["UNITS AND MEASUREMENT 37 2.20 The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun ? 2.21 Precise measurements of physical quantities are a need of science. For example, to ascertain the speed of an aircraft, one must have an accurate method to find its positions at closely separated instants of time. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precise measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed. 2.22 Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity) : (a) the total mass of rain-bearing clouds over India during the Monsoon (b) the mass of an elephant (c) the wind speed during a storm (d) the number of strands of hair on your head (e) the number of air molecules in your classroom. 2.23 The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases ? Check if your guess is correct from the following data : mass of the Sun = 2.0 \u00d7 1030 kg, radius of the Sun = 7.0 \u00d7 108 m. 2.24 When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35.72\\\" of arc. Calculate the diameter of Jupiter. Additional Exercises 2.25 A man walking briskly in rain with speed v must slant his umbrella forward making an angle \u03b8 with the vertical. A student derives the following relation between \u03b8 and v : tan \u03b8 = v and checks that the relation has a correct limit: as v \u21920, \u03b8 \u2192 0, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct ? If not, guess the correct relation. 2.26 It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s ? 2.27 Estimate the average mass density of a sodium atom assuming its size to be about 2.5 \u00c5. (Use the known values of Avogadro\u2019s number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase : 970 kg m\u20133. Are the two densities of the same order of magnitude ? If so, why ? 2.28 The unit of length convenient on the nuclear scale is a fermi : 1 f = 10\u201315 m. Nuclear sizes obey roughly the following empirical relation : r = r0 A1\/3 where r is the radius of the nruulceleimusp,liAesittshmatansus cnlueamrbmera,sasnddenrosiistyaicsonnesatralnytceoqnusatal ntot about, 1.2 f. Show that the for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27. 2.29 A LASER is a source of very intense, monochromatic, and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to 2020-21","38 PHYSICS return after reflection at the Moon\u2019s surface. How much is the radius of the lunar orbit around the Earth ? 2.30 A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR, the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water = 1450 m s\u20131). 2.31 The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us ? 2.32 It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon. 2.33 A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of ). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants? 2020-21","CHAPTER THREE MOTION IN A STRAIGHT LINE 3.1 Introduction 3.1 INTRODUCTION 3.2 Position, path length and Motion is common to everything in the universe. We walk, displacement run and ride a bicycle. Even when we are sleeping, air moves into and out of our lungs and blood flows in arteries and 3.3 Average velocity and average veins. We see leaves falling from trees and water flowing down a dam. Automobiles and planes carry people from one speed place to the other. The earth rotates once every twenty-four hours and revolves round the sun once in a year. The sun 3.4 Instantaneous velocity and itself is in motion in the Milky Way, which is again moving within its local group of galaxies. speed Motion is change in position of an object with time. How 3.5 Acceleration does the position change with time ? In this chapter, we shall 3.6 Kinematic equations for learn how to describe motion. For this, we develop the concepts of velocity and acceleration. We shall confine uniformly accelerated motion ourselves to the study of motion of objects along a straight line, also known as rectilinear motion. For the case of 3.7 Relative velocity rectilinear motion with uniform acceleration, a set of simple equations can be obtained. Finally, to understand the relative Summary nature of motion, we introduce the concept of relative velocity. Points to ponder Exercises In our discussions, we shall treat the objects in motion as Additional exercises point objects. This approximation is valid so far as the size Appendix 3.1 of the object is much smaller than the distance it moves in a reasonable duration of time. In a good number of situations in real-life, the size of objects can be neglected and they can be considered as point-like objects without much error. In Kinematics, we study ways to describe motion without going into the causes of motion. What causes motion described in this chapter and the next chapter forms the subject matter of Chapter 5. 3.2 POSITION, PATH LENGTH AND DISPLACEMENT Earlier you learnt that motion is change in position of an object with time. In order to specify position, we need to use a reference point and a set of axes. It is convenient to choose 2020-21","40 PHYSICS a rectangular coordinate system consisting of with the path of the car\u2019s motion and origin of three mutually perpenducular axes, labelled X-, the axis as the point from where the car started Y-, and Z- axes. The point of intersection of these moving, i.e. the car was at x = 0 at t = 0 (Fig. 3.1). three axes is called origin (O) and serves as the Let P, Q and R represent the positions of the car reference point. The coordinates (x, y. z) of an at different instants of time. Consider two cases object describe the position of the object with of motion. In the first case, the car moves from respect to this coordinate system. To measure O to P. Then the distance moved by the car is time, we position a clock in this system. This OP = +360 m. This distance is called the path coordinate system along with a clock constitutes length traversed by the car. In the second a frame of reference. case, the car moves from O to P and then moves back from P to Q. During this course of motion, If one or more coordinates of an object change the path length traversed is OP + PQ = + 360 m with time, we say that the object is in motion. + (+120 m) = + 480 m. Path length is a scalar Otherwise, the object is said to be at rest with quantity \u2014 a quantity that has a magnitude respect to this frame of reference. only and no direction (see Chapter 4). The choice of a set of axes in a frame of Displacement reference depends upon the situation. For example, for describing motion in one dimension, It is useful to define another quantity we need only one axis. To describe motion in two\/three dimensions, we need a set of two\/ displacement as the change in position. Let three axes. x1 and x2 be the positions of an object at time t Description of an event depends on the frame and t . Then 1 of reference chosen for the description. For its displacement, denoted by \u2206x, example, when you say that a car is moving on in a road, you are describing the car with respect time 2\u2206t = (t - t ), is given by the difference to a frame of reference attached to you or to the 21 ground. But with respect to a frame of reference between the final and initial positions : attached with a person sitting in the car, the car is at rest. \u2206x = x \u2013 x 21 To describe motion along a straight line, we (We use the Greek delta (\u2206) to denote a can choose an axis, say X-axis, so that it letter coincides with the path of the object. We then measure the position of the object with reference change in a quantity.) to a conveniently chosen origin, say O, as shown in Fig. 3.1. Positions to the right of O are taken If x > x , \u2206x is positive; and if x < x , \u2206x is as positive and to the left of O, as negative. negat2ive. 1 21 Following this convention, the position coordinates of point P and Q in Fig. 3.1 are +360 Displacement has both magnitude and m and +240 m. Similarly, the position coordinate of point R is \u2013120 m. direction. Such quantities are represented by Path length vectors. You will read about vectors in the next Consider the motion of a car along a straight chapter. Presently, we are dealing with motion line. We choose the x-axis such that it coincides along a straight line (also called rectilinear motion) only. In one-dimensional motion, there are only two directions (backward and forward, upward and downward) in which an object can move, and these two directions can easily be specified by + and \u2013 signs. For example, displacement of the car in moving from O to P is : \u2206x = x \u2013 x = (+360 m) \u2013 0 m = +360 m 21 The displacement has a magnitude of 360 m and is directed in the positive x direction as indicated by the + sign. Similarly, the displacement of the car from P to Q is 240 m \u2013 360 m = \u2013 120 m. The Fig. 3.1 x-axis, origin and positions of a car at different times. 2020-21","MOTION IN A STRAIGHT LINE 41 negative sign indicates the direction of then returns to O, the final position coincides displacement. Thus, it is not necessary to use with the initial position and the displacement vector notation for discussing motion of objects is zero. However, the path length of this journey in one-dimension. is OP + PO = 360 m + 360 m = 720 m. The magnitude of displacement may or may Motion of an object can be represented by a not be equal to the path length traversed by position-time graph as you have already learnt an object. For example, for motion of the car about it. Such a graph is a powerful tool to from O to P, the path length is +360 m and the represent and analyse different aspects of displacement is +360 m. In this case, the motion of an object. For motion along a straight magnitude of displacement (360 m) is equal to line, say X-axis, only x-coordinate varies with the path length (360 m). But consider the motion time and we have an x-t graph. Let us first of the car from O to P and back to Q. In this consider the simple case in which an object is case, the path length = (+360 m) + (+120 m) = + stationary, e.g. a car standing still at x = 40 m. 480 m. However, the displacement = (+240 m) \u2013 The position-time graph is a straight line parallel (0 m) = + 240 m. Thus, the magnitude of to the time axis, as shown in Fig. 3.2(a). displacement (240 m) is not equal to the path length (480 m). If an object moving along the straight line covers equal distances in equal intervals of The magnitude of the displacement for a time, it is said to be in uniform motion along a course of motion may be zero but the straight line. Fig. 3.2(b) shows the position-time corresponding path length is not zero. For graph of such a motion. example, if the car starts from O, goes to P and Fig. 3.2 Position-time graph of (a) stationary object, and (b) an object in uniform motion. x (m) t (s) Fig. 3.3 Position-time graph of a car. 2020-21","42 PHYSICS Now, let us consider the motion of a car that Consider the motion of the car in Fig. 3.3. The starts from rest at time t = 0 s from the origin O and picks up speed till t = 10 s and thereafter portion of the x-t graph between t = 0 s and t = 8 moves with uniform speed till t = 18 s. Then the brakes are applied and the car stops at s is blown up and shown in Fig. 3.4. As seen t = 20 s and x = 296 m. The position-time graph for this case is shown in Fig. 3.3. We shall refer from the plot, the average velocity of the car to this graph in our discussion in the following sections. between time t = 5 s and t = 7 s is : 3.3 AVERAGE VELOCITY AND AVERAGE v = x2 \u2212 x1 = (27.4 \u2212 10.0) m = 8.7 m s \u20131 SPEED t2 \u2212 t1 (7 \u2212 5) s When an object is in motion, its position Geometrically, this is the slope of the straight changes with time. But how fast is the position changing with time and in what direction? To line P1P2 connecting the initial position P1 to describe this, we define the quantity average velocity. Average velocity is defined as the the final position P2 as shown in Fig. 3.4. change in position or displacement (\u2206x) divided The average velocity can be positive or negative by the time intervals (\u2206t), in which the displacement occurs : depending upon the sign of the displacement. It is zero if the displacement is zero. Fig. 3.5 shows the x-t graphs for an object, moving with positive velocity (Fig. 3.5a), moving with negative velocity (Fig. 3.5b) and at rest (Fig. 3.5c). v = x 2 \u2212 x1 = \u2206x (3.1) t2 \u2212 t1 \u2206t where x2 and x1 are the positions of the object Fig. 3.5 Position-time graph for an object (a) moving at time t2and t1, respectively. Here the bar over with positive velocity, (b) moving with the symbol for velocity is a standard notation negative velocity, and (c) at rest. used to indicate an average quantity. The SI unit for velocity is m\/s or m s\u20131, although km h\u20131 Average velocity as defined above involves is used in many everyday applications. only the displacement of the object. We have seen earlier that the magnitude of displacement may Like displacement, average velocity is also a be different from the actual path length. To vector quantity. But as explained earlier, for describe the rate of motion over the actual path, motion in a straight line, the directional aspect we introduce another quantity called average of the vector can be taken care of by + and \u2013 speed. signs and we do not have to use the vector notation for velocity in this chapter. Average speed is defined as the total path length travelled divided by the total time interval during which the motion has taken place : Average speed = Total path length (3.2) Total time interval Fig. 3.4 The average velocity is the slope of line P P . Average speed has obviously the same unit 12 (m s\u20131) as that of velocity. But it does not tell us in what direction an object is moving. Thus, it is always positive (in contrast to the average velocity which can be positive or negative). If the motion of an object is along a straight line and in the same direction, the magnitude of displacement is equal to the total path length. In that case, the magnitude of average velocity 2020-21","MOTION IN A STRAIGHT LINE 43 is equal to the average speed. This is not always 3.4 INSTANTANEOUS VELOCITY AND SPEED the case, as you will see in the following example. The average velocity tells us how fast an object Example 3.1 A car is moving along a has been moving over a given time interval but straight line, say OP in Fig. 3.1. It moves does not tell us