ='> U = 20.5 mls al a2

4.48 Physics for JIT-JEE: Mechanics I r- Yfdi;i-'(2.!.-I-t~z I1m Jai v +a2t HI .fiil v+a,1 =} Jaiv + a, Jait = v.fiil' + a2.fiil1 11 =} lJ = (.ja,a2) I 'f 23. I, = .j2\"/g -+ for first stone Fig. 4.136 =u (fJ -n) ~g (ff-nr21 , u2 (20.5)' -h = -U[I, - nl - g(l, - n)- -+ for second stone H=-=--=21m h+ 2g 2 x 10 20. t= y(-;I;i;Ji, h - ~= ut _ ~ gl2 n 2 simplify to get: 8h(u - gn)2 = gn 2(2u - gnl' u= ~ = Inkg a24. d - = u(3n) + ~\"'(3n)2 = 3un + 9 (~\"'n2) (i) I2 hln ••• a I~O (\"\"II l \"\"' 2n t = 311 • lv, • •1 c 1 b .1 tV2 h-\"~ -I d .1 lit n Fig, 4.138 Fig. 4.137 c- a = u(2n) + 1 (ii) 1)2 u - gf u. (iii) 2\",(2n)2 (iv) VI = gt, 112 = u - gl, - = - - = -, - 1 \"I gl gt b - a ;::; un + -1-an2 =} -\"2 = -ght -2 1 =h-ng- 1 = }VI - = -2 - 2 g2h V2 n - 2 VI + 3 I21, a, I = 2k = (iii) =} c - b= un 2\"'n2 g+a (ii) - 2 x 2.5 2 s = - from (i) and (iv), d - a = 3 (c - b) 10 + 1.25 3 an(ii) - 2 x (3), c - a - 2(b - a) = 2 b. velocity of lift after 1 s: V =0 + 1.25 x 1 = 1.25 m/s =5/4 mls c+a - 2b This .will be the initial velocity of bolt n2 2 +25. I, 12 = 2 =} S + S = 2 =} S = 40 km, Distance moved up by lift in - s: 30 60 3 total distance = 80 km x = ~ x ~ + ~ 1.25 (~)2 = 10 m 30 km/h 60 km/h 432 3 9 Car 1 s\" = =Displacement of bolt 2.5 - x 2.5 - 910 = 1258 m Car 2 A,_/:..-~_·O~_ _ _ _ _ _ _~B ,,' 211 Maximum height attained by bolt above point of dropping V2 (5/4)' 5 Fig. 4.139 =-=--=-m 80 =~a(2)', a =40 kmlb', I, = 40 = ~ h, t2 = 40 = ~ h 2g 2xlO 64 2 303603 for t < 34h, h: If their speeds arc equal, So . travelled = 2 x 5 + -25 = -445 m 3 dIstance - 30=at=}t=-h 64 18 288 2 If one overtakes the other: 30t = 21at2 =} t = 23 h V, V,22, II =t2 - I,= + v, , = 12 s= 12 34 , S 2a,I\" 2a2t, for I > h: If theIr speeds are equal: =VI all), V2:;:;:; a2/2 :;:;;} V2 + v =-a\\t\\ +=> a2/2 v = alII = a1l2 - alt v + alt 12= --- aJ - a2

Motion in -One Dimension 4.49 3 v. as the slope is positive and increasing, so positive in- 60 = at =} t = - h 2 creasing acceleration. (t -If one over takes other: 40 + 60 ~) =~at2 vi. as the slope is positive and decreasing, so positive de- 60t - t 3t40 + creasing acceleration. = =+80 20t2 =} 2 - 28. a. Cannot represent one dimensional motion, because there 20 = =(t - l)(t - 2) 0 =} t I h (not valid), 2 h cannot be two values of displacements at one time. Also 1 ~(:0:2:~_0:)::a):~ we don't consider time to be negative. =S:: h. Cannot represent one dimensional motion, because dis- 26, SI 1:1.5 m tance cannot be negative. Also we don't consider time to d = S2- SI = 108 m be negative. c. represents one dimen.;;ional motion, as displacement can be negative. d. Cannot represent one dimensional motion, because there I---d_! cannot be two values of displacements at one time. I'-~- - S 2 .1_ ·x--l e. Cannot represent one dimensional motion, because speed • •v • cannot be negative. f. represents one'dimensional motion, as velocity can be neg- ative. Fig, 4.140 g. Cannot represent one dimensional motion, because there a. 350 m, with respect to train cannot be two values of velocity at one time, x = I - d = 350 - 108 = 242 m, with respeet to earth h. Cannot represent one dimensional motion, because total 242 path length does not decrease with time. - b. v= 60 = 4.03 mls 29. a, True, From graph, we see thaUor time t = 0 to t = 2 s, displacement of the particle remains same as 3 m. It means 27. a. No, if position-time graph is a straight line parallel to position axis, its slope is infinite which indicates infinite particle was at rest for this time intervaL velocity. h. True, Velocity is maximum when the slope is maximum. pO'WOOLL, And slope is maximum f(,r time t = 5 s to t = 7 s. velocity = =-. 0-- 5 2.5 mls during this time slope -5 30. 3.- true, average velocity = Total displacement . Total tnTIe Fig. 4.141 = -3---8- = - 2.5 m/s It is not possible practically 4 - 1.5 b. Displacement c. Distance h. False, slope at t = 2 sis, d. speed Slope = -1-3 = -3.7 mls ~~~t 3.5 Fig. 4.142 c. True, since slope is zero at t =: 4 s, so velocity is zero at e. Velocity f, Speed t = 4 s. g. No, its slopc is infinite which indicates infinite velocity. It 31. Slope for C is greatest so C will have greater velocity and is not possible practically. slope for A is least so A will have least velocity. h. (c), its slope is negative 32, a. 0 - 20 s, 140 - 180 s, 240 - 280 s, 380 - 440 s: For these • . Q_ X2 - -XI -_ -to-ta:l-d-i-sp-l'a'c-e;m-e-n-t time intervals, the graph is parallel to time axis. I. Slope of P . - time taken - - b. 20 - 140 s, 280 - 380 s: For these time intervals, the value 12 - tJ of position (x) is increasing. = Average velocity c. 180 - 240 s: For this time interval, the value of position j. We know that slope of velocity-time graph is equal to (x) is decreasing. aeceleration. So 33. a. At point IV velocity is zero because slope is zero at this i. as slope is zero, so zero acceleration. point. ii. as slope is positive and constant, so constant positive h. At point I slope is positive and constant. acceleration. c. At point V slope is negative and constant iii. as slope is infinite, so infinite acceleration. d. At point II slope suddenly increases. iv. as slope is negative and constant, so constant negative e. At point III slope is positive but decreasing. acceleration. 34. a. At t= 3 s the graph is horizontal so the acceleration is O. = =From t 5 s to t 9 s, the acceleration is constant (from