how fast it moves at different from O to P in 18 s and returns from P to Q instants of time during that interval. For this, in 6.0 s. What are the average velocity we define instantaneous velocity or simply and average speed of the car in going (a) velocity v at an instant t. from O to P ? and (b) from O to P and back to Q ? The velocity at an instant is defined as the limit of the average velocity as the time interval Answer (a) \u2206t becomes infinitesimally small. In other words, Average velocity = Displacement \u2206x (3.3a) Time interval v = lim \u2206t \u2192 0 \u2206t = dx (3.3b) dt v = + 360 m = + 20 m s\u22121 where the symbol lim stands for the operation 18 s \u2206t \u21920 of taking limit as \u2206t 0 of the quantity on its Average speed = Path length Time interval right. In the language of calculus, the quantity on the right hand side of Eq. (3.3a) is the differential coefficient of x with respect to t and = 360 m = 20 m s\u22121 dx 18 s is denoted by (see Appendix 3.1). It is the Thus, in this case the average speed is equal to the magnitude of the average velocity. dt (b) In this case, rate of change of position with respect to time, Average velocity = Displacement = +240 m s Time interval at that instant. (18 + 6.0) We can use Eq. (3.3a) for obtaining the value =+10 m s-1 of velocity at an instant either graphically or numerically. Suppose that we want to obtain Path length OP + PQ graphically the value of velocity at time t = 4 s Time interval \u2206t (point P) for the motion of the car represented in Fig. 3.3. The figure has been redrawn in Fig. 3.6 choosing different scales to facilitate the Average speed = = (360 +120) m 20 m s-1 = = 24 s Thus, in this case the average speed is not equal Fig. 3.6 Determining velocity from position-time to the magnitude of the average velocity. This graph. Velocity at t = 4 s is the slope of the happens because the motion here involves tangent to the graph at that instant. change in direction so that the path length is greater than the magnitude of displacement. This shows that speed is, in general, greater than the magnitude of the velocity. If the car in Example 3.1 moves from O to P and comes back to O in the same time interval, average speed is 20 m\/s but the average velocity is zero ! 2020-21","44 PHYSICS calculation. Let us take \u2206t = 2 s centred at instant for motion of the car shown in Fig. 3.3. For this case, the variation of velocity with time t = 4 s. Then, by the definition of the average is found to be as shown in Fig. 3.7. velocity, the slope of line P1P2 ( Fig. 3.6) gives Fig. 3.7 Velocity\u2013time graph corresponding to motion shown in Fig. 3.3. the value of average velocity over the interval 3 s to 5 s. Now, we decrease the value of \u2206t from The graphical method for the determination of the instantaneous velocity is always not a 2 s to 1 s. Then line P1P2 becomes Q1Q2 and its convenient method. For this, we must carefully plot the position\u2013time graph and calculate the slope gives the value of the average velocity over value of average velocity as \u2206t becomes smaller the interval 3.5 s to 4.5 s. In the limit \u2206t \u2192 0, and smaller. It is easier to calculate the value of velocity at different instants if we have data the line P1P2 becomes tangent to the position- of positions at different instants or exact time curve at the point P and the velocity at t = expression for the position as a function of time. Then, we calculate \u2206x\/\u2206t from the data for 4 s is given by the slope of the tangent at that decreasing the value of \u2206t and find the limiting value as we have done in Table 3.1 or use point. It is difficult to show this process differential calculus for the given expression and graphically. But if we use numerical method dx to obtain the value of the velocity, the calculate at different instants as done in meaning of the limiting process becomes dt clear. For the graph shown in the following example. Fig. 3.6, x = 0.08 t3. Table 3.1 gives the value of \u2206x\/\u2206t calculated for \u2206t equal to 2.0 s, 1.0 s, 0.5 s, 0.1 s and 0.01 s centred at t = 4.0 s. The second and third columns give the value of t1= \uf8eb \u2212 \u2206t \uf8f6 \uf8eb + \u2206t \uf8f6 \uf8f7 = \uf8ed\uf8ect \uf8f7 \uf8ec t and t2 and the fourth and \uf8ed 2\uf8f8 2\uf8f8 the fifth columns give the corresponding values (s\u2206otift1x,)xtia,h.eni..cdeto.htlhxueem(atlv1an)es=rltais0cgto.es0luv8temhltoe1n3cdaigtiniyfvfdeecrsoxertn(rhtec2es)epr=a\u2206o0txni.od0=io8nxfgt\u2206(23ttx.2o)aTt\u2013nhhdexe value of \u2206t listed in the first column. We see from Table 3.1 that as we decrease the value of \u2206t from 2.0 s to 0.010 s, the value of the average velocity approaches the limiting value 3.84 m s\u20131 which is the value of velocity at dx t = 4.0 s, i.e. the value of at t = 4.0 s. In this dt manner, we can calculate velocity at each \u2206x Table 3.1 Limiting value of \u2206t at t = 4 s 2020-21","MOTION IN A STRAIGHT LINE 45 Example 3.2 The position of an object This led to the concept of acceleration as the rate moving along x-axis is given by x = a + bt2 of change of velocity with time. where a = 8.5 m, b = 2.5 m s\u20132 and t is measured in seconds. What is its velocity at The average acceleration a over a time t = 0 s and t = 2.0 s. What is the average interval is defined as the change of velocity velocity between t = 2.0 s and t = 4.0 s ? divided by the time interval : Answer In notation of differential calculus, the a = v2 \u2013 v1 = \u2206v (3.4) t2 \u2013 t1 \u2206t velocity is where v2 and v1 are the instantaneous velocities or simply velocities at time t2 and t1 . It is the ( )v = dx = d a + bt 2 = 2b t = 5.0 t m s-1 average change of velocity per unit time. The SI dt dt unit of acceleration is m s\u20132 . At t = 0 s, v = 0 m s\u20131 and at t = 2.0 s, On a plot of velocity versus time, the average v = 10 m s-1 . acceleration is the slope of the straight line x (4.0) \u2212 x (2.0) Average velocity = connecting the points corresponding to (v2, t2) 4.0 \u2212 2.0 and (v1, t1). The average acceleration for velocity-time graph shown in Fig. 3.7 for = a + 16b \u2013 a \u2013 4b = 6.0 \u00d7 b 2.0 different time intervals 0 s - 10 s, 10 s \u2013 18 s, = 6.0 \u00d7 2.5 = 15 m s-1 and 18 s \u2013 20 s are : From Fig. 3.7, we note that during the period 0 s - 10 s a = (24 \u2013 0) m s\u20131 = 2.4 m s\u20132 t =10 s to 18 s the velocity is constant. Between (10 \u2013 0)s period t =18 s to t = 20 s, it is uniformly 10 s - 18 s a = (24 \u2013 24) m s\u20131 = 0 m s\u20132 decreasing and during the period t = 0 s to t (18 \u2013 10) s = 10 s, it is increasing. Note that for uniform motion, velocity is the same as the average 18 s - 20 s a = (0 \u2013 24) m s\u20131 = \u2013 12 m s\u20132 velocity at all instants. (20 \u2013 18) s Instantaneous speed or simply speed is the a (m s\u20132) magnitude of velocity. For example, a velocity of + 24.0 m s\u20131 and a velocity of \u2013 24.0 m s\u20131 \u2014 both have an associated speed of 24.0 m s-1. It should be noted that though average speed over a finite interval of time is greater or equal to the magnitude of the average velocity, instantaneous speed at an instant is equal to the magnitude of the instantaneous velocity at that instant. Why so ? 3.5 ACCELERATION The velocity of an object, in general, changes Fig. 3.8 Acceleration as a function of time for motion during its course of motion. How to describe this represented in Fig. 3.3. change? Should it be described as the rate of change in velocity with distance or with time ? Instantaneous acceleration is defined in the same This was a problem even in Galileo\u2019s time. It was way as the instantaneous velocity : first thought that this change could be described by the rate of change of velocity with distance. a = lim \u2206v = dv (3.5) But, through his studies of motion of freely falling \u2206t dt objects and motion of objects on an inclined \u2206t \u21920 plane, Galileo concluded that the rate of change of velocity with time is a constant of motion for The acceleration at an instant is the slope of all objects in free fall. On the other hand, the change in velocity with distance is not constant the tangent to the v\u2013t curve at that instant. For \u2013 it decreases with the increasing distance of fall. the v\u2013t curve shown in Fig. 3.7, we can obtain acceleration at every instant of time. The resulting a \u2013 t curve is shown in Fig. 3.8. We see 2020-21","46 PHYSICS that the acceleration is nonuniform over the (b) An object is moving in positive direction period 0 s to 10 s. It is zero between 10 s and with a negative acceleration, for example, 18 s and is constant with value \u201312 m s\u20132 motion of the car in Fig 3.3 between between 18 s and 20 s. When the acceleration t = 18 s and 20 s. is uniform, obviously, it equals the average acceleration over that period. (c) An object is moving in negative direction with a negative acceleration, for example Since velocity is a quantity having both the motion of a car moving from O in Fig. magnitude and direction, a change in velocity 3.1 in negative x-direction with may involve either or both of these factors. increasing speed. Acceleration, therefore, may result from a change in speed (magnitude), a change in (d) An object is moving in positive direction direction or changes in both. Like velocity, till time t1, and then turns back with the acceleration can also be positive, negative or same negative acceleration, for example zero. Position-time graphs for motion with the motion of a car from point O to point positive, negative and zero acceleration are Q in Fig. 3.1 till time t1 with decreasing shown in Figs. 3.9 (a), (b) and (c), respectively. speed and turning back and moving with Note that the graph curves upward for positive the same negative acceleration. acceleration; downward for negative acceleration and it is a straight line for zero An interesting feature of a velocity-time graph acceleration. As an exercise, identify in Fig. 3.3, for any moving object is that the area under the the regions of the curve that correspond to these curve represents the displacement over a three cases. given time interval. A general proof of this Although acceleration can vary with time, our study in this chapter will be restricted to motion with constant acceleration. In this case, the average acceleration equals the constant value of acceleration during the interval. If the velocity of an object is v at t = 0 and v at time t, o we have a = v \u2212 v0 or, v = v0 + a t (3.6) t \u22120 Fig. 3.9 Position-time graph for motion with Fig. 3.10 Velocity\u2013time graph for motions with (a) positive acceleration; (b) negative constant acceleration. (a) Motion in positive acceleration, and (c) zero acceleration. direction with positive acceleration, Let us see how velocity-time graph looks like for some simple cases. Fig. 3.10 shows velocity- (b) Motion in positive direction with time graph for motion with constant acceleration negative acceleration, (c) Motion in negative for the following cases : direction with negative acceleration, (a) An object is moving in a positive direction with a positive acceleration, for example (d) Motion of an object with negative the motion of the car in Fig. 3.3 between acceleration that changes direction at time t = 0 s and t = 10 s. t1. Between times 0 to t1, its moves in positive x - direction and between t1 and t2 it moves in the opposite direction. 2020-21","MOTION IN A STRAIGHT LINE 47 statement requires use of calculus. We can, = 1 (v \u2013v0 ) t + v0t however, see that it is true for the simple case of 2 an object moving with constant velocity u. Its velocity-time graph is as shown in Fig. 3.11. Fig. 3.11 Area under v\u2013t curve equals displacement of the object over a given time interval. The v-t curve is a straight line parallel to the Fig. 3.12 Area under v-t curve for an object with time axis and the area under it between t = 0 uniform acceleration. and t = T is the area of the rectangle of height u and base T. Therefore, area = u\u00d7T = uT which As explained in the previous section, the area is the displacement in this time interval. How under v-t curve represents the displacement. come in this case an area is equal to a distance? Therefore, the displacement x of the object is : Think! Note the dimensions of quantities on the two coordinate axes, and you will arrive at x = 1 (v \u2013v0 )t + v0t (3.7) the answer. 2 (3.8) Note that the x-t, v-t, and a-t graphs shown But v \u2212 v0 = a t (3.9a) in several figures in this chapter have sharp kinks at some points implying that the Therefore, x = 1 a t2 + v0t functions are not differentiable at these 2 points. In any realistic situation, the functions will be differentiable at all points or, x = v0t + 1 at 2 and the graphs will be smooth. 2 What this means physically is that Equation (3.7) can also be written as acceleration and velocity cannot change values abruptly at an instant. Changes are x = v + v0 t = v t always continuous. 2 where, 3.6 KINEMATIC EQUATIONS FOR v = v + v0 (constant acceleration only) UNIFORMLY ACCELERATED MOTION 2 (3.9b) For uniformly accelerated motion, we can derive Equations (3.9a) and (3.9b) mean that the object some simple equations that relate displacement has undergone displacement x with an average velocity equal to the arithmetic average of the (x), time taken (t), initial velocity (v0), final initial and final velocities. velocity (v) and acceleration (a). Equation (3.6) From Eq. (3.6), t = (v \u2013 v0)\/a. Substituting this in Eq. (3.9a), we get already obtained gives a relation between final and initial velocities v and v0 of an object moving with uniform acceleration a : v = v0 + at (3.6) This relation is graphically represented in Fig. 3.12. x = v t = \uf8ebv + v0 \uf8f6 \uf8ebv \u2212v 0 \uf8f6 = v2 \u2212 v02 The area under this curve is : \uf8ec\uf8ed 2 \uf8f8\uf8f7 \uf8ed\uf8ec a \uf8f7\uf8f8 2a Area between instants 0 and t = Area of triangle ABC + Area of rectangle OACD v2 = v02 + 2ax (3.10) 2020-21","48 PHYSICS This equation can also be obtained by = \u222b (t + at ) dt substituting the value of t from Eq. (3.6) into Eq. (3.8). Thus, we have obtained three 0 v0 important equations : x \u2013 x0 = v0 t + 1a t2 v = v0 + at 2 = + 1 at 2 x = x0 + v0 t + 1 a t2 2 2 x v0t We can write v2 = v02 + 2ax (3.11a) a = dv = dv dx = v dv dt dx dt dx connecting five quantities v0, v, a, t and x. These are kinematic equations of rectilinear motion or, v dv = a dx for constant acceleration. Integrating both sides, The set of Eq. (3.11a) were obtained by \u222b v v dv = \u222b x a dx assuming that at t = 0, the position of the v0 x0 particle, x is 0. We can obtain a more general equation if we take the position coordinate at t v 2 \u2013 v02 = a (x \u2013 x0 ) = 0 as non-zero, say x0. Then Eqs. (3.11a) are 2 modified (replacing x by x \u2013 x0 ) to : v2 = v02 + 2a (x \u2013 x 0 ) v = v0 + at The advantage of this method is that it can be x = x0 + v0t + 1 at 2 (3.11b) used for motion with non-uniform acceleration 2 also. Now, we shall use these equations to some v2 = v02 + 2a(x \u2212 x 0 ) (3.11c) important cases. Example 3.3 Obtain equations of motion Example 3.4 A ball is thrown vertically for constant acceleration using method of upwards with a velocity of 20 m s\u20131 from calculus. the top of a multistorey building. The height of the point from where the ball is Answer By definition thrown is 25.0 m from the ground. (a) How high will the ball rise ? and (b) how long a = dv will it be before the ball hits the ground? dt Take g = 10 m s\u20132. dv = a dt Answer (a) Let us take the y-axis in the Integrating both sides vertically upward direction with zero at the \u222b v dv \u222b t = dt ground, as shown in Fig. 3.13. a v0 0 Now v = + 20 m s\u20131, = a \u222b t dt o 0 (a is constant) a = \u2013 g = \u201310 m s\u20132, v = 0 m s\u20131 v \u2013 v0 = at If the ball rises to height y from the point of v = v0 + at ( )launch, then using the equation Further, v = dx v2 = v02 + 2 a y \u2013 y0 dt we get dx = v dt 0 = (20)2 + 2(\u201310)(y \u2013 y0) Solving, we get, (y \u2013 y0) = 20 m. Integrating both sides \u222b x dx = \u222b t dt (b) We can solve this part of the problem in two ways. Note carefully the methods used. v x0 0 2020-21","MOTION IN A STRAIGHT LINE 49 0 = 25 +20 t + (\u00bd) (-10) t2 Or, 5t2 \u2013 20t \u2013 25 = 0 Solving this quadratic equation for t, we get t = 5s Note that the second method is better since we do not have to worry about the path of the motion as the motion is under constant acceleration. Example 3.5 Free-fall : Discuss the motion of an object under free fall. Neglect air resistance. Fig. 3.13 Answer An object released near the surface of the Earth is accelerated downward under the FIRST METHOD : In the first method, we split influence of the force of gravity. The magnitude the path in two parts : the upward motion (A to of acceleration due to gravity is represented by B) and the downward motion (B to C) and g. If air resistance is neglected, the object is calculate the corresponding time taken t1 and said to be in free fall. If the height through t2. Since the velocity at B is zero, we have : which the object falls is small compared to the earth\u2019s radius, g can be taken to be constant, v = vo + at equal to 9.8 m s\u20132. Free fall is thus a case of 0 = 20 \u2013 10t1 motion with uniform acceleration. Or, t1 = 2 s We assume that the motion is in y-direction, This is the time in going from A to B. From B, or more correctly in \u2013y-direction because we the point of the maximum height, the ball falls choose upward direction as positive. Since the freely under the acceleration due to gravity. The acceleration due to gravity is always downward, ball is moving in negative y direction. We use it is in the negative direction and we have equation a = \u2013 g = \u2013 9.8 m s\u20132 The object is released from rest at y = 0. Therefore, v0 = 0 and the equations of motion become: v= 0\u2013gt = \u20139.8 t m s\u20131 1 at 2 y = 0 \u2013 \u00bd g t2 = \u20134.9 t 2 m 2 y = y0 + v0t + v2 = 0 \u2013 2 g y = \u201319.6 y m2 s\u20132 We have, y0 = 45 m, y = 0, v0 = 0, a = \u2013 g = \u201310 m s\u20132 These equations give the velocity and the 0 = 45 + (\u00bd) (\u201310) t22 distance travelled as a function of time and also the variation of velocity with distance. The Solving, we get t2 = 3 s variation of acceleration, velocity, and distance, Therefore, the total time taken by the ball before with time have been plotted in Fig. 3.14(a), (b) it hits the ground = t1 + t2 = 2 s + 3 s = 5 s. and (c). SECOND METHOD : The total time taken can also be calculated by noting the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation y = y0 + v0t + 1 at 2 2 Now y0 = 25 m y=0m vo = 20 m s-1, a = \u201310m s\u20132, t=? (a) 2020-21","50 PHYSICS traversed during successive intervals of time. Since initial velocity is zero, we have y = \u2212 1 gt 2 2 Using this equation, we can calculate the position of the object after different time intervals, 0, \u03c4, 2\u03c4, 3\u03c4\u2026 which are given in second column of Table 3.2. If we take (\u20131\/ 2 ) g\u03c42 as y0 \u2014 the position coordinate after first time interval \u03c4, then third column gives (b) the positions in the unit of yo. The fourth column gives the distances traversed in successive \u03c4s. We find that the distances are in the simple ratio 1: 3: 5: 7: 9: 11\u2026 as shown in the last column. This law was established by Galileo Galilei (1564-1642) who was the first to make quantitative studies of free fall. Example 3.7 Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is (c) an important factor for road safety and depends on the initial velocity (v0) and the Fig. 3.14 Motion of an object under free fall. braking capacity, or deceleration, \u2013a that (a) Variation of acceleration with time. is caused by the braking. Derive an (b) Variation of velocity with time. expression for stopping distance of a vehicle (c) Variation of distance with time in terms of vo and a. Example 3.6 Galileo\u2019s law of odd Answer Let the distance travelled by the vehicle numbers : \u201cThe distances traversed, during before it stops be ds. Then, using equation of equal intervals of time, by a body falling motion v2 = vo2 + 2 ax, and noting that v = 0, we from rest, stand to one another in the same have the stopping distance ratio as the odd numbers beginning with unity [namely, 1: 3: 5: 7\u2026...].\u201d Prove it. ds = \u2013 v02 2a Answer Let us divide the time interval of motion of an object under free fall into many Thus, the stopping distance is proportional to equal intervals \u03c4 and find out the distances the square of the initial velocity. Doubling the Table 3.2 2020-21","MOTION IN A STRAIGHT LINE 51 initial velocity increases the stopping distance Or, by a factor of 4 (for the same deceleration). Given d = 21.0 cm and g = 9.8 m s\u20132 the reaction For the car of a particular make, the braking time is distance was found to be 10 m, 20 m, 34 m and 50 m corresponding to velocities of 11, 15, 20 3.7 RELATIVE VELOCITY and 25 m\/s which are nearly consistent with the above formula. You must be familiar with the experience of travelling in a train and being overtaken by Stopping distance is an important factor another train moving in the same direction as considered in setting speed limits, for example, you are. While that train must be travelling faster in school zones. than you to be able to pass you, it does seem slower to you than it would be to someone Example 3.8 Reaction time : When a standing on the ground and watching both the situation demands our immediate trains. In case both the trains have the same action, it takes some time before we velocity with respect to the ground, then to you really respond. Reaction time is the the other train would seem to be not moving at time a person takes to observe, think all. To understand such observations, we now and act. For example, if a person is introduce the concept of relative velocity. driving and suddenly a boy appears on the road, then the time elapsed before Consider two objects A and B moving he slams the brakes of the car is the uniformly with average velocities vA and vB in reaction time. Reaction time depends one dimension, say along x-axis. (Unless on complexity of the situation and on otherwise specified, the velocities mentioned in an individual. this chapter are measured with reference to the ground). If xA (0) and xB (0) are positions of objects You can measure your reaction time A and B, respectively at time t = 0, their positions by a simple experiment. Take a ruler xA (t) and xB (t) at time t are given by: and ask your friend to drop it vertically through the gap between your thumb xA (t ) = xA (0) + vA t (3.12a) and forefinger (Fig. 3.15). After you xB (t) = xB (0) + vB t (3.12b) catch it, find the distance d travelled by the ruler. In a particular case, d was Then, the displacement from object A to object found to be 21.0 cm. Estimate reaction B is given by time. xBA(t) = xB (t) \u2013 xA (t) (3.13) Fig. 3.15 Measuring the reaction time. = [ xB (0) \u2013 xA (0) ] + (vB \u2013 vA) t. Answer The ruler drops under free fall. Equation (3.13) is easily interpreted. It tells us Therefore, vo = 0, and a = \u2013 g = \u20139.8 m s\u20132. The distance travelled d and the reaction time tr are that as seen from object A, object B has a related by velocity vB \u2013 vA because the displacement from A to B changes steadily by the amount vB \u2013 vA in each unit of time. We say that the velocity of object B relative to object A is vB \u2013 vA : vBA = vB \u2013 vA (3.14a) Similarly, velocity of object A relative to object B is: vAB = vA \u2013 vB (3.14b) 2020-21","52 PHYSICS Fig. 3.16 Position-time graphs of two objects with equal velocities. This shows: vBA = \u2013 vAB (3.14c) Fig. 3.17 Position-time graphs of two objects with unequal velocities, showing the time of Now we consider some special cases : meeting. (a) If vB = vA, vB \u2013 vA = 0. Then, from Eq. (3.13), xB t(s) (t) \u2013 xA (t) = xB (0) \u2013 xA (0). Therefore, the two objects stay at a constant distance (xB (0) \u2013 xA Fig. 3.18 Position-time graphs of two objects with (0)) apart, and their position\u2013time graphs are velocities in opposite directions, showing the time of meeting. straight lines parallel to each other as shown Example 3.9 Two parallel rail tracks run in Fig. 3.16. The relative velocity vAB or vBA is north-south. Train A moves north with a zero in this case. speed of 54 km h\u20131, and train B moves south with a speed of 90 km h\u20131. What is the (b) If vA > vB, vB \u2013 vA is negative. One graph is (a) velocity of B with respect to A ?, steeper than the other and they meet at a (b) velocity of ground with respect to B ?, common point. For example, suppose vA = 20 m s-1 and and xA (0) = 10 m; and vB = 10 m s-1, xB (0) = 40 (c) velocity of a monkey running on the m; then the time at which they meet is t = 3 s roof of the train A against its motion (Fig. 3.17). At this instant they are both at a (with a velocity of 18 km h\u20131 with respect to the train A) as observed by position xA (t) = xB (t) = 70 m. Thus, object A a man standing on the ground ? overtakes object B at this time. In this case,vBA = 10 m s\u20131 \u2013 20 m s\u20131 = \u2013 10 m s\u20131= \u2013 vAB. Answer Choose the positive direction of x-axis (c) Suppose vA and vB are of opposite signs. For to be from south to north. Then, example, if in the above example object A is moving with 20 m s\u20131 starting at xA(0) = 10 m and object B is moving with \u2013 10 m s\u20131 starting at xB (0) = 40 m, the two objects meet at t = 1 s (Fig. 3.18). The velocity of B relative to A, vBA = [\u201310 \u2013 (20)] m s\u20131 = \u201330 m s\u20131 = \u2013 vAB. In this case, the magnitude of vBA or vAB ( = 30 m s\u20131) is greater than the magnitude of velocity of A or that of B. If the objects under consideration are two trains, then for a person sitting on either of the two, the other train seems to go very fast. Note that Eq. (3.14) are valid even if vA and vB represent instantaneous velocities. 2020-21","MOTION IN A STRAIGHT LINE 53 vA = + 54 km h\u20131 = 15 m s\u20131 B = 0 \u2013 vB = 25 m s\u20131. vB = \u2013 90 km h\u20131 = \u2013 25 m s\u20131 Relative velocity of B with respect to A = vB \u2013 vA= In (c), let the velocity of the monkey with respect \u2013 40 m s\u20131 , i.e. the train B appears to A to move to ground be vM. Relative velocity of the monkey with a speed of 40 m s\u20131 from north to south. with respect to A, vMA = vM \u2013 vA = \u201318 km h\u20131 = \u20135 m s\u20131. Therefore, Relative velocity of ground with respect to vM = (15 \u2013 5) m s\u20131 = 10 m s\u20131. SUMMARY 1. An object is said to be in motion if its position changes with time. The position of the object can be specified with reference to a conveniently chosen origin. For motion in a straight line, position to the right of the origin is taken as positive and to the left as negative. 2. Path length is defined as the total length of the path traversed by an object. 3. Displacement is the change in position : \u2206x = x2 \u2013 x1. Path length is greater or equal to the magnitude of the displacement between the same points. 4. An object is said to be in uniform motion in a straight line if its displacement is equal in equal intervals of time. Otherwise, the motion is said to be non-uniform. 5. Average velocity is the displacement divided by the time interval in which the displacement occurs : v = \u2206x \u2206t On an x-t graph, the average velocity over a time interval is the slope of the line connecting the initial and final positions corresponding to that interval. 6. Average Speed is the ratio of total path length traversed and the corresponding time interval. The average speed of an object is greater or equal to the magnitude of the average velocity over a given time interval. 7. Instantaneous velocity or simply velocity is defined as the limit of the average velocity as the time interval \u2206t becomes infinitesimally small : v = lim v = lim \u2206x = dx \u2206t dt \u2206t \u21920 \u2206t \u21920 The velocity at a particular instant is equal to the slope of the tangent drawn on position-time graph at that instant. 