4.50 Physics for IJT-JEE: Mechanics I 45 - 20 = 6.25 m/s2 x the graph) and equal to 9-5 d. .4/3m/s:b From I = 9 s to I = 13 s the acceleration is constant and 0-45 equal to - - - = -11.25 mis' 13 - 9 b. In the first five seconds, the area under the graph is the (a) (b) area of the rectangle, 20 x 5 = 100 m. Fig. 4.145 Between I = 5 sand t = 9 s, the area under the trapezoid is 36. To find, a, we apply the relation a = dv!dt, which is the same as a = .6. v/ /':;.,.t over time intervals in which v varies (112)(45 + 20)(4) = 130 m and so the total distance in the linearly with I. For the time interval between 10 = 0 and I, = 2 first 9 s is 100 + 130 = 230 m. s, the slope of the graph is positive and constant, and is equal to Between I = 9 s and I = 13 s, the area under the triangle is (1/2)(45)(4) = 90 m, and so the total distance in the first vol 13 s is 230 + 90 = 320 m. ai = -L'.v = (VI - = (2m!s - 0) 2 .= , . V2 - VI L'.t (t, - to) 2s - 0 1 m/s 35. a. Slope of graph IS constant, I.e., ---= constant. 12 - II Between II ;::::; 2 sand t2 = 4 s, the slope is zero, so the acceleration is zero for thi,s time interval. Between t2 ;::::; 4 s To find velocity at t = 4 s: and t3 ;::::; 5 s, the slope is negative and constant, equal to \"*8-0 8-v v·= 8 crnls Slope: 0 _ 6 = 0 - 4' ~ To find velocity at I = 7 s: 3 L'.v (V3 - v,) (0 - 2 m/s) , a2 = - = - - - - = = -2 m/s' L'.I (13 - I,) 5s - 4s \"*8-0 8-v v = 4/3 cmls Slope: --- = - - 0-6 0-7 (m/s2) a 8.0 . 3 b. a= slope of v - I graph=- - = - 4/3 emls which is can· 6.0 stant. c. For first 4.5 s: Let us first find the velocity at I = 4.5 s o-j--'_...L.......JL..-!--_!---,-I 2 3 4 ): (s) \"*8-0 8-v v = 2 emls ,, ,, () - 6 = () - 4.5 _________ __ L....-...l Now L'.x = at'ea under v-I graph for first 4.5 s -- '2 +;::;;;;} L'lx = ARcctanglc ATriang!e Fig. 4.146 g cm!s Figure 4.146 is a graph of the instantaneous acceleration for the motion. 2 cm/sl---t-:---,~ 37. We know that displacement is area under v-I graph. Area o 4.5 s from 0 to 2 s: 4 x 2 = 8 m, area from 2 to 4 s : -2 x 2 = 4 m, areafrom4t06s:2 x 2=4m Fig. 4.143 So displacement = 8 - 4 + 4 = 8 m When we find distance, we consider all areas to be +vc. = 4.5 x 2+ I x 4.5 x 6 = 22.5 . So distance = 8 + 4 + 4 = 16 m 2 em 38. Att=5sVI =u+al=O+2x 5= 10mls (Pig. 4.147) From I = 0 to t = 7.5 s: att = 15 S: VI = lOmis Let us first find velocity at I = 7.5 s att = 25 s: V2 = VI + al = 10- 2 x 10 = -10 m/s \"*8-0 8-v v = -2cmls at I = 35 S, V2 = -10 m/s 0-6 = 0-7.5 =att = 40 s: V3 = \"2 + al = -10 + 2 x 5 0 mls 8 Cll1/s To find x fOl. 0 to 5 S, x = I2 x 2 ,x 5 = 25 m 7.5 s f.or 5 to 15 S, x = 25 + 10.x 10 =1 125 m. for 15 to 20 s, x = 125 + v,l + 2al' 6$ = 125+ lO x 5+ 1 -2) 52 = 150m 2 cm/s 2( Fig. 4.144 I 2 for20t025s,x= 150+0 x 5+ ( - 2)52 = 125m The dislance is the sum of the magnitudes of the areas. for 25 t035s,x= 125+v21= 125-10 x 1O=25m d = -r (6s) (8cm-) + -1 (1.5s) ( 2cm- ) = 25.5 em I 2 s2 s 25 + v2t + 2al2 for 35 to 40 s, x =

Motion in One Dimension 4.51 = 25 - 10 x 5 + 21: x 2 X 52 = 0 d. False, while jogging Vav = L':.x (XI - xo) 6- 0 - = 30 - 0 L':./ (tl - to) '\\,(m/s) x (m) 150 = 0.200 km/min 100 e. True, while walking 50 L':.x (Xl -xil = (4km-6km) =-= V ,av L':.t (tl - tl) (50 min -30 min) 10 20 30 40 = --2-k.m- = -0.1 00 km / m.m (walk1.l1)g I (s) 20mm Fig. 4.147 The negative value indicates a velocity in the -x direction. A word of caution: Please note that the runner's av- 2.5 1 39. Time taken to reach the office = 5 h =30min erage velocity for the entire trip is not simply the average of her jogging and walking velocities, because the times (Fig. 4.148). 2 she spends in these two different motions are not equal. x(km) 41. a. The truck's position as a function of time is given by XI' = 2.5 VTI, with Vr being the truck's constant speed, and the car's position is given by Xc = (1/2) act2 Equating the two 9,00 9,]0 5,00 5,06 expressions and dividing by a factor of t (this reflects the am am pm pm fact that the car and the truck are at the same place at t = 0) and solving for t yields Fig. 4.148 t = 2VT = 2(20.0 m/s) = 12.5 s 2.5 = 1I0 h = 6 min ac 3.20 mis' Time taken to return home = 25 and at this time Xl' = Xc = 250 m. 40. We nrslsketch Fig. 4.149 (a) to depict the motion and indicate =b, v, ~ acl= (3.20 m/s2)(12.5 s) 40.0 mls our choice of locating the origin of the coordinate system where'the runner starts. We also show significant positions x(m) of the runner during the motion. c. 40of·,·~·,,,,,·,,,c',\"-c-~~'''~'~--~'-~'C-' 360 Position at to= 0 (km) x 240 200 ~ 6 ------ --,1---- I----!----~~ X t60 4 o 2 4 6 (km) t20 80 I' x, .1 2 40 _ I. .1 o+ - '-_ '-- ' -- '-_' -- _ 1. - 'o k:::.~'::\":'~~-'--'--'--'--'--'--'-:..J_ 1(8) 10 20 ]0 40 50 (mm) I 2 3 4 5 6 7 8 9 10 II 12131415 Fig. 4,150 Ca) (b) d, Let us see when the velocities of both become same Fig. 4.149 Fig. 4.151: for this v, = v, =:> ael = v, Position Time 3.2t = 20 =:> / = 6.25 s xo=O /0=0 tl = 30 min 40 - - vt(t) . x,Jogging XI = 6 km t2 = 50 min - - - Ve(t) , Walking = 4 km v,(mls) a. True,L':.x = (x, - xo) = (4 km - 0) = 4 km 20~-'--'--'-+'~/---'--'----'-~ Total distance· 8 km b. True, average speed = ---- . = Total tnne 50 min = 0.160 km/min c. False, the average velocity for the entire trip is L':.x = (Xl - xo) 4- 0 Vav =- - --- o5 L':.t (t, - 50 - 0 t5 to) 1 (8) = 0.080 km/min Fig. 4.151