8. Average acceleration is the change in velocity divided by the time interval during which the change occurs : a = \u2206v \u2206t 9. Instantaneous acceleration is defined as the limit of the average acceleration as the time interval \u2206t goes to zero : a = lim a = lim \u2206v = dv \u2206t dt \u2206t \u21920 \u2206t \u21920 The acceleration of an object at a particular time is the slope of the velocity-time graph at that instant of time. For uniform motion, acceleration is zero and the x-t graph is a straight line inclined to the time axis and the v-t graph is a straight line 2020-21","54 PHYSICS parallel to the time axis. For motion with uniform acceleration, x-t graph is a parabola while the v-t graph is a straight line inclined to the time axis. 10. The area under the velocity-time curve between times t1 and t2 is equal to the displacement of the object during that interval of time. 11. For objects in uniformly accelerated rectilinear motion, the five quantities, displacement x, time taken t, initial velocity v0, final velocity v and acceleration a are related by a set of simple equations called kinematic equations of motion : v = v0 + at x = v0t + 1 at 2 2 v2 = v02 + 2ax if the position of the object at time t = 0 is 0. If the particle starts at x = x0 , x in above equations is replaced by (x \u2013 x0). 2020-21","MOTION IN A STRAIGHT LINE 55 POINTS TO PONDER 1. The path length traversed by an object between two points is, in general, not the same as the magnitude of displacement. The displacement depends only on the end points; the path length (as the name implies) depends on the actual path. In one dimension, the two quantities are equal only if the object does not change its direction during the course of motion. In all other cases, the path length is greater than the magnitude of displacement. 2. In view of point 1 above, the average speed of an object is greater than or equal to the magnitude of the average velocity over a given time interval. The two are equal only if the path length is equal to the magnitude of displacement. 3. The origin and the positive direction of an axis are a matter of choice. You should first specify this choice before you assign signs to quantities like displacement, velocity and acceleration. 4. If a particle is speeding up, acceleration is in the direction of velocity; if its speed is decreasing, acceleration is in the direction opposite to that of the velocity. This statement is independent of the choice of the origin and the axis. 5. The sign of acceleration does not tell us whether the particle\u2019s speed is increasing or decreasing. The sign of acceleration (as mentioned in point 3) depends on the choice of the positive direction of the axis. For example, if the vertically upward direction is chosen to be the positive direction of the axis, the acceleration due to gravity is negative. If a particle is falling under gravity, this acceleration, though negative, results in increase in speed. For a particle thrown upward, the same negative acceleration (of gravity) results in decrease in speed. 6. The zero velocity of a particle at any instant does not necessarily imply zero acceleration at that instant. A particle may be momentarily at rest and yet have non-zero acceleration. For example, a particle thrown up has zero velocity at its uppermost point but the acceleration at that instant continues to be the acceleration due to gravity. 7. In the kinematic equations of motion [Eq. (3.11)], the various quantities are algebraic, i.e. they may be positive or negative. The equations are applicable in all situations (for one dimensional motion with constant acceleration) provided the values of different quantities are substituted in the equations with proper signs. 8. The definitions of instantaneous velocity and acceleration (Eqs. (3.3) and (3.5)) are exact and are always correct while the kinematic equations (Eq. (3.11)) are true only for motion in which the magnitude and the direction of acceleration are constant during the course of motion. EXERCISES 3.1 In which of the following examples of motion, can the body be considered approximately a point object: (a) a railway carriage moving without jerks between two stations. (b) a monkey sitting on top of a man cycling smoothly on a circular track. (c) a spinning cricket ball that turns sharply on hitting the ground. (d) a tumbling beaker that has slipped off the edge of a table. 3.2 The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct entries in the brackets below ; (a) (A\/B) lives closer to the school than (B\/A) (b) (A\/B) starts from the school earlier than (B\/A) (c) (A\/B) walks faster than (B\/A) (d) A and B reach home at the (same\/different) time (e) (A\/B) overtakes (B\/A) on the road (once\/twice). 2020-21","56 PHYSICS 3.3 Fig. 3.19 3.4 A woman starts from her home at 9.00 am, walks with a speed of 5 km h\u20131 on a 3.5 straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and 3.6 returns home by an auto with a speed of 25 km h\u20131. Choose suitable scales and 3.7 plot the x-t graph of her motion. 3.8 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, 3.9 followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically 3.10 and otherwise how long the drunkard takes to fall in a pit 13 m away from the start. A jet airplane travelling at the speed of 500 km h\u20131 ejects its products of combustion at the speed of 1500 km h\u20131 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ? A car moving along a straight highway with speed of 126 km h\u20131 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop ? Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h\u20131 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s\u20132. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them ? On a two-lane road, car A is travelling with a speed of 36 km h\u20131. Two cars B and C approach car A in opposite directions with a speed of 54 km h\u20131 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident ? Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h\u20131 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road? A player throws a ball upwards with an initial speed of 29.4 m s\u20131. (a) What is the direction of acceleration during the upward motion of the ball ? (b) What are the velocity and acceleration of the ball at the highest point of its motion ? (c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion. (d) To what height does the ball rise and after how long does the ball return to the player\u2019s hands ? (Take g = 9.8 m s\u20132 and neglect air resistance). 2020-21","MOTION IN A STRAIGHT LINE 57 3.11 Read each statement below carefully and state with reasons and examples, if it is 3.12 true or false ; 3.13 A particle in one-dimensional motion (a) with zero speed at an instant may have non-zero acceleration at that instant 3.14 (b) with zero speed may have non-zero velocity, (c) with constant speed must have zero acceleration, 3.15 (d) with positive value of acceleration must be speeding up. 3.16 A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s. Explain clearly, with examples, the distinction between : (a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval; (b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only]. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h\u20131. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h\u20131. What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !] In Exercises 3.13 and 3.14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why ? Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle. Fig. 3.20 2020-21","58 PHYSICS Fig. 3.21 3.17 Figure 3.21 shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t >0 ? If not, suggest a suitable physical context for this graph. 3.18 A police van moving on a highway with a speed of 3.19 30 km h\u20131 fires a bullet at a thief\u2019s car speeding away in the same direction with a speed of 192 km h\u20131. If the muzzle speed of the bullet is 150 m s\u20131, with what speed does the bullet hit the thief\u2019s car ? (Note: Obtain that speed which is relevant for damaging the thief\u2019s car). Suggest a suitable physical situation for each of the following graphs (Fig 3.22): 3.20 Fig. 3.22 Figure 3.23 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, \u2013 1.2 s. Fig. 3.23 3.21 Figure 3.24 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval. Fig. 3.24 2020-21","MOTION IN A STRAIGHT LINE 59 3.22 Figure 3.25 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time Fig. 3.25 are shown. In which interval is the average acceleration greatest in magnitude ? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ? Additional Exercises 3.23 A three-wheeler starts from rest, accelerates uniformly with 1 m s\u20132 on a straight 3.24 road for 10 s, and then moves with uniform velocity. Plot the distance covered by 3.25 the vehicle during the nth second (n = 1,2,3\u2026.) versus n. What do you expect this plot to be during accelerated motion : a straight line or a parabola ? A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s\u20131. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s-1 and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands ? On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed 9 km h\u20131 (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h\u20131. For an observer on a stationary platform outside, what is the (a) speed of the child running in the direction of motion of the belt ?. (b) speed of the child running opposite to the direction of motion of the belt ? (c) time taken by the child in (a) and (b) ? Which of the answers alter if motion is viewed by one of the parents ? 3.26 Fig. 3.26 Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m s\u20131 and 30 m s\u20131. Verify that the graph shown in Fig. 3.27 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 m s\u20132. Give the equations for the linear and curved parts of the plot. 2020-21","60 PHYSICS 3.27 Fig. 3.27 The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s. Fig. 3.28 3.28 What is the average speed of the particle over the intervals in (a) and (b) ? The velocity-time graph of a particle in one-dimensional motion is shown in Fig. 3.29 : Fig. 3.29 Which of the following formulae are correct for describing the motion of the particle over the time-interval t to t : 12 (a) x(t2 ) = x(t1) + v (t1) (t2 \u2013 t1) +(\u00bd) a (t2 \u2013 t1)2 (b) v(t2 ) = v(t1) + a (t2 \u2013 t1) (c) vaverage = (x(t2) \u2013 x(t1))\/(t2 \u2013 t1) (d) aaverage = (v(t2) \u2013 v(t1))\/(t2 \u2013 t1) (e) x(t2 ) = x(t1) + vaverage (t2 \u2013 t1) + (\u00bd) aaverage (t2 \u2013 t1)2 (f) x(t2 ) \u2013 x(t1) = area under the v-t curve bounded by the t-axis and the dotted line shown. 2020-21","MOTION IN A STRAIGHT LINE 61 APPENDIX 3.1 : ELEMENTS OF CALCULUS Differential Calculus Using the concept of \u2018differential coefficient\u2019 or \u2018derivative\u2019, we can easily define velocity and acceleration. Though you will learn in detail in mathematics about derivatives, we shall introduce this concept in brief in this Appendix so as to facilitate its use in describing physical quantities involved in motion. Suppose we have a quantity y whose value depends upon a single variable x, and is expressed by an equation defining y as some specific function of x. This is represented as: y = f (x) (1) This relationship can be visualised by drawing a graph of function y = f (x) regarding y and x as Cartesian coordinates, as shown in Fig. 3.30 (a). (a) (b) Fig. 3.30 Consider the point P on the curve y = f (x) whose coordinates are (x, y) and another point Q where coordinates are (x + \u2206x, y + \u2206y). The slope of the line joining P and Q is given by: tan\u03b8 = \u2206y = (y + \u2206y) \u2212 y (2) \u2206x \u2206x Suppose now that the point Q moves along the curve towards P. In this process, \u2206y and \u2206x decrease and approach zero; though their ratio \u2206y will not necessarily vanish. What happens \u2206x to the line PQ as \u2206y\u2192 0, \u2206x\u2192 0. You can see that this line becomes a tangent to the curve at point P as shown in Fig. 3.30(b). This means that tan \u03b8 approaches the slope of the tangent at P, denoted by m: m = lim \u2206y = lim (y + \u2206y) \u2212 y (3) \u2206x \u2206x \u2206x \u21920 \u2206x \u21920 The limit of the ratio \u2206y\/\u2206x as \u2206x approaches zero is called the derivative of y with respect to x and is written as dy\/dx. It represents the slope of the tangent line to the curve y = f (x) at the point (x, y). Since y = f (x) and y + \u2206y = f (x + \u2206x), we can write the definition of the derivative as: dy = df ( x) = \u2206lxim\u21920 \u2206y = \u2206lxim\u21920 \uf8ee f (x + \u2206x) \u2013 f (x )\uf8f9 dx dx \u2206x \uf8f0\uf8ef \u2206x \uf8fb\uf8fa Given below are some elementary formulae for derivatives of functions. In these u (x) and v (x) represent arbitrary functions of x, and a and b denote constant quantities that are independent of x. Derivatives of some common functions are also listed . 2020-21","62 PHYSICS d (a u) = a du ; du = du . dx dx dx dt dx dt d(uv) = u dv + v du ; d (u \/v) = 1 du \u2013 u dv dx dx dx v2 dx dx dx du = du dx dv dv dx d (sin x) = cos x ; d (cos x ) = \u2013 sin x dx dx d (tan x ) = sec2 x ; d (cot x ) = \u2013 cos ec 2 x dx dx d (sec x) = tan x sec x ; d (cosec2x ) = \u2013 cot x co sec x d x dx d (u )n = n un \u20131 du ; d ( ln u) = 1 dx dx du u d (eu ) = eu du In terms of derivatives, instantaneous velocity and acceleration are defined as v = lim \u2206x = dx \u2206t dt \u2206t \u21920 a = lim \u2206v = dv = d2x \u2206t \u21920 \u2206t dt dt 2 Integral Calculus You are familiar with the notion of area. The formulae for areas of simple geometrical figures are also known to you. For example, the area of a rectangle is length times breadth and that of a triangle is half of the product of base and height. But how to deal with the problem of determination of area of an irregular figure? The mathematical notion of integral is necessary in connection with such problems. Let us take a concrete example. Suppose a variable force f (x) acts on a particle in its motion along x - axis from x = a to x = b. The problem is to determine the work done (W) by the force on the particle during the motion. This problem is discussed in detail in Chapter 6. Figure 3.31 shows the variation of F(x) with x. If the force were constant, work would be simply the area F (b-a) as shown in Fig. 3.31(i). But in the general case, force is varying . Fig. 3.31 2020-21","MOTION IN A STRAIGHT LINE 63 To calculate the area under this curve [Fig. 3.31 (ii)], let us employ the following trick. Divide the interval on x-axis from a to b into a large number (N) of small intervals: x0(=a) to x1, x1 to x2 ; x2 to x3, ................................ xN-1 to xN (=b). The area under the curve is thus divided into N strips. Each strip is approximately a rectangle, since the variation of F(x) over a strip is negligible. The area of the ith strip shown [Fig. 3.31(ii)] is then approximately \u2206Ai = F(xi )(xi \u2013 xi \u20131) = F(xi )\u2206x where \u2206x is the width of the strip which we have taken to be the same for all the strips. You may wonder whether we should put F(xi-1) or the mean of F(xi) and F(xi-1) in the above expression. If we take N to be very very large (N\u2192\u221e ), it does not really matter, since then the strip will be so thin that the difference between F(xi) and F(xi-1) is vanishingly small. The total area under the curve then is: \u2211 \u2211N N A = \u2206Ai = F (xi )\u2206x i =1 i =1 The limit of this sum as N\u2192\u221e is known as the integral of F(x) over x from a to b. It is given a special symbol as shown below: b \u222bA = F(x )dx a The integral sign \u222b looks like an elongated S, reminding us that it basically is the limit of the sum of an infinite number of terms. A most significant mathematical fact is that integration is, in a sense, an inverse of differentiation. Suppose we have a function g (x) whose derivative is f (x), i.e. f (x ) = dg(x ) dx The function g (x) is known as the indefinite integral of f (x) and is denoted as: \u222bg(x) = f (x )dx An integral with lower and upper limits is known as a definite integral. It is a number. Indefinite integral has no limits; it is a function. A fundamental theorem of mathematics states that b \u222b f (x ) dx = g(x ) b \u2261 g(b) \u2013 g(a ) a a As an example, suppose f (x) = x2 and we wish to determine the value of the definite integral from x =1 to x = 2. The function g (x) whose derivative is x2 is x3\/3. Therefore, \u222b2 x3 2 8 1 7 x 2 dx 3 3 3 3 = = \u2013 = 11 Clearly, to evaluate definite integrals, we need to know the corresponding indefinite integrals. Some common indefinite integrals are 2020-21","64 PHYSICS \u222b ( 1 )dx = ln x (x > 0) x This introduction to differential and integral calculus is not rigorous and is intended to convey to you the basic notions of calculus. 2020-21","CHAPTER FOUR MOTION IN A PLANE 4.1 Introduction 4.1 INTRODUCTION 4.2 Scalars and vectors In the last chapter we developed the concepts of position, 4.3 Multiplication of vectors by displacement, velocity and acceleration that are needed to describe the motion of an object along a straight line. We real numbers found that the directional aspect of these quantities can be taken care of by + and \u2013 signs, as in one dimension only two 4.4 Addition and subtraction of directions are possible. But in order to describe motion of an object in two dimensions (a plane) or three dimensions vectors \u2014 graphical method (space), we need to use vectors to describe the above- mentioned physical quantities. Therefore, it is first necessary 4.5 Resolution of vectors to learn the language of vectors. What is a vector ? How to 4.6 Vector addition \u2014 analytical add, subtract and multiply vectors ? What is the result of multiplying a vector by a real number ? We shall learn this method to enable us to use vectors for defining velocity and acceleration in a plane. We then discuss motion of an object 4.7 Motion in a plane in a plane. As a simple case of motion in a plane, we shall 4.8 Motion in a plane with discuss motion with constant acceleration and treat in detail the projectile motion. Circular motion is a familiar class of constant acceleration motion that has a special significance in daily-life situations. We shall discuss uniform circular motion in some detail. 4.9 Relative velocity in two The equations developed in this chapter for motion in a dimensions plane can be easily extended to the case of three dimensions. 4.10 Projectile motion 4.2 SCALARS AND VECTORS 4.11 Uniform circular motion In physics, we can classify quantities as scalars or Summary vectors. Basically, the difference is that a direction is Points to ponder associated with a vector but not with a scalar. A scalar Exercises quantity is a quantity with magnitude only. It is specified Additional exercises completely by a single number, along with the proper unit. Examples are : the distance between two points, mass of an object, the temperature of a body and the time at which a certain event happened. The rules for combining scalars are the rules of ordinary algebra. Scalars can be added, subtracted, multiplied and divided 2020-21","66 PHYSICS just as the ordinary numbers*. For example, represented by another position vector, OP\u2032 denoted by r\u2032. The length of the vector r if the length and breadth of a rectangle are represents the magnitude of the vector and its 1.0 m and 0.5 m respectively, then its direction is the direction in which P lies as seen perimeter is the sum of the lengths of the from O. If the object moves from P to P\u2032, the four sides, 1.0 m + 0.5 m +1.0 m + 0.5 m = vector PP\u2032 (with tail at P and tip at P\u2032) is called 3.0 m. The length of each side is a scalar the displacement vector corresponding to and the perimeter is also a scalar. Take motion from point P (at time t) to point P\u2032 (at time t\u2032). another example: the maximum and minimum temperatures on a particular day Fig. 4.1 (a) Position and displacement vectors. are 35.6 \u00b0C and 24.2 \u00b0C respectively. Then, (b) Displacement vector PQ and different the difference between the two temperatures courses of motion. is 11.4 \u00b0C. Similarly, if a uniform solid cube of aluminium of side 10 cm has a mass of It is important to note that displacement 2.7 kg, then its volume is 10\u20133 m3 (a scalar) vector is the straight line joining the initial and and its density is 2.7\u00d7103 kg m\u20133 (a scalar). final positions and does not depend on the actual path undertaken by the object between the two A vector quantity is a quantity that has both positions. For example, in Fig. 4.1(b), given the a magnitude and a direction and obeys the initial and final positions as P and Q, the triangle law of addition or equivalently the displacement vector is the same PQ for different parallelogram law of addition. So, a vector is paths of journey, say PABCQ, PDQ, and PBEFQ. specified by giving its magnitude by a number Therefore, the magnitude of displacement is and its direction. Some physical quantities that either less or equal to the path length of an are represented by vectors are displacement, object between two points. This fact was velocity, acceleration and force. emphasised in the previous chapter also while discussing motion along a straight line. To represent a vector, we use a bold face type in this book. Thus, a velocity vector can be 4.2.2 Equality of Vectors represented by a symbol v. Since bold face is Two vectors A and B are said to be equal if, and difficult to produce, when written by hand, a only if, they have the same magnitude and the vector is often represrented by an arrow placerd same direction.** over a letter, say v . Thus, both v and v Figure 4.2(a) shows two equal vectors A and represent the velocity vector. The magnitude of B. We can easily check their equality. Shift B a vector is often called its absolute value, parallel to itself until its tail Q coincides with that indicated by |v| = v. Thus, a vector is of A, i.e. Q coincides with O. Then, since their represented by a bold face, e.g. by A, a, p, q, r, ... tips S and P also coincide, the two vectors are x, y, with respective magnitudes denoted by light said to be equal. In general, equality is indicated face A, a, p, q, r, ... x, y. 4.2.1 Position and Displacement Vectors To describe the position of an object moving in a plane, we need to choose a convenient point, say O as origin. Let P and P\u2032 be the positions of the object at time t and t\u2032, respectively [Fig. 4.1(a)]. We join O and P by a straight line. Then, OP is the position vector of the object at time t. An arrow is marked at the head of this line. It is represented by a symbol r, i.e. OP = r. Point P\u2032 is * Addition and subtraction of scalars make sense only for quantities with same units. However, you can multiply and divide scalars of different units. ** In our study, vectors do not have fixed locations. So displacing a vector parallel to itself leaves the vector unchanged. Such vectors are called free vectors. However, in some physical applications, location or line of application of a vector is important. Such vectors are called localised vectors. 2020-21","MOTION IN A PLANE 67 Fig. 4.2 (a) Two equal vectors A and B. (b) Two The factor \u03bb by which a vector A is multiplied could be a scalar having its own physical vectors A\u2032 and B\u2032 are unequal though they dimension. Then, the dimension of \u03bb A is the product of the dimensions of \u03bb and A. For are of the same length. example, if we multiply a constant velocity vector by duration (of time), we get a displacement as A = B. Note that in Fig. 4.2(b), vectors A\u2032 and vector. B\u2032 have the same magnitude but they are not equal because they have different directions. 4.4 ADDITION AND SUBTRACTION OF Even if we shift B\u2032 parallel to itself so that its tail VECTORS \u2014 GRAPHICAL METHOD Q\u2032 coincides with the tail O\u2032 of A\u2032 , the tip S\u2032 of B\u2032 does not coincide with the tip P\u2032 of A\u2032 . As mentioned in section 4.2, vectors, by 4.3 MULTIPLICATION OF VECTORS BY REAL definition, obey the triangle law or equivalently, the parallelogram law of addition. We shall now NUMBERS describe this law of addition using the graphical Multiplying a vector A with a positive number \u03bb method. Let us consider two vectors A and B that gives a vector whose magnitude is changed by lie in a plane as shown in Fig. 4.4(a). The lengths the factor \u03bb but the direction is the same as that of the line segments representing these vectors of A : are proportional to the magnitude of the vectors. To find the sum A + B, we place vector B so that \uf8e6 \u03bbA\uf8e6 =\u03bb\uf8e6 A\uf8e6 if\u03bb>0. its tail is at the head of the vector A, as in Fig. 4.4(b). Then, we join the tail of A to the head For example, if A is multiplied by 2, the resultant of B. This line OQ represents a vector R, that is, vector 2A is in the same direction as A and has the sum of the vectors A and B. Since, in this a magnitude twice of |A| as shown in Fig. 4.3(a). procedure of vector addition, vectors are Multiplying a vector A by a negative number \u2212\u03bb gives another vector whose direction is opposite to the direction of A and whose magnitude is \u03bb times |A|. Multiplying a given vector A by negative numbers, say \u20131 and \u20131.5, gives vectors as shown in Fig 4.3(b). (c) (d) Fig. 4.3 (a) Vector A and the resultant vector after multiplying A by a positive number 2. (b) Vector A and resultant vectors after Fig. 4.4 (a) Vectors A and B. (b) Vectors A and B multiplying it by a negative number \u20131 added graphically. (c) Vectors B and A and \u20131.5. added graphically. (d) Illustrating the associative law of vector addition. 2020-21","68 PHYSICS arranged head to tail, this graphical method is What is the physical meaning of a zero vector? called the head-to-tail method. The two vectors Consider the position and displacement vectors and their resultant form three sides of a triangle, in a plane as shown in Fig. 4.1(a). Now suppose so this method is also known as triangle method that an object which is at P at time t, moves to of vector addition. If we find the resultant of P\u2032 and then comes back to P. Then, what is its B + A as in Fig. 4.4(c), the same vector R is displacement? Since the initial and final obtained. Thus, vector addition is commutative: positions coincide, the displacement is a \u201cnull vector\u201d. A+B=B+A (4.1) The addition of vectors also obeys the associative Subtraction of vectors can be defined in terms law as illustrated in Fig. 4.4(d). The result of of addition of vectors. We define the difference adding vectors A and B first and then adding of two vectors A and B as the sum of two vectors vector C is the same as the result of adding B A and \u2013B : and C first and then adding vector A : A \u2013 B = A + (\u2013B) (4.5) (A + B) + C = A + (B + C) (4.2) It is shown in Fig 4.5. The vector \u2013B is added to vector A to get R2 = (A \u2013 B). The vector R1 = A + B What is the result of adding two equal and is also shown in the same figure for comparison. opposite vectors ? Consider two vectors A and We can also use the parallelogram method to \u2013A shown in Fig. 4.3(b). Their sum is A + (\u2013A). find the sum of two vectors. Suppose we have Since the magnitudes of the two vectors are the two vectors A and B. To add these vectors, we same, but the directions are opposite, the bring their tails to a common origin O as resultant vector has zero magnitude and is shown in Fig. 4.6(a). Then we draw a line from represented by 0 called a null vector or a zero the head of A parallel to B and another line from vector : the head of B parallel to A to complete a parallelogram OQSP. Now we join the point of A\u2013A=0 |0|= 0 (4.3) the intersection of these two lines to the origin O. The resultant vector R is directed from the Since the magnitude of a null vector is zero, its common origin O along the diagonal (OS) of the direction cannot be specified. parallelogram [Fig. 4.6(b)]. In Fig.4.6(c), the triangle law is used to obtain the resultant of A The null vector also results when we multiply and B and we see that the two methods yield the same result. Thus, the two methods are a vector A by the number zero. The main equivalent. properties of 0 are : A+0=A \u03bb0=0 0A=0 (4.4) Fig. 4.5 (a) Two vectors A and B, \u2013 B is also shown. (b) Subtracting vector B from vector A \u2013 the result is R2. For comparison, addition of vectors A and B, i.e. R1 is also shown. 