4.52 Physics for IIT-JEE: Mechanics I 42. a. f>XOI = x j - Xi = -2 - 0 = -2m 15. d. s = ktl/2 =} a = d's = _~kt-3/2 f>X\\3=Xj-Xi =6-(-2)=8m dt' 4 As t increases, retardation decreases. _ f>XOI -2m b. VOl = - - = - - = -2 m/s 16. a. 17. a. The only force acting on bofh will be gravity which will f>t Is produce same acceleration g in both. Further, both the balls f>X\\3 8m are dropped simultaneously from same height, hence both 1)\\3 = - - = - = 4 m/s will come together on the ground. f>t 2s 18. d. s = 2I .at 2 =} t ()( vrs: c. v= dx = d (-4t + 2t2) = 4 (-I + t) mls - - dt dt =Therefore, att 2.5 s, V = 4 (-I + 2.5) = 6 m/s 43. a = 2t - 2 19. b. When a body possesses constant velocity, then both its t t2)4J 2t)~a. [\" dv =' }[o' adt =} v = (2t 2/2 - = t2 - 2t magnitude (Le., speed) and direction must remain constant. o On the other hand, if the speed of a body is constant, then =b. S 12 vdt = (t:33 - 22 2 = 320 m its velocity mayor may not remain constant. For example, in uniform circular motion, though the speed of body re- mains constant but velocity changes from point to point due Objective Type to change in direction, A body moving with a constant speed along a circular 1. b. If the displacement is zero, the distance moved mayor may path constantly experiences a centripetal acceleration. not be zero. For example, if a particle returns to its initial 20. c. x = at2 - bt3 position after moving through a distance, displacement will dx , velocity = - = 2at - 3bt- be zero but distance ~ovcred will not be zero. dt (~t ~:~)and acceleration = ( 2. a. =' 2a - 6bt. . Displacement 3. d. We know that average velOCIty = . and TIme Acceleration will be zero if Distance 2a - 6bt = 0 =} t = 2a a Average speed = -cecc.-- = 6b 3b TIme Since displacement can be less than or equal to distance, 21. a. V\"\" = VI (t12) ~ v,(t 12) = VI ~ V2 so average velocity can be less than or equal to average speed. 22. d. tl = S12 , t2 = S12 , V\" = _s_ = 2vI V2 4. b. At any instant velocity and speed will be equal. + +VI V2 II 1.2 Vj V2 S. c. 6. a. If the location of a particle changes, then both distance and 23. b. tl Sj3 S/3 S/3 -,/2 = -,/3 = - displacement must have some value. VI V2 v) 7. a. During retarded motion, acceleration and velocity are in opposite directions. S 3VIV2V3 + + + +Vav = = 8. b. tl t2 13 VI V2 V2 V3 V3 VI 9. c. To have distance equal to magnitude of displacement, the particle has to move in the same direction. The velocity may 2\"12 I 60t or may not be constant. 24. c. SI = 2. at = 2.(at)t = = 30t 10. d. s = kt 2 S, S3 __ a __ -a ds =} v = - = 2kt Fig. 4.152 st S2 = 60 X 8t = 480t, S3 = SI =30t dv =}a=-=2k st From above, acceleration is independent of time, hence v\" = SI + ++ S3 = 54 km/h acceleration is constant. t + 8t t 11. b. 25. a. tl = :x132. = 6X ' XI = 4.5 t2, X2 = 7.5 t2 12. c. Also XI + X2 = 2x. = (4.5 + 7.5)t2 =} x 13. b. s = kt, differentiating s twice to get acceleration, we see t2 = that acceleration comes out to be zero. 24 6x + 2x 14. c. Since maximum velocity is more than average velocity, +Total time: t = II 2/2 = 24 = 4X therefore ratio of the average velocity to maximum velocity has to be less than one, = - =X Vav 4 mls I

Motion in One Dimension 4.53 T J2; T26. c. Time of flight: = zatNow s = ut + I 2 = 0 x 3 + Z1 x 2 x (3)' = 9 m =} ex ,fii 37. d. v2 - u2 = 2as, v = 0 27. a. As both balls are dropped from same height. hence both Maximum retardation, S ex: u2 . When the initial velocity is made n times, then the dis- will come together on the earth. ittance over which can be stopped becomes n2 times. 28. b. Distance travelled in 4th seeond = J6 - 9 = 7 m 38. d. Relative velocity of policeman w.r.1. the thief is 10 - 9 = Now Dn = U + a - . 1 m/s. Since the relative separation between them is 100 m, Z(211 1), Glver\" = 0 hence, the time taken will be = relative separationlrelative velocity =100/1 = 100 s. =} 7 = 0 + a x 4 - 1) =} a = 2 mis' Z(2 39. c. We know that slope ofdisplacement-time graph is equal to velocity. SO VA = tan30° = 11.)3, 29. a, 30 = \" + a x 2, 60 = u + a x 4 VB = tan 60° = .)3, hence \"A/VB = 1/3 Solve to get\" = 0 40. d. Distance covered by the object in first 2 s: 30. b. x = 2 - 5t + 6t' =} dx 11 v = - = -5+ 12t hi = Zgt2 = Z x 10 X 22 = 20 m dt \",_0 = -5 + 12 x 0 = -5 mls 31. d.2ax = (50)2 - (lo? and 2(-ale-x) = ,,' - (50)' This gives 11' - (SO)' = (SO)' - (10)', Le., \" = 70 mls 32. a. v' = 108 - 9x' (i) dv dv dx d(y'lOs - 9x') dx Similarly distance covered by the object in next 2 s will a=-= -.-= also be 20 m, hence the required height = H - 20 - 20 = dt dx dt dx dt H -40m a= 1(-18x) .J]Os-9x'=-9xm/s2 I1 2.)108 - 9x' 41. c. h = Zgl2 and h - 20 = zg(t - 1)2 1, Solving them we get, 1 = 2.5 sand h = 31.25 m. 33. c. We have h = ZgT- 42. a. Relative velocity of overtaking = 40 - 30 = 10 ms- l. T. ~g (~)2 h Total relative distance covered with this relative velocity dur- 9 ing overtaking = 100 + 200 = 300 m In - s, the distance fallen = 3 23 So position of the ball from the ground So time taken = 300/10 = 30 s u h 8h =11--= -m 43. a. Time of ascent = Is=} - = 1 =} II = 10 mls 99 g u2 102 dx d'x 34. b.x=acost, dt = -asint; dt2 = -acost Maximum height attained = - = - - - = 5 m 2g 2 x 10 35. b. Beeause one taxi leaves every 10 min, hence at any instant there will be 11 taxies on the way towards each station, otie 44. c. Maximum height attained ex u 2 will be arriving and another leaving the other station. Figure shows the location of taxies going from X to Y at the instant 45. b.Given7x = 1i.(2n-1)andx = ~g(l)2 2.00 PM. The taxi which left station X at 0.00 PM has just arrived at station y, Consider the taxi leaving the station Y 22 Solving these two equations: n = 4 s at 2.00 PM. 46. c. v' - ,,' = 2as Suppose velocity of the middle part = Vm Then v;! - u2 = 2as x :12 = as or + + +v2 = U Z as = uZ m 0000000000000 JII' 2=:} VII! = --+2v- v2 _ u2 u2 VZ O < n \" < ! \" r r l 0 ! ...... O<n\"<!\"tr>0!,....,O 2 = - -2 - N~...;,.....;......;......;,.....;OOOOOO til Ej:U:j;U;II;Uj;U+, til til til til til til X 1 2 3 4 5 6 7 8 9 10 II Y Fig. 4.153 47. a. t = ax' + fJx = x(ax + fJ) It will meet all the 11 taxies marked 1 to 11 as well as 12 dx dx other taxies which would leave the station X from 2.00 PM 1 =2a-.x+fJ- to 3.50 PM. When it arrives at the station X at 4.00 PM, there will be one more taxi leaving that station. However, it will dt dt not be counted among the taxies crossed by taxi under con- sideration. That is, it will cross 23 taxies leaving the station dx dv -2av = -2av', X from 0.10 PM to 3.50 PM. -(fJ - -____ -V - dt -- .- +2ax)2 fJ+2a-,,' dt - 48. a. t = ,jX + 3, differentiating with respect to I, we get, 1 = 36. a. Given that u = 0 (the electron starts from rest) I dx dx --+Oor-=2,[X . d- =v2 m,ls 2,jX dt dt A t any tIme t: v = kt = 21. a = (constant dt when velocity is zero, then 2,jX = 0 or x = O. acceleration)

4.54 Physics for IIT-JEE: Mechanics I 49. d.Ifspeeds are comparable to the velocity oflightc, according 2S to theory of relativity, velocity of B relative to A (when both 25 = Vmaxt2 or t'2 = are moving along the same line in opposite directions) is gI.ven by: VBA = [ IVB ++V ~VVA ]' From thlS',\"It IS CIear that'1f Vmax c2 1 lOS v+c 5S = -2Vmax t3 or t3 = -- VA is equal to c, VBA = - - v = c Vmax 1+- Total displacement S + 2S + 5S c v\" = Total time = 2S 2S lOS .~o~: Spt~~ij'liiJh,i!Jln4epeh4tiitM~~k/iip.e.~WtF/I/P~f· -+-+- 1J9\"i!it$~!tI'CI{IIII/i:ulJ,$ery.Ilr,<· · · \" : S ! Vmax Vmax Vmax 8S = 4 - Vm\" 14S 7 Alternatively: V a, Total displacement d2x d2y Vmax Total diSPlacement) (DiSPlacement) _. - 50. b. ax = = 8 and a, = = 0 2 during acceleration + during uniform Ja; a;dt 2 . dt 2 ,-__ ( and retardation + = 8 m/s2 veloclty Hence, net acceleration = Vav = 8S = 8 4 14 7 51. b. The required ratio is 1:3:5: ... so on 2(S+5S)+2S 52. c. Relative acceleration of both will be zero w.r.t. each other. 57. b.When a body slides on an inclined plane, component of SO, S\", = ur\"t or lOO = lOOt or ( = 1 s weight along the plane produces an acceleration. 53. a. Since the last five steps covering 5 m land the drunkard emgsine fell into the pit, the displacement prior to this is (11 - 5) m = a = = g sin = constant. If s be the length of m 6m. Time taken for first eight steps (displacement in first the inclined plane, then eight steps = 5 - 3 =2 m) = 8 s. Then Time taken to cover =0+ I2 = 21: g . e x t 2 ~first six meters of journey = x 8 = 24 s sm S 2:at s' 1'2 S t2 = -- 0[ - = -,2 S' t t2 Time taken to cover last 5 m \"'\" 5 s i,S Given, 1 = 4 sand s' = Total time = 24 + 5 = 29 s t' = Iff = 4{f =~ = 2 s 54. c. Here h = 21: x 10 X (5)2 = 125 m In 3 s it falls through: 58. a. Bomb B, will have less velocity upward on dropping, so h, = 21: x 10 x (3)2 = 45 m it will reach ground first. Rest 80 m is covered in 4 s. Hence, total time taken = 3 59. c. Time taken to cover first n m is given by s + 4 s = 7 sec. n= ~gt; or til = y[2git =55. b. 200 u x 2 - (l/2)a(2)2 2 oru -a = 100 (i) Time taken to cover first (n + l)m is given by /2(n:t,,+1 = 200 + 220 = u(2 + 4) - (1/2)(2 + 4)2a I) oru-~=m 00 • Solving equations (i) and (ii), we get a = 15 cmls2 and t(n+l) fin flSo time taken to cover (n + I)thm is given by ~g =U 115 em/s. Further, V = u - al = 115 - 15 x 7 = 10 cmls. 1,,+1-1\"= - -g = -[v n +l-v'iIl g 56. c. Graphically, area of (v-t) curve represents displacement: This gives the required ratio as: 1 2S Jj, (h -VI), (v'3 - h), . .. etc. (starting from S = -2Vmaxtl or tl = n =0) Vmax i 60. c.AB=30m,BC=20m,CD=30hm /:,BCE is isosceles (Fig. 4,155), so v BE=BC=20m AE=AB-BE I, =30-20=lOm EC =20hm Fig. 4.154 ED=CD EC=30h-20h= IOhm =/:, ADE is isosceles, so AD = AE 10 m

Motion in One Dimension 4.55 Br-20=m~~C 69. c. Given u = O. Let during three phases time taken are t. 81 = =and I, respectively. Vm\" at 60 kmlh 30 m \"I2 at2 + v max 8t + I2 at 2 + 8vmax t +at 8vmax E lilt 10 {) A Vav = 1 + 8t + t = lOt Fig. 4.155 60+8x60 = 10 = 54km/h 70. c. According to 3rd cquation of motion. o.61. a. u = V = 27.5 mis, t = 10 s v2 - u2 = 2as v - u 27.5 , Given V = 3v, u = V and a = g a = - - ' = - = 2.75 mls\" t 10 4v2 Or (3v)\" - v' = 2gs or s = - In the first 10 s, distance travelled: I .• g .\\', = 0 x 10 + 2: x 2.75 X 102 = 137.5 m y. . f2gh 71. a. Tnne of fall = In the first 20 s, distance travelled: h Time taken by the sound to come out = - S2 = 0 x 20 + -I x 2.75 X 202 = 550 m c 2 +'Total time = Vf2gh J:. = h [ y/fi2h+ ~] Required distance = 550 - 137.5 = 412.5 m c c 62. b. Distance covered in first three seconds = Distance covered ss s in last second 72. c. tl = - , t2 = - •... , t/1 = - =:>~g(3i = 1£(21 - 1) =:> 1 = 5 s VI V2 Vn =Average speed (total distance)l(total time) 22 63. a. S = ul + I 2 v,s I s I S -at VI VII 2 -76 = 4 x 6 - ~ x 10 x (6)\" =:> u = 52 mls I, 12 I\" 23 Fig. 4.156 64. a. Suppose v be the velocity of the body after falling through ns =:> V = - - - - - - half the distanee. Then I, + 12 + ... + In s = -392.2 = 19.6m, U =Oandg =9.8 mls2 . ns n =:> V= s S S II 1 -+-+ ... +- -+-+\".+- v2 = u2 + 2gh = 02 + 2 x 9.8 x 19.6 Vj V2 VII Vj V2 Vn . ' I 1 1)taking reciprocal, we get = = -1(1- + - + ... - =:> v = 19.6 mls. V n Vt V2 Vn When the acceleration due to gravity ceases to act, the body travels with the uniform velocity of 19.6 mls. So it hits 73, c. H =- =u2 the ground with velocity 19.6 mls. 2g Given V2 2Vl (1) 65, c. Distance travelled by first train, vrA to S: = u2 - 2gh (2) vi uA to c: = 2 - 2g(-h) (3) s,= -u 2= (15)2 --=112.5m 2a 2 x I Solving (1), (2), and (3), find the value of u2 and then u2 Distance travelled by second train, get the value of H by using H = -2g (Fig.. 4.157) (20)' V, H S2 = -2 x-I = 200m B~ I - Total distance travelled =112.5 + 200 = 312.5 m Distance of separation = SOO - 312.S =187.S m uh Sco' 4 S =:> V2 = 4S kmlh A +66. d. t = - =:> 60 = 0 h Vrel 3 V2 67, a. 3VA = VB, Sco' = v\",t =:> 100 + 60 = (VA + VB) x 4 .j, Solve to get, VA = 10 mls and VB = 30 mls C ~'- 68. d. Relative speed of trains = 37.5 + 37.S = 75 kmlh , II Time taken by the trains to meet = 90175 = 61S h Since speed of bird = 60 kmlh, So distance travelled by Fig. 4.157 the bird = 60x61S = 72 km.