2020-21","MOTION IN A PLANE 69 Fig. 4.6 (a) Two vectors A and B with their tails brought to a common origin. (b) The sum A + B obtained using the parallelogram method. (c) The parallelogram method of vector addition is equivalent to the triangle method. Example 4.1 Rain is falling vertically with 4.5 RESOLUTION OF VECTORS a speed of 35 m s\u20131. Winds starts blowing after sometime with a speed of 12 m s\u20131 in Let a and b be any two non-zero vectors in a east to west direction. In which direction plane with different directions and let A be should a boy waiting at a bus stop hold another vector in the same plane(Fig. 4.8). A can his umbrella ? be expressed as a sum of two vectors \u2014 one obtained by multiplying a by a real number and Fig. 4.7 the other obtained by multiplying b by another real number. To see this, let O and P be the tail Answer The velocity of the rain and the wind and head of the vector A. Then, through O, draw are represented by the vectors vr and vw in Fig. a straight line parallel to a, and through P, a 4.7 and are in the direction specified by the straight line parallel to b. Let them intersect at problem. Using the rule of vector addition, we Q. Then, we have see that the resultant of vr and vw is R as shown in the figure. The magnitude of R is A = OP = OQ + QP (4.6) R = vr2 + vw2 = 352 + 122 m s\u22121 = 37 m s\u22121 But since OQ is parallel to a, and QP is parallel to b, we can write : The direction \u03b8 that R makes with the vertical is given by OQ = \u03bb a, and QP = \u00b5 b (4.7) tan \u03b8 = vw = 12 = 0.343 where \u03bb and \u00b5 are real numbers. vr 35 Therefore, A = \u03bb a + \u00b5 b (4.8) Or, \u03b8 = tan-1( 0.343) = 19\u00b0 Therefore, the boy should hold his umbrella Fig. 4.8 (a) Two non-colinear vectors a and b. (b) Resolving a vector A in terms of vectors in the vertical plane at an angle of about 19o a and b. with the vertical towards the east. We say that A has been resolved into two component vectors \u03bb a and \u00b5 b along a and b respectively. Using this method one can resolve 2020-21","70 PHYSICS a given vector into two component vectors along and A2 is parallel to j , we have : (4.11) a set of two vectors \u2013 all the three lie in the same A1= Ax i , A2 = Ay j (4.12) plane. It is convenient to resolve a general vector along the axes of a rectangular coordinate where Ax and Ay are real numbers. system using vectors of unit magnitude. These are called unit vectors that we discuss now. A Thus, A = Ax i + Ay j unit vector is a vector of unit magnitude and points in a particular direction. It has no This is represented in Fig. 4.9(c). The quantities dimension and unit. It is used to specify a Ax and Ay are called x-, and y- components of the direction only. Unit vectors along the x-, y- and vector A. Note that Ax is itself not a vector, but z-axes of a rectangular coordinate system are Ax i is a vector, and so is Ay j . Using simple denoted by i , j and k\u02c6 , respectively, as shown trigonometry, we can express Ax and Ay in terms in Fig. 4.9(a). of the magnitude of A and the angle \u03b8 it makes Since these are unit vectors, we have with the x-axis : \uf8e6 \u02c6i \uf8e6 = \uf8e6 \u02c6j \uf8e6 = \uf8e6 k\u02c6 \uf8e6=1 (4.9) Ax = A cos \u03b8 (4.13) Ay = A sin \u03b8 These unit vectors are perpendicular to each As is clear from Eq. (4.13), a component of a other. In this text, they are printed in bold face with a cap (^) to distinguish them from other vector can be positive, negative or zero vectors. Since we are dealing with motion in two dimensions in this chapter, we require use of depending on the value of \u03b8. only two unit vectors. If we multiply a unit vector, Now, we have two ways to specify a vector A say n\u02c6 by a scalar, the result is a vector in a plane. It can be specified by : \u03bb = \u03bbn\u02c6 . In general, a vector A can be written as (i) its magnitude A and the direction \u03b8 it makes with the x-axis; or (ii) its components Ax and Ay A = |A| n\u02c6 (4.10) If A and \u03b8 are given, Ax and Ay can be obtained using Eq. (4.13). If Ax and Ay are given, A and \u03b8 where n\u02c6 is a unit vector along A. can be obtained as follows : We can now resolve a vector A in terms Ax2 + Ay2 = A2cos2\u03b8 + A2sin2\u03b8 of component vectors that lie along unit vectors = A2 \u02c6i and j. Consider a vector A that lies in x-y Or, A= A 2 + Ay2 (4.14) plane as shown in Fig. 4.9(b). We draw lines from x the head of A perpendicular to the coordinate axes as in Fig. 4.9(b), and get vectors A1 and A2 And tan \u03b8 = Ay , \u03b8= tan\u2212 1 Ay (4.15) Ax Ax such that A1 + A2 = A. Since A1 is parallel to i Fig. 4.9 (a) Unit vectors i , j and k lie along the x-, y-, and z-axes. (b) A vector A is resolved into its components Ax and Ay along x-, and y- axes. (c) A1 and A2 expressed in terms of i and j . 2020-21","MOTION IN A PLANE 71 So far we have considered a vector lying in B = Bx i + By j (4.19a) an x-y plane. The same procedure can be used Let R be their sum. We have to resolve a general vector A into three components along x-, y-, and z-axes in three R=A+B dimensions. If \u03b1, \u03b2, and \u03b3 are the angles* ( ) ( )= Ax i + Ay j + Bx i + By j between A and the x-, y-, and z-axes, respectively Since vectors obey the commutative and [Fig. 4.9(d)], we have associative laws, we can arrange and regroup the vectors in Eq. (4.19a) as convenient to us : ( )( )R = Ax + Bx i + Ay + By j (4.19b) SinceR = Rx i + Ry j (4.20) we have, Rx = Ax + Bx , Ry = Ay + By (4.21) Thus, each component of the resultant vector R is the sum of the corresponding components of A and B. In three dimensions, we have A = Ax i + Ay j + Azk B = Bx i + By j + Bzk (d) R = A + B = Rx i + Ry j + Rzk Fig. 4.9 (d) A vector A resolved into components along with Rx = Ax + Bx x-, y-, and z-axes A x = A cos \u03b1, A y = A cos \u03b2, A z = A cos \u03b3 (4.16a) Ry = Ay + By In general, we have Rz = Az + Bz (4.22) A = Ax\u02c6i + Ay\u02c6j + Az k\u02c6 (4.16b) This method can be extended to addition and The magnitude of vector A is subtraction of any number of vectors. For example, if vectors a, b and c are given as A = Ax2 + Ay2 + Az2 (4.16c) A position vector r can be expressed as a = axi + ayj + azk r= x i+yj+zk (4.17) where x, y, and z are the components of r along b = bx i + by j + bzk x-, y-, z-axes, respectively. c = cxi + cyj + czk (4.23a) 4.6 VECTOR ADDITION \u2013 ANALYTICAL then, a vector T = a + b \u2013 c has components : METHOD Tx = a x + bx \u2212 cx (4.23b) Although the graphical method of adding vectors Ty = a y + by \u2212 cy helps us in visualising the vectors and the resultant vector, it is sometimes tedious and has Tz = a z + bz \u2212 cz . limited accuracy. It is much easier to add vectors by combining their respective components. Example 4.2 Find the magnitude and Consider two vectors A and B in x-y plane with direction of the resultant of two vectors A components Ax, Ay and Bx, By : and B in terms of their magnitudes and angle \u03b8 between them. A = Ax i + Ay j (4.18) * Note that angles \u03b1, \u03b2, and \u03b3 are angles in space. They are between pairs of lines, which are not coplanar. 2020-21","72 PHYSICS Example 4.3 A motorboat is racing towards north at 25 km\/h and the water current in that region is 10 km\/h in the direction of 60\u00b0 east of south. Find the resultant velocity of the boat. Fig. 4.10 Answer The vector vb representing the velocity of the motorboat and the vector vc representing Answer Let OP and OQ represent the two vectors the water current are shown in Fig. 4.11 in A and B making an angle \u03b8 (Fig. 4.10). Then, directions specified by the problem. Using the using the parallelogram method of vector parallelogram method of addition, the resultant addition, OS represents the resultant vector R : R is obtained in the direction shown in the figure. R=A+B SN is normal to OP and PM is normal to OS. From the geometry of the figure, OS2 = ON2 + SN2 but ON = OP + PN = A + B cos \u03b8 SN = B sin \u03b8 OS2 = (A + B cos \u03b8)2 + (B sin \u03b8)2 or, R2 = A2 + B2 + 2AB cos \u03b8 R = A2 + B2 + 2AB cos \u03b8 (4.24a) In \u2206 OSN, SN = OS sin\u03b1 = R sin\u03b1, and in \u2206 PSN, SN = PS sin \u03b8 = B sin \u03b8 Therefore, R sin \u03b1 = B sin \u03b8 or, R=B (4.24b) sin \u03b8 sin \u03b1 (4.24c) Fig. 4.11 Similarly, PM = A sin \u03b1 = B sin \u03b2 We can obtain the magnitude of R using the Law of cosine : or, A =B sin \u03b2 sin \u03b1 Combining Eqs. (4.24b) and (4.24c), we get R= v 2 + vc2 + 2v bvccos120o b RAB (4.24d) = 252 +102 + 2 \u00d7 25 \u00d710 (-1\/2) \u2245 22 km\/h == sin \u03b8 sin \u03b2 sin \u03b1 Using Eq. (4.24d), we get: To obtain the direction, we apply the Law of sines B (4.24e) R = vc or, sin \u03c6 = vc sin \u03b8 sin \u03b1 = sin \u03b8 sin \u03b8 sin \u03c6 R R where R is given by Eq. (4.24a). 10 \u00d7 sin120 10 3 tan \u03b1 = SN B sin \u03b8 = = \u2245 0.397 OP + PN A + B cos\u03b8 or, = (4.24f) 21.8 2 \u00d7 21.8 Equation (4.24a) gives the magnitude of the \u03c6 \u2245 23.4 resultant and Eqs. (4.24e) and (4.24f) its direction. Equation (4.24a) is known as the Law of cosines 4.7 MOTION IN A PLANE and Eq. (4.24d) as the Law of sines. In this section we shall see how to describe motion in two dimensions using vectors. 2020-21","MOTION IN A PLANE 73 4.7.1 Position Vector and Displacement Suppose a particle moves along the curve shown The position vector r of a particle P located in a plane with reference to the origin of an x-y by the thick line and is at P at time t and P\u2032 at reference frame (Fig. 4.12) is given by time t\u2032 [Fig. 4.12(b)]. Then, the displacement is : r = xi+yj where x and y are components of r along x-, and \u2206r = r\u2032 \u2013 r (4.25) y- axes or simply they are the coordinates of the object. and is directed from P to P\u2032 . (a) We can write Eq. (4.25) in a component form: (b) ( ) ( )\u2206r = x' i + y' j \u2212 x i + y j Fig. 4.12 (a) Position vector r. (b) Displacement \u2206r and = i\u2206x + j\u2206y average velocity v of a particle. where \u2206x = x \u2032 \u2013 x, \u2206y = y\u2032 \u2013 y (4.26) Velocity The average velocity (v) of an object is the ratio of the displacement and the corresponding time interval : \u2206 r \u2206x i + \u2206y j \u2206x \u2206y v= = =i +j (4.27) \u2206t \u2206t \u2206t \u2206t Or, v = vx \u02c6i + vy j \u2206r Since v = , the direction of the average velocity \u2206t is the same as that of \u2206r (Fig. 4.12). The velocity (instantaneous velocity) is given by the limiting value of the average velocity as the time interval approaches zero : v = lim \u2206r = dr (4.28) \u2206t\u21920 \u2206t dt The meaning of the limiting process can be easily understood with the help of Fig 4.13(a) to (d). In these figures, the thick line represents the path of an object, which is at P at time t. P1, P2 and P3 represent the positions of the object after times \u2206t1,\u2206t2, and \u2206t3. \u2206r1, \u2206r2, and \u2206r3 are the displacements of the object in times \u2206t1, \u2206t2, and Fig. 4.13 As the time interval \u2206t approaches zero, the average velocity approaches the velocity v. The direction of v is parallel to the line tangent to the path. 2020-21","74 PHYSICS \u2206t3, respectively. The direction of the average Fig. 4.14 The components v and vy of velocity v and velocity v is shown in figures (a), (b) and (c) for x three decreasing values of \u2206t, i.e. \u2206t1,\u2206t2, and \u2206t3, (\u2206t1 > \u2206t2 > \u2206t3). As \u2206t \u2192 0, \u2206r \u2192 0 and is along the tangent to the path [Fig. 4.13(d)]. Therefore, the direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of motion. We can express v in a component form : dr the angle \u03b8 it makes with x-axis. Note that v= v = v cos \u03b8, v = v sin \u03b8. dt x y \uf8eb \u2206x \u2206y \uf8f6 (4.29) The acceleration (instantaneous acceleration) = lim \uf8ec i + j\uf8f7 is the limiting value of the average acceleration as the time interval approaches zero : \u2206t \u21920\uf8ed \u2206t \u2206t \uf8f8 \u2206x \u2206y \u2206v (4.32a) = i lim + j lim a = lim \u2206t \u21920 \u2206t \u2206t \u21920 \u2206t \u2206t \u2192 0 \u2206t Or, dx dy = vx i + vy j. Since \u2206v = \u2206vx i + \u2206vy j,we have v=i + j dt dt a = i lim \u2206v x + j lim \u2206v y dx dy (4.30a) where vx = dt ,vy = dt \u2206t \u2192 0 \u2206t \u2206t \u2192 0 \u2206t So, if the expressions for the coordinates x and Or, a = a x i + a y j y are known as functions of time, we can use (4.32b) these equations to find vx and vy. where, ax = dv x , ay = dvy (4.32c)* The magnitude of v is then dt dt v= v 2 + vy2 (4.30b) As in the case of velocity, we can understand x and the direction of v is given by the angle \u03b8 : graphically the limiting process used in defining \uf8eb \uf8f6 acceleration on a graph showing the path of the \u22121\uf8ec \uf8f7 tan\u03b8 = v y , \u03b8 = tan v y \uf8f7 object\u2019s motion. This is shown in Figs. 4.15(a) to vx \uf8ec v x \uf8f8 \uf8ed (4.30c) (d). P represents the position of the object at time t and P , P , P positions after time \u2206t , \u2206t , 123 12 vx, vy and angle \u03b8 are shown in Fig. 4.14 for a \u2206t , respectively (\u2206t > \u2206t >\u2206t ). The velocity vectors velocity vector v at point p. 3 1 23 at points P, P , P , P are also shown in Figs. 4.15 123 (a), (b) and (c). In each case of \u2206t, \u2206v is obtained Acceleration using the triangle law of vector addition. By The average acceleration a of an object for a time interval \u2206t moving in x-y plane is the change definition, the direction of average acceleration in velocity divided by the time interval : is the same as that of \u2206v. We see that as \u2206t decreases, the direction of \u2206v changes and ( )\u2206v \u2206 vx i + vy j = \u2206v x i + \u2206vy j consequently, the direction of the acceleration a= = (4.31a) changes. Finally, in the limit \u2206t 0 [Fig. 4.15(d)], \u2206t \u2206t \u2206t \u2206t the average acceleration becomes the Or, a = a x i + a y j . (4.31b) instantaneous acceleration and has the direction as shown. * In terms of x and y, a and a can be expressed as xy 2020-21","MOTION IN A PLANE 75 x (m) Fig. 4.15 The average acceleration for three time intervals (a) \u2206t1, (b) \u2206t2, and (c) \u2206t3, (\u2206t1> \u2206t2> \u2206t3). (d) In the limit \u2206t 0, the average acceleration becomes the acceleration. Note that in one dimension, the velocity and \u03b8 = tan-1 \uf8eb vy \uf8f6 = tan\u22121 \uf8eb 4 \uf8f6 \u2245 53\u00b0 with x-axis. the acceleration of an object are always along \uf8ec vx \uf8f7 \uf8ed\uf8ec 3 \uf8f7\uf8f8 the same straight line (either in the same \uf8ed \uf8f8 direction or in the opposite direction). However, for motion in two or three 4.8 MOTION IN A PLANE WITH CONSTANT dimensions, velocity and acceleration vectors ACCELERATION may have any angle between 0\u00b0 and 180\u00b0 between them. Suppose that an object is moving in x-y plane and its acceleration a is constant. Over an Example 4.4 The position of a particle is interval of time, the average acceleration will given by equal this constant value. Now, let the velocity of the object be v0 at time t = 0 and v at time t. r = 3.0t \u02c6i + 2.0t 2\u02c6j + 5.0 k\u02c6 Then, by definition where t is in seconds and the coefficients have the proper units for r to be in metres. a = v \u2212 v0 = v \u2212 v0 (a) Find v(t) and a(t) of the particle. (b) Find t\u22120 t the magnitude and direction of v(t) at t = 1.0 s. Or, v = v0 + at (4.33a) In terms of components : Answer vx = vox + a xt ( )v(t ) = dr = d 3.0 t i + 2.0t 2 j + 5.0 k dt dt vy = voy + a yt (4.33b) = 3.0i + 4.0t j Let us now find how the position r changes with time. We follow the method used in the one- a (t ) = dv = +4.0 j dimensional case. Let ro and r be the position vectors of the particle at time 0 and t and let the dt velocities at these instants be vo and v. Then, over this time interval t, the average velocity is a = 4.0 m s\u20132 along y- direction (vo + v)\/2. The displacement is the average At t = 1.0 s, v = 3.0\u02c6i + 4.0\u02c6j velocity multiplied by the time interval : It\u2019s magnitude is v = 32 + 42 = 5.0 m s-1 r \u2212 r0 = \uf8ebv + v0 \uf8f6 t = \uf8eb (v0 + at ) + v0 \uf8f6 t and direction is \uf8ec\uf8ed 2 \uf8f7\uf8f8 \uf8f8\uf8f7 \uf8ed\uf8ec 2 2020-21","76 PHYSICS = v0t + 1 at 2 Given x (t) = 84 m, t = ? 