4.56 Physics for IIT-JEE: Mechanics I 74. b. Time taken by same ball to retum to the hands of juggler D G)- h = - ( g(11 - 12)12 - gli =Ii ---w-=2u 2 x 20 = 4 s. So he is throwing the balls after each I s. Let at some instant he is throwing ball number 4. Before j (iv) (v) Is of it he throws ball 3. So height of ball 3: 2=} h = gtl t2 h3 =20 x 1- I = IS m For third case, u = 0, t = ? G) G)- h = 0 x t - 21O(lj2 Before 2 s. he throws ball 2. So height of ball 2: I!t' or h = I!t 2 I Combining equation (iv) and equation (v), we get h2 = 20x2 - 210(2)' = 20 m Before 3 s. he throws ball 1. So height of ball I: j2 1. = ~ 2gtll2 OJ I = =I 21!t = h, 20x3 - -10(3)2 15 m 78. c. For no collision. the speed of car A should be reduced to 2 vB before the caJ's meet, i.e., final relative velocity of car A 75. b. Let initial relative velocity, relative acceleration and the with respect to cal\" B is zero, i.e., Vr = O. relative displacement of the coin with respect to the floor of Here Ur = initial relative velocity = VA ~ VB the lift be Un al\"' and s,., then Relative acceleration = ell\" = -a - 0 = -Q s, = u,t + (1/2)a,.t' Let relative displacement = SI\" and u, = u, - u, = !O - 10 = 0 Then using the equation, v;' = u; + 2ar s,. a, = ae - a, = (-9.8) - 0 = -9.8 m/s2 o -2 _ (VA - VB) 2 - 2as,. or...I.,. _ \"(V-A\"'- -VcB_~)' Sr = Sc - Sf = -2.45 m _ -2.45 = Oft) + (1/2)(-9.8)t2 2a ort2 = 112 or t = 1/.J2 s .. . (VA 2- aVB)2 colllStOn, I.e., 76. d. a = g sin a = g sin(90' - Ii) (Fig. 4.158) :sFor no Sr :::: S, s = geosB 1= 2ReosO 79. a. From considerations of symmetry, the four persons meet at. the centre of the square. The displacement from the corner to the centre of the square for each person is given by Sr = Jel2 +el' -ti~ .J2 11\"\"0 = 2 The speed of each person can be resolved into two com- ponents: the radial component and the perpendicular compo- nent Throughout the journey, the radial component of veloc- , a ity towards the centre is given by Fig. 4.158 VI\" = V cos 45() = ~V-- T.Ilne = -S, = -el/.-J'i = ti .J2' V, V/.J2 V Now usm. g s = ut + 21ut2 80. d. Here til! = -kv'3 =} I = 0 x t + 2I g cos Iit 2 - I 1\" 11tit =} 2ReosO = 2geosOt2 or -tiv, = -kelt or v -dv, = () -k tit vov' =>t=j!f or [- _I_J\" = -kt or __1_ + _1_ = -kt 2v' ~ 2v2 2v(2} This is independent of Ii. or v2 = ---,vO2'-2~ or v = ~=V~o = 77. e. Suppose the body be projected vertically upwards from A 1+ 2vokt J2v&kt + I with a speed uo. 81. b. Given v = 3x2 - 2x, differentiating v, we get Using equation, s = ut + (~) at 2 til! tix - = (6x - 2)- = (6x - 2)1! (~)For first case, -h = uoll - ell glf (i) tit (~) gl~For second case, -h = -uOt2 - =} u = (6x - 2)(3x' - 2x). Now put, x = 2 m G) get? - til+ +(i) - (ii) =} 0 = UO(t2 t,) (ii) =} a = (6 x 2 - 2)(3(2)2 - 2 x 2) = 80 mis' (iii) DUo = ( g(tl - t2) 82. a. Suppose h be the height of each storey, then 25h = 0 + 21 x 10 x (' = 21 x 10 X 52 or h = 5 m Put the value of Uo in (ii), In first second, let the stone passes through n storey, so I n x 5 = ;; x 10 x (1)' or n = I

Motion;n One Dimension 4.57 83. b. Suppose v be the velocity attained by the body after time The third water drop will be at a height of = 5 -1.25 = 3.75 m. II.Thenv=u-gli (I) 91. b,SI +S,+S3+S4= 16m, Let the body reaches the same point at time t2. Now S, : S, : S3 : S4 = 1 : 3 : 5 : 7 Solve to get S, = I m, S2 = 3 m, S3 = 5 m, S4 = 7 m velocity will be downward with same magnitude v, then - v = u - gi2 (2) (I) - (2) =} 2v = g(/2 - II) 2;or/,-/I= = ~(II-g/')=2(~-tI) I .2 S4 1 =84. a, -h = 4 x 4 - 2:10(4)' =} h 64 m Iiiiiiillliiim 85, a, The velocity v acquired by the parachutist after lOs: Fig. 4.159 v = II + gl = 0 + 10 x 10 = 100 m/s 92. c. x 2 = 1 + I' or x = (1 + 1')'/' Then, SI = ul + 1 1x 10 X 102 = ' 2: gl2 = 0 + 2: 500 m -dx = -I(1 + (2)-1/221 = 1(1 + (2)-1/2 The distance travelled by the parachutist under retarda- dl 2 tion, S2 = 2495 - 500 = 1995 m = x3 Let Vg be this velocity on reaching the ground. .X v; -Then v; - v2 = 2as2 93. b. Suppose u be the initial velocity. or (100)2 = 2 x (-2.5) x 1995 or Vg = 5 m/s 86. c.lf police is able to catch the dacoit after time t, then I 2 . . a2 vt C': x + 2:\"t . ThIs£lves 2:1 - vI +X = 0 or t = v ± -.lv2 - 2ax Velocity after time I,: VII = U + aI, a +Velocity after time II + 12: V22 = It a(t, + I,) For t to be real, v2 ::: 2ax and Velocity atIertime II + 12 + 13: V33 = It + a(ti + 12 + (3) 87. d. Here relative velocity of the train W.r.t. other train U+VII u+u+at, 1 = V v. Hence, 0 - (V - vp = 2ax --2- = llt ] iVI = Now = U+ (V - v)2 VII +V22 u + all + 1 2: a12 or a=- 2 = 2x Minimum retardation = (V - V)2 +1)22 V:n 1 2 .. = It + aI, + aI, + 2:a13 2x 88. b. distance covered = S = Vav x time For first second: SI = 5 x 1 = 5 m v, - 1 For second second: S2 = 10 x 1 = 10. m So V2 = - 2:a(II + (2) For third second: S3 = 15 x 1 = 15 m +I Total distance travelled v, - V3 = -2:a(t, (3) S = S, + S2 + S3 = 5 + 10 + 15 = 30 m v+u ++(VI - V2) : (v, - V3) = (I, I,) : (I, 13) 89. b. Given v\" = -2- = 0.34 and v - u = 0.18 Solving thcse two equations, we get 94. c. Let the man starts crossing the road at an angle iJ with the u = 0.25 mis, v = 0.43 m/s. Given s = 3.06 m roadside. For safe crossing, the condition is that the man must Now usc v2 - u2 = 2as to find a. cross the road by the time truck describes the distance (4 + 2 90. d. By the time 5th watcr drop starts falling, the first water cot 8) drop reaches the ground. 4 + 2cot8 21 sin I) 8 So, 8 = - -v - or v = ;2:s-icn-i:J:+-c--o-s;e: 12 21: x II = 0, h = 2:g1 =} 5 = 10 x I2 =} I = Is For mm. imum v, dv = 0 - d8 Hence, the interval of falling of each water drop -8(2 cos iJ - sin e) . Is or . , =00r2cose-sml)=0 = - = 0.25 s (2sine +cos8) 4 e e2 ' I When the 5th drop starts its joumey towards ground, the or tan = 2, so sin 8 = 1<' cos = I< v5 v5 third drop travels in air for 0.25 + 0.25 = 0.5 s 88 ... Height (distance) covered by 3rd drop in air is Vrn;, = ( 2 ) 1 = -J5 = 3.57 m/s 2 -J5 + -J5 h, = 2I:g1 2 = 2I: x 10 x (0.5)2 = 5 x 0.25 = 1.25 m

4.58 Physics for IIT·JEE: Mechanics I Graphical . . 1/)3 1 Concepts ReqUIred ratIO = /0 = - 1. h. Because in options (a), (c) and (d), we can find from the . ,,3 3 graphs that at a single time there can be more than one ve~ loeities, which is not possible practically. 12. d. Velocity given by OA =' tan 60' =)3 mls I 2. h. We know that area under v-t graph is displacement. I Velocity given by A lJ = - tan 30\" = - /0 m/s v3 Area from 0 to 6 s = 2: x 6 x 2 = 6 m . . )3 I ReqUIred ratIO = /0 = 3: I Area from 6 to 8 s = 2: x 2 x (-2)= - 2 m 1/,,3 Area from 8 to 10 s = 2 x 1 = 2 m 13~ d. From 0 to 11, velocity is positive and constant as indicated So Net displacement = 6 - 2 + 2 = 6 m by positive and constant slope. From tt to 13, slope is zero, hence velocity is zero, 3. c. Displacement = area under the graph From t:, to (4, velocity is negative and constant as indicated 1 .I I by negative and constant slope. = 2 x 2 + -(2 + 6) x I + - x 1 x 6 - - Option d satisfies all these observations, 2 22 14. d. At 1= 0, velocity is positive and maximum. As the particle xlx6-lx6+2x4=lOm I goes up, velocity decreases and becomes zero at the high- est point. When the particle starts coming down, velocity 4. ll. Area from 0 to 10 5 = 2:llO + 4Jx5 = 35 m increases in negative direction, 15. a. At time {, let displacement of first stone is Sl I . 2I\" g(2 Arcafrom 10to 125= 2: X 2 x (-2.5)= -2.5m that of second stone IS S2 = ul - Distance travelled = 35 + 2.5 = 37.5 m Distance bctween the two stones at time t: 5. c. Maximum height will be attained at 110 s. Because after +S = S, S2 = ul =} S = ul 110 s, velocity becomes negative and rocket will start coming down. Area from 0 to 110 s I =2: x IIOx IOOO=55,000m=55km 6. d. Displacement in 12 s;::::: area under v-t graph I = 2: x (12+5) x4= 34m V.,W = Displacement 34 17 Fig. 4.160 = -12 = -6 m/s Time So the graph should be a straight line passing through origin as shown by option a, Hence (a) is incorrect. 16. b. Let the particle is thrown up with in-itial velocity u. Dis- Option b is incorrect because during first 3 s velocity I placement (S) at any time 1 is S = ul _. 2: gl' increases [rom 0 to 4 mls The graph should be parabolic downward as shown by Option c is incorrect, because in part AB, velocity is option b. constant. 17. b. In (I). slope is negativc and its magnitude is decreasing with time. It means slope is increasing numerically. So ve- 7. c. Acceleration bctween 8 to 10 s (or at 1= 9 5): locity is increasing towards right, that means acceleration is positive, a= v, - v, 5 - 15 -5 m/s2 In (IV), slope is positive and its magnitude is increasing with -\"-- = --- = time, So velocity is increasing towards right, that means ac- 12 - I, 10 - 8 celeration is positive, 18. a. We know that for a body thrown up, its displacement is 8. c. Near B velocity decreases with time, Hence there is retar- givenasS = ul - 2I\" gl2,Sothes-f graph is parabolic down- dation or there is opposing force acting on the body. ward. 9. 3. Maximum acceleration will be from 30 to 40 s, because Also the ball collides inelastically, so it will rebound to less height every time as shown by the graph, slope in this interval is maximum t[, f2, t3 arc the instants when the ball collides with the ground, v, - v, 60 - Here slope of s-t graph is suddenly changing from nega- 20. 4mIs2 a=---= 12 - I, 40 - 30 10. a, In option a., there is some part of the graph which is .F to t axis. It indicates infinite acceleration which is not possible practically. ~11. c. During 0 A, acceleration = tan 30° = Ill/S2 ,,3 During AB, acceleration = - tan 60\" = -)3 m/s2

Motion in One Dimension 4.59 < tive to positive. It means velocity before collision is negative 2. b., d. Freely falling block will reach the ground first, because (downward) and after collision is positive (upward). it has to travel less distance and with greater acceleration in comparison to the other block. 19. 3. From 0 to 11, acceleration is increasing linearly with time, hence v-t graph should be parabolic upwards. Both the blocks will reach the ground with same speed, be- From f\\ to f2, acceleration is dereasing linearly with time, cause potential energy of both decrease, by same amount hence v-t graph should be parabolic downwards. which gets converted into KE. 20. a. Att = 0, slope of x-I graph iszero, hence velocity is zero at 3. 3., c. al = 2 ms2, el2 = -4 m/s2 t ::::::: O. As time increases, slope increases in negative direction, Vo = alt! = 211, Vo = a2t2 = 4t2 hence velocity increases in negative direction. At point '1' slope changes suddenly from negative to positive value, hence I, + 12 = 6 => -V2o +. -V4o = 6 => Vo = 8 mls velocity changes suddenly from negative to positive and then velocity starts decreasing and becomes zero at '2'. Option ea) a,A ~ ~ B represents all these clearly. C \"2 • •I 21. d. Before the second ball is dropped, the tirst ball would u=O I, VO I, V\"' 0 . = \"I2 gto2. Arter dropping Fig. 4.162, have travelled some distance say So second ball, relative acceleration of both balls becomes zero. Total distance travelled So distances between them increascs linearly. After some S = AC + C13 = I '+ -2I x V(h time, the first ball will collide with the ground and the distance -2Vol, bctween them will start decreasing and magnitude of relative => S = 1 + 12l = I, velocity will be increasing for this time. Option (d) represents -2vo[l, -2 x 8 x 6'= 24 m all these clearly. 4. a., b., c. Let they meet at height h after time I. 4 => a=-21+4 h = 1001 - -I gl 2 -+ for first arrow 22.b.a=-2t=4 2 V= jadl+C=- j(21+4)+C=-12 +4t+C = lOO(1 - 5) - ~g(1 - 5)2 -+for second arrow => t = 12.5 (after solving). So a is correct. Hence graph will be parabolic .. = .M- = -2x'l-OO- = 20s. 23. c. Particle will acquire the initial velocity when areas A I and Tnneotfiightoffirstarrow: l' Ii 10 A2 arc equal Fig. 4.161). For this 10 = 8 s Second arrow will reach after 5 s of reaching first, so (b) is correct a v( = 100 -10 x 20 = -lOOmis to 10 V2= 100--IOx 15=-50m/s A, . ~VI = 2 : 1, so C . correct. ratIO : IS 112 Fig. 4.161 Maximum height attained 24. a. For 0 to 5 s, acceleration is positive; for 5 to 15 s acceler- H = -u2 = -(100-)2 = 500 m. Hence d is incorrect ation'is negative; for 15 to 20 s, acceleration is positive. 2g 2 x 10 25. c. For 0 to 3 s, acceleration is positive; from 3 to 5 s, accel- 5. a., c. I = Vf'2gh' v = -,J2gh eration is zero; from 5 to 7 s, acceleration is negative. 6. a., c. Ball A will return to the top of tower after 26. d. The slope of the graph is negative at this point. 27. 3. Uniform motion involves equal distances covered in equal 2u 2 x 10 time intervals or the velocity is constant. 1'=g=1O=2s Multipfe Correct with a speed of 10 m/s downward. And this time 13 is also projected downward with 10 m/s. Answers Type So both reach ground simultaneously. Also they will hit the 1. b.,c.,d. A body having a constant speed can have a varying ground with same speed. velocity due to change in direction of velocity. Thus a body having constant speed can have an acceleration. I If velocity and acceleration are in same direction, then dis- 7. a., d. From s = 2at2, U = 0 tance is equal to displacement, because then there is no change in direction of motion. The body will continuously s ex t 2• Since particle starts from rest and acceleration is travel in one direction only. constant, so there is no change in direction of velocity and particle moves in a straight line always. 8. a., b., d. a = (/v, if velocity changes, definitely there will be <It acceleration. If speed changes, then velocity also changes, so