2 5.0 t + 1.5 t 2 = 84 \u21d2 t = 6 s Or, r = r0 + v0t + 1 at 2 (4.34a) At t = 6 s, y = 1.0 (6)2 = 36.0 m 2 Now, the velocity v = dr = (5.0 + 3.0t )\u02c6i + 2.0 t \u02c6j It can be easily verified that the derivative of dt Eq. (4.34a), i.e. dr gives Eq.(4.33a) and it also At t = 6 s, v = 23.0i + 12.0j dt speed = v = 232 + 122 \u2245 26 m s\u22121 . satisfies the condition that at t=0, r = ro. Equation (4.34a) can be written in component 4.9 RELATIVE VELOCITY IN TWO form as DIMENSIONS x = x0 + voxt + 1 a t 2 The concept of relative velocity, introduced in 2 section 3.7 for motion along a straight line, can x be easily extended to include motion in a plane or in three dimensions. Suppose that two objects y = y0 + voyt + 1 a t 2 (4.34b) A and B are moving with velocities vA and vB 2 (each with respect to some common frame of y reference, say ground.). Then, velocity of object A relative to that of B is : One immediate interpretation of Eq.(4.34b) is that the motions in x- and y-directions can be treated vAB = vA \u2013 vB (4.35a) independently of each other. That is, motion in a plane (two-dimensions) can be treated as two and similarly, the velocity of object B relative to separate simultaneous one-dimensional motions with constant acceleration along two that of A is : perpendicular directions. This is an important result and is useful in analysing motion of objects vBA = vB \u2013 vA (4.35b) in two dimensions. A similar result holds for three Therefore, vAB = \u2013 vBA dimensions. The choice of perpendicular directions is convenient in many physical and, vAB = vBA (4.35c) situations, as we shall see in section 4.10 for projectile motion. Example 4.6 Rain is falling vertically with a speed of 35 m s\u20131. A woman rides a bicycle Example 4.5 A particle starts from origin with a speed of 12 m s\u20131 in east to west at t = 0 with a velocity 5.0 \u00ee m\/s and moves direction. What is the direction in which in x-y plane under action of a force which she should hold her umbrella ? produces a constant acceleration of Answer In Fig. 4.16 vr represents the velocity (3.0i+2.0j) m\/s2. (a) What is the of rain and vb , the velocity of the bicycle, the woman is riding. Both these velocities are with y-coordinate of the particle at the instant respect to the ground. Since the woman is riding its x-coordinate is 84 m ? (b) What is the a bicycle, the velocity of rain as experienced by speed of the particle at this time ? Answer From Eq. (4.34a) for r0 = 0, the position of the particle is given by r (t ) = v0t + 1 at 2 2 ( )= 5.0\u02c6it + (1\/2) 3.0\u02c6i + 2.0\u02c6j t 2 ( )= 5.0t + 1.5t 2 \u02c6i + 1.0t 2\u02c6j Fig. 4.16 Therefore, x (t ) = 5.0t + 1.5t 2 her is the velocity of rain relative to the velocity of the bicycle she is riding. That is vrb = vr \u2013 vb y (t ) = +1.0t 2 2020-21","MOTION IN A PLANE 77 This relative velocity vector as shown in Fig. 4.16 makes an angle \u03b8 with the vertical. It is given by tan \u03b8 = vb 12 = 0.343 = vr 35 Or, \u03b8 \u2245 19 Therefore, the woman should hold her umbrella at an angle of about 19\u00b0 with the vertical towards the west. Note carefully the difference between this Fig 4.17 Motion of an object projected with velocity Example and the Example 4.1. In Example 4.1, the boy experiences the resultant (vector v at angle \u03b80. sum) of two velocities while in this example, o the woman experiences the velocity of rain relative to the bicycle (the vector difference If we take the initial position to be the origin of of the two velocities). the reference frame as shown in Fig. 4.17, we have : 4.10 PROJECTILE MOTION xo = 0, yo = 0 As an application of the ideas developed in the Then, Eq.(4.34b) becomes : previous sections, we consider the motion of a x = vox t = (vo cos \u03b8o ) t y = (vo sin \u03b8o ) t \u2013 ( \u00bd )g t2 projectile. An object that is in flight after being and (4.38) thrown or projected is called a projectile. Such The components of velocity at time t can be obtained using Eq.(4.33b) : a projectile might be a football, a cricket ball, a baseball or any other object. The motion of a projectile may be thought of as the result of two vx = vox = vo cos \u03b8o separate, simultaneously occurring components vy = vo sin \u03b8o \u2013 g t (4.39) of motions. One component is along a horizontal Equation (4.38) gives the x-, and y-coordinates direction without any acceleration and the other of the position of a projectile at time t in terms of along the vertical direction with constant two parameters \u2014 initial speed vo and projection angle \u03b8o. Notice that the choice of mutually acceleration due to the force of gravity. It was perpendicular x-, and y-directions for the Galileo who first stated this independency of the horizontal and the vertical components of analysis of the projectile motion has resulted in projectile motion in his Dialogue on the great a simplification. One of the components of world systems (1632). velocity, i.e. x-component remains constant In our discussion, we shall assume that the throughout the motion and only the air resistance has negligible effect on the motion y- component changes, like an object in free fall of the projectile. Suppose that the projectile is in vertical direction. This is shown graphically launched with velocity vo that makes an angle at few instants in Fig. 4.18. Note that at the point \u03b8 with the x-axis as shown in Fig. 4.17. of maximum height, vy= 0 and therefore, o After the object has been projected, the \u03b8 = tan\u22121 vy = 0 acceleration acting on it is that due to gravity vx which is directed vertically downward: Equation of path of a projectile a = \u2212g j What is the shape of the path followed by the projectile? This can be seen by eliminating the Or, ax = 0, ay = \u2013 g (4.36) time between the expressions for x and y as given in Eq. (4.38). We obtain: The components of initial velocity vo are : vox = vo cos \u03b8o voy= vo sin \u03b8o (4.37) 2020-21","78 PHYSICS v0 sin\u03b8 0 2 2g y = (tan \u03b8o ) x \u2212 g x2 ( )Or, )2 (4.40) hm = (4.42) 2 (vocos\u03b8o Now, since g, \u03b8o and vo are constants, Eq. (4.40) Horizontal range of a projectile is of the form y = a x + b x2, in which a and b are constants. This is the equation of a parabola, The horizontal distance travelled by a projectile i.e. the path of the projectile is a parabola from its initial position (x = y = 0) to the position (Fig. 4.18). where it passes y = 0 during its fall is called the horizontal range, R. It is the distance travelled during the time of flight Tf . Therefore, the range R is R = (vo cos \u03b8o) (Tf ) =(vo cos \u03b8o) (2 vo sin \u03b8o)\/g Or, R = v02 sin 2\u03b80 (4.43a) g Equation (4.43a) shows that for a given projection velocity vo , R is maximum when sin 2\u03b80 is maximum, i.e., when \u03b80 = 450. The maximum horizontal range is, therefore, Rm = v 2 (4.43b) 0 g Fig. 4.18 The path of a projectile is a parabola. Example 4.7 Galileo, in his book Two new sciences, stated that \u201cfor elevations which Time of maximum height exceed or fall short of 45\u00b0 by equal amounts, the ranges are equal\u201d. Prove this How much time does the projectile take to reach statement. the maximum height ? Let this time be denoted by tm. Since at this point, vy= 0, we have from Answer For a projectile launched with velocity Eq. (4.39): vo at an angle \u03b8o , the range is given by vy = vo sin\u03b8o \u2013 g tm = 0 (4.41a) R = v02 sin 2\u03b80 Or, tm = vo sin\u03b8o \/g g The total time Tf during which the projectile is Now, for angles, (45\u00b0 + \u03b1 ) and ( 45\u00b0 \u2013 \u03b1), 2\u03b8o is in flight can be obtained by putting y = 0 in (90\u00b0 + 2\u03b1 ) and ( 90\u00b0 \u2013 2\u03b1 ) , respectively. The values of sin (90\u00b0 + 2\u03b1 ) and sin (90\u00b0 \u2013 2\u03b1 ) are Eq. (4.38). We get : the same, equal to that of cos 2\u03b1. Therefore, ranges are equal for elevations which exceed or Tf = 2 (vo sin \u03b8o )\/g (4.41b) fall short of 45\u00b0 by equal amounts \u03b1. Tf is known as the time of flight of the projectile. We note that Tf = 2 tm , which is expected because of the symmetry of the parabolic path. Maximum height of a projectile Example 4.8 A hiker stands on the edge of a cliff 490 m above the ground and The maximum height hm reached by the throws a stone horizontally with an initial projectile can be calculated by substituting speed of 15 m s-1. Neglecting air resistance, find the time taken by the stone to reach t = tm in Eq. (4.38) : the ground, and the speed with which it hits the ground. (Take g = 9.8 m s-2 ). ( )y = hm \uf8eb v0 sin\u03b8 0 \uf8f6 g \uf8eb v0sin\u03b8 0 \uf8f62 = v0 sin\u03b8 0 \uf8ed\uf8ec\uf8ec g \uf8f7\uf8f7\uf8f8 \u2212 2 \uf8ed\uf8ec\uf8ec g \uf8f8\uf8f7\uf8f7 2020-21","MOTION IN A PLANE 79 Answer We choose the origin of the x-,and y- Neglecting air resistance - what does the assumption really mean? axis at the edge of the cliff and t = 0 s at the While treating the topic of projectile motion, instant the stone is thrown. Choose the positive we have stated that we assume that the air resistance has no effect on the motion direction of x-axis to be along the initial velocity of the projectile. You must understand what the statement really means. Friction, force and the positive direction of y-axis to be the due to viscosity, air resistance are all dissipative forces. In the presence of any of vertically upward direction. The x-, and y- such forces opposing motion, any object will lose some part of its initial energy and components of the motion can be treated consequently, momentum too. Thus, a projectile that traverses a parabolic path independently. The equations of motion are : would certainly show deviation from its idealised trajectory in the presence of air x (t) = xo + vox t resistance. It will not hit the ground with y (t) = yo + voy t +(1\/2) ay t2 the same speed with which it was projected Here, xo = yo = 0, voy = 0, ay = \u2013g = \u20139.8 m s-2, from it. In the absence of air resistance, the vox = 15 m s-1. x-component of the velocity remains The stone hits the ground when y(t) = \u2013 490 m. constant and it is only the y-component that undergoes a continuous change. However, \u2013 490 m = \u2013(1\/2)(9.8) t2. in the presence of air resistance, both of these would get affected. That would mean This gives t =10 s. that the range would be less than the one given by Eq. (4.43). Maximum height The velocity components are vx = vox and attained would also be less than that vy = voy \u2013 g t predicted by Eq. (4.42). Can you then, so that when the stone hits the ground : anticipate the change in the time of flight? vox = 15 m s\u20131 In order to avoid air resistance, we will voy = 0 \u2013 9.8 \u00d7 10 = \u2013 98 m s\u20131 have to perform the experiment in vacuum Therefore, the speed of the stone is or under low pressure, which is not easy. When we use a phrase like \u2018neglect air v 2 + vy2 = 152 + 982 = 99 m s\u22121 resistance\u2019, we imply that the change in x parameters such as range, height etc. is much smaller than their values without air Example 4.9 A cricket ball is thrown at a resistance. The calculation without air speed of 28 m s\u20131 in a direction 30\u00b0 above resistance is much simpler than that with the horizontal. Calculate (a) the maximum air resistance. height, (b) the time taken by the ball to return to the same level, and (c) the 4.11 UNIFORM CIRCULAR MOTION distance from the thrower to the point where the ball returns to the same level. When an object follows a circular path at a constant speed, the motion of the object is called Answer (a) The maximum height is given by uniform circular motion. The word \u201cuniform\u201d refers to the speed, which is uniform (constant) hm = (v0 sin\u03b8o )2 = (28 sin 30\u00b0)2 m throughout the motion. Suppose an object is 2 (9.8) moving with uniform speed v in a circle of radius 2g R as shown in Fig. 4.19. Since the velocity of the object is changing continuously in direction, the 14 \u00d714 object undergoes acceleration. Let us find the = 2 \u00d7 9.8 = 10.0 m magnitude and the direction of this acceleration. (b) The time taken to return to the same level is Tf = (2 vo sin \u03b8o )\/g = (2\u00d7 28 \u00d7 sin 30\u00b0 )\/9.8 = 28\/9.8 s = 2.9 s (c) The distance from the thrower to the point where the ball returns to the same level is ( )R = vo2sin 2\u03b8o = 28 \u00d7 28 \u00d7 sin 60o = 69 m g 9.8 2020-21","80 PHYSICS Fig. 4.19 Velocity and acceleration of an object in uniform circular motion. The time interval \u2206t decreases from (a) to (c) where it is zero. The acceleration is directed, at each point of the path, towards the centre of the circle. Let r and r\u2032 be the position vectors and v and r\u2032 be \u2206\u03b8. Since the velocity vectors v and v\u2032 are v\u2032 the velocities of the object when it is at point P always perpendicular to the position vectors, the and P \u2032 as shown in Fig. 4.19(a). By definition, angle between them is also \u2206\u03b8 . Therefore, the velocity at a point is along the tangent at that triangle CPP\u2032 formed by the position vectors and point in the direction of motion. The velocity vectors v