4.60 Physics for IIT-JEE: Mechanics I definitely there will be acceleration. Options c and dare eorrcct and options a and bare Acceleration may be due to change in direction of velocity wrong. only and not magnitude. 14. a., d. Since the graph is a straight line, its slope is constant, If body has acceleration, its speed may change if accel- it means acceleration of the particle is constant. eration is due to change in magnitude of velocity, Velocity ofthe particle changes from positive to negative at t = lOs, so particle changes direction at this time, 9. a., d. The body will speed up if angle between velocity and The particle has zero displacement up to 20 s, but not for the entire motion. acceleration is acute. The average speed in thc interval () to lOs is the same as the average speed in the interval lOs to 20 s because distance 10. c., d. Since average acceleration = change in velocity/time, so covered in both time intervals is .same. average acceleration is in the direction of change in velocity. 15. a., b., c., d. Particle changes direction of motion at / = T. Also if initial velocity is zero then final velocity and change in velocity will be il1 same direction. Acceleration remains constant, because velocity-time graph is a straight line. Displacement is zero, because net area is 11. b., c., d. In circular motion speed is constant but velocity is zero. Initial and final speeds are equaL varying. For a body moving with constant velocity, accelera- 16. a,b,c Initially at origin, slope is not zero, so the particle has tion is zero. At highest point of vertical path, velocity is zero but acceleration is finite. A body cannot have varying speed some initial velocity but with time we see that slope is de- without having varying velocity. creasing, and finally the slope becomes zero, so the particle stops finally. 12. a., b., c. If a vertieally projected body, returns to the starting As the magnitude of velocity is decreasing, so veloeity and acceleration will be in opposite directions. point, its displacement and average velocity become zero. As 17. a., c., d. Maximum value of position coordinate = initial co- ordinate + area under the graph upto I = 24 s (As upto I = 24 s acceleration is constant, so average speed during upward or the displaeement of the particle will be positive (Fig. 4.163» downward motion is (u + 0)/2 = u!2. The same will be the average speed for the whole motion. f f5 5 vdl (41 - (2)dl 13. c., d. - -4 = 0 0 Average velocity v --- = \"----- f f5 5 dl dl 6 o0 t, [ ']5212 -\"13- 0 , ,2 - --I 30 40 125 0 10 1824 !(sec) 50-- = _~32- = 25 5 -2 5 5 3x5 3 , ··6 -------------1 For average speed, let us put v = 0, which gives Fig. 4,163 I = () and I = 4 s. =-16+ [(2 x 10)+ C~6) x(18-1O) f f4 5 vdl + vdl .. average speed = o .-'-'.4_ - - ' f5 + 1 ; 6 x (24 _ 18)] dt o = -16 + [20 + 32 + 18] = 54 m f f4 5 At time t = 18 s (41 - (2)dl + vdl Position = - 16 + Area of graph upto / = 18 s o4 = -16 + [20 + 32J = 36 m 5 2/2 _ t2]4 + [2/2 _ t2]5 At time I = 30 s =[ 3 0 Position = - 16 + Area of graph upto I = 30 s 34 ~ x x = 36 m [70 - 6 6]= -16 + 5 ¥)(32 - + [2(25 - 16) - ~(125 - 64)] 5 18. a.,b.,d. Four slopes are +2, 4 -3, 2. For . accel- = 3 --, maXImum 3 5 eration, option; a is correct. For maximum magnitude of ac- Foracce1eratl·On: a = -dv = -d(4/-1 2 )=4-21 celeration, option b is correct For displacement, from\"t = 0 dt dl to I = 7 s, it is increasing then falling, so option d is also At I = 0, a = 4 m/s2 correct.

Motion in One Dimension 4.61 Assertion-Reasoning For Problems 8-9 Type 8. b., 9. b. Sol. 1. a. When the body returns to its initial point, its displacement is zero, but distance travelled is not zero. 8. b. 2. b. Two different physical quantities may have same dimen- u=O a=:(X I •v, a ~-f3 v=o sions, B S2 C 3. c. Average velocity of the body may be equal to its instanta- neous velocity, because instantaneous velocity can take any 12 value. Fig. 4.164 For a given tfme interval of a given motion, both average velocity and average speed can have only one value as dis- From A to B: apply v = II + al placement and distance will have single valucs. +=} vo = 0 Cit, =} I, = voa 4. a. When body reverses the direction of motion it is momen- From B to C: again apply v = u + al tarily at rest, but it still possesses acceleration. Velocity zero =} 0 = va - fJI2 =} 12 = vo/fJ does not mean that acceleration is also zero. . I, + 12 = + -~ ~ = I =} Va = afJI 5. a. If direction of velocity changes (magnitude mayor may GIven --- not change), we say that velocity changes. If velocity changes I=}- then definitely there will be acceleration. a fJ a+fJ 6. a. When there is retardation, velocity decreases. So retarda- Vo is the maximum velocity attained. tion is equal to the t.ime rate of decrease of velocity. This retardation will bc oppositely directed to velocity. 9. b. From A to B: apply s = lit + I 2:a12 7. a. If velocity is constant, then displacement and distance will he equal, then magnitude of average velocity is equal to av- =} SI = () x I, + I2 = 1 I (I) erage speed, 2:atl 2:(al,)I, = 2:Voll 8. b. Such a body can move along any curved path including Similarly from B to C: circular path. =} S2 = vol2 + I2 = vot2 - I = 1 9. a. In kinematical equations, mass does not appear, 2:( -fJ)12 2:(fJ I 2)12 2: Vo12 10. b. Acceleration depends upon the force applied. Now total distance travelled: = S, + S2 = 1 + I = I + 12) 2:Votl 2VOt2 2vo(l, 1 afJI afJt 2 + += - - - I = c--'----cc- 2 a fJ 2(a fJ) lZompretiensive Also. we can find (I) and (2) by using the formula s = -u+-v I Type 2 For Problems 10-11 10. b., 11. d. Sol. J;'or Problems 1-2 10. b. Let they meet after time I. then distance travelled by both 1. d., 2. a. in time t should be same Sol. S = 8t = -I412 =} I=4s 2 2. a. \\'- M +2N14 V = -ds = 2NI\"3 dl 11. d. s = 8t = 8 x 4 = 32 m 4 putting I = I s, we get v = 2 N. For Problems 12-14 For Problems 3-5 12. b., 13. a., 14. b. 3. a., 4. c., 5. b. Sol. 12. b. s = 2I:gl~ Sol. S = I2 =} 2: gn t12 -- 50 x 2 100 10 or -ortl =- (I g g ~ S = 2: (2n - I) =} SIX (2n -I). and 100 = I 2 orl IO~ V = gn::::} v ex 11 _gt =-- For Problems 6-7 2~ 6. b., 7. a. 10 Sol. t2 =t -I, = -(v'z-l) =O.4tI 6. h. Here acceleration is constant. So we can use ~ s = ut + iat2. s-t graph will be parabolic. 13. a. t, : t2 = -~10 x --'0~::-- = - = ]0 5 1O(v'z - I) 0.4 2 7. a. Using v = u + at, we find that velocity-time graph will be 4 straight.