and v\u2032 are as shown in Fig. 4.19(a1). the triangle GHI formed by the velocity vectors \u2206v is obtained in Fig. 4.19 (a2) using the triangle v, v\u2032 and \u2206v are similar (Fig. 4.19a). Therefore, law of vector addition. Since the path is circular, the ratio of the base-length to side-length for v is perpendicular to r and so is v\u2032 to r\u2032. one of the triangles is equal to that of the other Therefore, \u2206v is perpendicular to \u2206r. Since triangle. That is : average acceleration is along \u2206v \uf8eb = \u2206v \uf8f6 , the \u2206v \u2206r \uf8ec\uf8eda \uf8f7 \u2206t \uf8f8 = vR average acceleration a is perpendicular to \u2206r. If \u2206r we place \u2206v on the line that bisects the angle between r and r\u2032, we see that it is directed towards Or, \u2206v = v the centre of the circle. Figure 4.19(b) shows the same quantities for smaller time interval. \u2206v and R hence a is again directed towards the centre. Therefore, In Fig. 4.19(c), \u2206t 0 and the average a = lim \u2206v v \u2206r v \u2206r = lim = lim acceleration becomes the instantaneous \u2206t \u2192 0 \u2206t \u2206t \u2192 0 R\u2206t R \u2206t \u2192 0 \u2206t acceleration. It is directed towards the centre*. If \u2206t is small, \u2206\u03b8 will also be small and then arc PP\u2032 can be approximately taken to be|\u2206r|: Thus, we find that the acceleration of an object in uniform circular motion is always directed \u2206r \u2245 v\u2206t towards the centre of the circle. Let us now find the magnitude of the acceleration. \u2206r \u2206t \u2245 v The magnitude of a is, by definition, given by Or, lim \u2206r \u2206v =v a = \u2206ltim\u2192 0 \u2206t \u2206t \u2192 0 \u2206t Let the angle between position vectors r and Therefore, the centripetal acceleration ac is : * In the limit \u2206t 0, \u2206r becomes perpendicular to r. In this limit \u2206v\u2192 0 and is consequently also perpendicular to V. Therefore, the acceleration is directed towards the centre, at each point of the circular path. 2020-21","MOTION IN A PLANE 81 \uf8ebv \uf8f6 (4.44) ac = v2 = \u03c92R2 = \u03c92R ac = \uf8ed\uf8ec R \uf8f7\uf8f8 v = v2\/R R R Thus, the acceleration of an object moving with ac = \u03c92R (4.47) speed v in a circle of radius R has a magnitude The time taken by an object to make one revolution v2\/R and is always directed towards the centre. is known as its time period T and the number of revolution made in one second is called its This is why this acceleration is called centripetal frequency \u03bd (=1\/T ). However, during this time the acceleration (a term proposed by Newton). A distance moved by the object is s = 2\u03c0R. thorough analysis of centripetal acceleration was first published in 1673 by the Dutch scientist Therefore, v = 2\u03c0R\/T =2\u03c0R\u03bd (4.48) Christiaan Huygens (1629-1695) but it was In terms of frequency \u03bd, we have probably known to Newton also some years earlier. \u201cCentripetal\u201d comes from a Greek term which means \u03c9 = 2\u03c0\u03bd \u2018centre-seeking\u2019. Since v and R are constant, the v = 2\u03c0R\u03bd magnitude of the centripetal acceleration is also constant. However, the direction changes \u2014 ac = 4\u03c02 \u03bd2R (4.49) pointing always towards the centre. Therefore, a centripetal acceleration is not a constant vector. Example 4.10 An insect trapped in a circular groove of radius 12 cm moves along We have another way of describing the the groove steadily and completes 7 velocity and the acceleration of an object in revolutions in 100 s. (a) What is the uniform circular motion. As the object moves angular speed, and the linear speed of the from P to P\u2032 in time \u2206t (= t\u2032 \u2013 t), the line CP motion? (b) Is the acceleration vector a (Fig. 4.19) turns through an angle \u2206\u03b8 as shown constant vector ? What is its magnitude ? in the figure. \u2206\u03b8 is called angular distance. We define the angular speed \u03c9 (Greek letter omega) Answer This is an example of uniform circular as the time rate of change of angular motion. Here R = 12 cm. The angular speed \u03c9 is displacement : given by \u2206\u03b8 (4.45) \u03c9 = 2\u03c0\/T = 2\u03c0 \u00d7 7\/100 = 0.44 rad\/s \u03c9= The linear speed v is : \u2206t v =\u03c9 R = 0.44 s-1 \u00d7 12 cm = 5.3 cm s-1 Now, if the distance travelled by the object during the time \u2206t is \u2206s, i.e. PP\u2032 is \u2206s, then : The direction of velocity v is along the tangent to the circle at every point. The acceleration is \u2206s directed towards the centre of the circle. Since v = \u2206t this direction changes continuously, acceleration here is not a constant vector. but \u2206s = R \u2206\u03b8. Therefore : However, the magnitude of acceleration is constant: \u2206\u03b8 (4.46) v=R =R\u03c9 a = \u03c92 R = (0.44 s\u20131)2 (12 cm) \u2206t = 2.3 cm s-2 v= R\u03c9 We can express centripetal acceleration ac in terms of angular speed : 2020-21","82 PHYSICS SUMMARY 1. Scalar quantities are quantities with magnitudes only. Examples are distance, speed, mass and temperature. 2. Vector quantities are quantities with magnitude and direction both. Examples are displacement, velocity and acceleration. They obey special rules of vector algebra. 3. A vector A multiplied by a real number \u03bb is also a vector, whose magnitude is \u03bb times the magnitude of the vector A and whose direction is the same or opposite depending upon whether \u03bb is positive or negative. 4. Two vectors A and B may be added graphically using head-to-tail method or parallelogram method. 5. Vector addition is commutative : A+B=B+A It also obeys the associative law : (A + B) + C = A + (B + C) 6. A null or zero vector is a vector with zero magnitude. Since the magnitude is zero, we don\u2019t have to specify its direction. It has the properties : A+0=A \u03bb0 = 0 0A=0 7. The subtraction of vector B from A is defined as the sum of A and \u2013B : A \u2013 B = A+ (\u2013B) 8. A vector A can be resolved into component along two given vectors a and b lying in the same plane : A = \u03bba + \u00b5b where \u03bb and \u00b5 are real numbers. 9. A unit vector associated with a vector A has magnitude 1 and is along the vector A: n\u02c6 = A A The unit vectors i, j, k are vectors of unit magnitude and point in the direction of the x-, y-, and z-axes, respectively in a right-handed coordinate system. 10. A vector A can be expressed as A = Ax i + Ay j where Ax, Ay are its components along x-, and y -axes. If vector A makes an angle \u03b8 with the x-axis, then Ax = A cos \u03b8, Ay=A sin \u03b8 and A = A = A 2 + Ay2 , tan\u03b8 = Ay . x Ax 11. Vectors can be conveniently added using analytical method. If sum of two vectors A and B, that lie in x-y plane, is R, then : R = Rx i + Ry j, where, Rx = Ax + Bx, and Ry = Ay + By 12. The position vector of an object in x-y plane is given by r = x i + y j and the displacement from position r to position r\u2019 is given by \u2206r = r\u2032\u2212 r = (x \u2032 \u2212 x ) i + (y\u2032 \u2212 y) j = \u2206x i + \u2206y j 13. If an object undergoes a displacement \u2206r in time \u2206t, its average velocity is given by \u2206r v = . The velocity of an object at time t is the limiting value of the average velocity \u2206t 2020-21","MOTION IN A PLANE 83 as \u2206t tends to zero : \u2206r dr lim =. v = \u2206t \u2192 0 \u2206t dt It can be written in unit vector notation as : v = vxi + vy j + vz k where vx = dx ,vy = dy , vz = dz dt dt dt When position of an object is plotted on a coordinate system, v is always tangent to the curve representing the path of the object. 14. If the velocity of an object changes from v to v\u2032in time \u2206t, then its average acceleration is given by: a = v \u2212 v' \u2206v = \u2206t \u2206t The acceleration a at any time t is the limiting value of a as \u2206t 0 : lim \u2206v dv a= = \u2206t \u2192 0 \u2206t dt In component form, we have : a = a x i + a y j + a zk where, ax = dvx , ay = dvy , az = dvz dt dt dt 15. If an object is moving in a plane with constant acceleration a = a = a 2 + a 2 and x y its position vector at time t = 0 is ro, then at any other time t, it will be at a point given by: r = ro + vot + 1 at 2 2 and its velocity is given by : v = vo + a t where vo is the velocity at time t = 0 In component form : x = xo + vox t + 1 a t 2 2 x y = yo + voy t + 1 ay t 2 2 vx = vox + a xt vy = voy + a yt Motion in a plane can be treated as superposition of two separate simultaneous one- dimensional motions along two perpendicular directions 16. An object that is in flight after being projected is called a projectile. If an object is projected with initial velocity vo making an angle \u03b8o with x-axis and if we assume its initial position to coincide with the origin of the coordinate system, then the position and velocity of the projectile at time t are given by : x = (vo cos \u03b8o) t y = (vo sin \u03b8o) t \u2212 (1\/2) g t2 vx = vox = vo cos \u03b8o vy = vo sin \u03b8o \u2212 g t The path of a projectile is parabolic and is given by : y = (tan\u03b80 ) x \u2013 gx 2 )2 2 (vo cos\u03b8o The maximum height that a projectile attains is : 2020-21","84 PHYSICS hm = (vo sinqo )2 2g The time taken to reach this height is : tm = vo sin\u03b8o g The horizontal distance travelled by a projectile from its initial position to the position it passes y = 0 during its fall is called the range, R of the projectile. It is : R = vo2 sin 2\u03b8o g 17. When an object follows a circular path at constant speed, the motion of the object is called uniform circular motion. The magnitude of its acceleration is ac = v2 \/R. The direction of ac is always towards the centre of the circle. The angular speed \u03c9, is the rate of change of angular distance. It is related to velocity v by v = \u03c9 R. The acceleration is ac = \u03c9 2R. If T is the time period of revolution of the object in circular motion and \u03bd is its frequency, we have \u03c9 = 2\u03c0 \u03bd, v = 2\u03c0\u03bdR, ac = 4\u03c02\u03bd2R 2020-21","MOTION IN A PLANE 85 POINTS TO PONDER 1. The path length traversed by an object between two points is, in general, not the same as the magnitude of displacement. The displacement depends only on the end points; the path length (as the name implies) depends on the actual path. The two quantities are equal only if the object does not change its direction during the course of motion. In all other cases, the path length is greater than the magnitude of displacement. 2. In view of point 1 above, the average speed of an object is greater than or equal to the magnitude of the average velocity over a given time interval. The two are equal only if the path length is equal to the magnitude of displacement. 3. The vector equations (4.33a) and (4.34a) do not involve any choice of axes. Of course, you can always resolve them along any two independent axes. 4. The kinematic equations for uniform acceleration do not apply to the case of uniform circular motion since in this case the magnitude of acceleration is constant but its direction is changing. 5. An object subjected to two velocities v1 and v2 has a resultant velocity v = v1 + v2. Take care to distinguish it from velocity of object 1 relative to velocity of object 2 : v12= v1 \u2212 v2. Here v1 and v2 are velocities with reference to some common reference frame. 6. The resultant acceleration of an object in circular motion is towards the centre only if the speed is constant. 7. The shape of the trajectory of the motion of an object is not determined by the acceleration alone but also depends on the initial conditions of motion ( initial position and initial velocity). For example, the trajectory of an object moving under the same acceleration due to gravity can be a straight line or a parabola depending on the initial conditions. EXERCISES 4.1 State, for each of the following physical quantities, if it is a scalar or a vector : volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity. 4.2 Pick out the two scalar quantities in the following list : force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity. 4.3 Pick out the only vector quantity in the following list : Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge. 4.4 State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful : (a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions , (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector. 4.5 Read each statement below carefully and state with reasons, if it is true or false : (a) The magnitude of a vector is always a scalar, (b) each component of a vector is always a scalar, (c) the total path length is always equal to the magnitude of the displacement vector of a particle. (d) the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of average velocity of the particle over the same interval of time, (e) Three vectors not lying in a plane can never add up to give a null vector. 4.6 Establish the following vector inequalities geometrically or otherwise : (a) |a+b| < |a| + |b| (b) |a+b| > ||a| \u2212|b|| 2020-21","86 PHYSICS (c) |a\u2212b| < |a| + |b| Q Fig. 4.20 (d) |a\u2212b| > ||a| \u2212 |b|| When does the equality sign above apply? 4.7 Given a + b + c + d = 0, which of the following statements are correct : (a) a, b, c, and d must each be a null vector, (b) The magnitude of (a + c) equals the magnitude of ( b + d), (c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d, (d) b + c must lie in the plane of a and d if a and d are not collinear, and in the line of a and d, if they are collinear ? 4.8 Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 4.20. What is the magnitude of the displacement vector for each ? For which girl is this equal to the actual length of path skate ? 4.9 A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 4.21. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist ? Fig. 4.21 4.10 On an open ground, a motorist follows a track that turns to his left by an angle of 600 after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case. 4.11 A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity ? Are the two equal ? 4.12 Rain is falling vertically with a speed of 30 m s-1. A woman rides a bicycle with a speed of 10 m s-1 in the north to south direction. What is the direction in which she should hold her umbrella ? 4.13 A man can swim with a speed of 4.0 km\/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km\/h and he makes his 2020-21"]
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