4.62 Physics for IIT-JEE: Mechanics I 10v'z ~ . Displacement = v'J2 + 122 = .jI45 '\" 12 14. b. t : t, = - - x - = v'z ~ 10 For Problems 15-16 15. c., 16. a. Sol. 15. c. s = at + I = 4 x 5 - 21 x 9,8 X 52 = I + -634 x 7 = 1 + -448 '\" 150 2at' 3 = 20 - 122.5 = -102,5 m For Problems 22-23 This shows that the body is 102,5 m below the initial position, i.e., height of the body = 120.5 - ·102,5 = 18 m, 22. d., 23. d Sol. 16. a. The distance travelled by the balloon in 5 s upwards = 4 x 5 = 20 m. C1 Distance of separation = 20 + 102.5 = 122.5 m , VI For Problems 17-18 ~B.~s1\"1 -- 17. b., 18. b. Mo>--+- Sol. V2 t = t 17. b. Let at time t. the cyclist overtakes the bus, then Fig. 4.167 96 + (distance travelled by bus in time t) CM=50m = (distance travelled by cyclist in time t) v, =20 mis, v, = 15 mis, AC =v,t, MB =v2t =} 2I x 2 x t2 + 96 = 20 x t S = v'M A2 + M B2 = /(5O-=-~,t)2 + (V2t)2 =} t2 - 20t + 96 = 0 ds This gives, t = 8 s or 12 s, Hence, the cyclist will overtake From dt = 0 (for minima), find t and put t in s. the bus at 8 s. For Problems 24-28 18. b. It is quite clear from above explanation that the bus will overtake the cyclist after a time of 12 s, 24. a., 25. a., 26. c., 27. b., 28. c. For Problems 19-21 Sol. 19. a., 20. d., 21. a. Sol. 24. a. SI = at S2 = v,t, = V2t2 (1) --, - 19. a. Distance = n R, Displaeement = 2R 22 R .. nR 1f §. §. atIO IS 2R = 2 SI 2 2 U t VI V2 2 II 12 20. d. Sec Fig. 4.165 Fig. 4.168 N 2I = I, + t, (2) +j W·~-::-+-::--E From (1) and (2) -/ +i I I = :2;(-v;-IV--2+'t.-v-,;)- ut S v,'lV = S = SI+S2 2+ 2v ,11 Fig. 4.165 - -.- = -\"---- ~} -4- -} ----* I1 1 + +Put the value of 11 and get v\" = -U-('V-I-';:v-,c)-=-2'v-~I -V2\" S = Sl +S, +S3 +2 (VI v,) =40) + 301 - 30v'z (1 cos 45 +) sin45) = +10) 25. a. x,(t~O) = 4, X2(t~51 = 44 21. a. Fig, 4, 166 X2 - XI 44 - 4 16f-----\" Vav = - - - = - - = 8 rnJs < 12 - 11 5 - 0 t 26. c. F = -kV' or rna = -kv2 or dv -kv' y 4f---7f m-= dl xx=---I--..x=2 Vo k Fig. 4.166 v= voal + 1 • where a = - m ds s Put v = - and find s. Then Vav = - dl I 27. b. 28. c. Net displacement is zero, so average velocity is zero.

Motion in One Dimension 4.63 20 20 v (:'l/sdl__Z~1----;t\\ Vi = -4:= 5 r!l/s, v.{ =)\"-;:- = 4 mis, o~~I~----~2~O~(2~O~+~t) --+1 (5) aav = - Vi -4- 5 9 Fig. 4.170 = -1 mis' For Problems 29-30 29.' c., 30. b. Sol. 29. c. Fig. 4.169 '*Total distance Area of graph ' ---=---:--;-- = 20 20 + t = 20 Total time -}I> 10 wi's • '* '*2I (20 + I + 20 - t) x 5t = 20 (20 + t) t = 5 s --.,U' 0 40. b. Maximum velocity = 5 t = 5 x 5 = 25 m/s --)l>- a 2 mis: 41. n. 5t(20 - t) = lOOt - 5t 2 = 375 m Fig. 4.169 For Problems 42-43 42. d., 43. c. Distance between the particles will he minimum when Sol. '*'Velocity of B becomes equal to that of A, i.c., 10 mIs, 42. d. In graph (i) and (iii), magnitude of slope is greater at t, Apply v = 11 + al 10 = 0 + 21 \"* I = 5 s than that at f2. 30. h. Distance travelled by A in lime 5 s, SI = 10 x 5 = 50 m 43. c. In graph (iii), magnitude of slope of graph is decreasing. Distance travelled by B in time 5 5: Hence magnitude of velocity is decreasing but in -ve direc- tion, hence acceleration is positive. S, = I 2 = I - x 2 X 52 = 25 m For Problems 44-45 _a1 44, b., 45. c. 22 Sol. 44. b. For the graphs (i) and (iv), slope is constant hence the Minimqm distance = 40·+ S2 - SI = 15 m velocity becomes negative, For Problems 31-35 45. c. For the graph (iii) the particles velocity first decreases and 31. a., 32. d., 33. a., 34. b., 35. c. then increases in -vcdirection. It means negative accelerat.ion is involved in this motion. Sol. 31. a. We know that, Vav = X, -XI to-O 5 mls ---- = ----- = x, - x, 12 - I, 2- 0 50 5 32. d. Vw = - - - = -~- = - m/s , 12 - I, 4 - () 4 v,,,33. a. X2 - XI 5 - 10 = - - - = - - - = -2.5 mls 12 .. I, 4 - 2 x,-x, -5-5 10 34. b. Vav = - - - - = -----,-- = -- ~ m/s , I, .. II 7 - 4 3 Matolling .\\\"2 - Xl 0 -- 0 '€otumn [liRe 35. c. =V'IV - - - = - - - = Om/s , 12 - I, 8 - 0 For Problems 36-38 1. i. -+ h., ii. --+ a., iii. -+ d., iv. -+ c. 36. u., 37. d., 38. b. Sol. dS a= dl! = 2y, so, . b V= Tt=f!+2 y l, - 1-+ 36. a. Distance covered := area of the speed-time graph dl f3 = 2I x (4 + 2) x 4 + 2I (4 + 2) x 2 = 18 m vdt 37. d. Distance from the origin will be equal to the flnal displace- (v) = _,_ = f!(3-2)+y(9-4) =f!+5y, ment of the parlicle which is equal to the area of velocity-time graph f3 I dt = 2I x (4 + 2) x 4 - 2I (4 -1- 2) x 2 = 6 m 2 38. h. {fa\\\" = -V--2--'V''l~ = -2-2 - 1 m/s2 ---- = so, ii. -+ a. 6- 2 Velocity at t = I s: v = f! + 2y x I = f! + 2y, so iii -> d Initial displacement i.e., t = 0, S = <X, so, iv -> C 12 - II 2. i -> a., b., ii. -+ c., iii. -+ d., iv. --+ d. For Problems 39-41 For a round trip displacement is zero, hence V:v = 0, 39. a., 40. b., 41. a. \"-)- \"\"'r Sol. Also V-~-)l-V = Vj+V2 --} .. ,' -}. • 39. a. Average speed = 20 111S-- 1 VI --2-' when IS lll1twl, V2 IS final. Hence, I. --,,.. a.,b.

4.64 Physics for IIT-JEE: Mechanics I total distancc InAB,SO(I v= d-S-O(r\" Average speed (U<IV) = dl time of !light Le., velocity is uniform, i.e., constant or independent of time. In Be, body is retarded, i.c., velocity decreases with time. In = 22(vVAo/I2gg) = va .. CD, S ex to, i.e., v:::: zero i.e., body is at rest. -2' Hence n. -+ c. S. i. -> b.,d., ii. --+ a.,<I., iii. -+ c., iv. -+ a. 7·ascent = T dcsent = -V{). Hence I..l.l. -+ d\" 'IV. -+ d. 1 a. Area of v-I graph lies below time axis, so displacement is negative, but slope is positive, so acceleration is positive, Ii 3. i. -> a., ii. -)- c., d., iii. -+ b., iv. -7 b. b. Area of v-I graph lies above time axis, so displacement is positive, and slope is positive, so acceleration is also When the ball is above the point of projection. its displace- positive, mentis always positive, but its velocity may be positive (when c. Displacement is zero, because half area is above time axis and half below, Slope is negative, so acceleration is nega- moving up), zero (at top point) or negative (when coming tive, down). d. Area of v-I graph lies above time axis, so displacemen1 Acceleration is always directed downward, so it is al- is positive, and slope is negative, so acceleration is also negative, ways negative. 4. i. -+ a., ii. -+ c., iii. -+ b., iv. -+ d. In 0 A, SO(l 2 , v = -dS 0( 21. I..e., 1) 0( 1 dl i.e., velocity increases with time.

·CHAPTER ·..·5. ·.•.·. Motion in Two Dimensions .» Re1ativ~veloclty .·?·~..QN\\b)lll1t.r~i~pi~ftoir\\ilU~cc~i~rI~~trii!\\aM~Q~~~·/t·ior.t .>- MotionwithUniformAccelel:auoriinaPlalle .·>:.k¢I~ti~~A~~4I!l!'y~JgdWy ,\" \" .>- Kinematics otCircularMotitm 5.1