Motion in a Straight Line PDF

CHAPTER THREE MOTION IN A STRAIGHT LINE 3.1 INTRODUCTION Motion is common to everything in the universe. We walk, run and ride a bicycle. Even when we are sleep...

CHAPTER THREE MOTION IN A STRAIGHT LINE 3.1 INTRODUCTION Motion is common to everything in the universe. We walk, run and ride a bicycle. Even when we are sleeping, air moves 3.1 Introduction into and out of our lungs and blood flows in arteries and 3.2 Position, path length and veins. We see leaves falling from trees and water flowing displacement down a dam. Automobiles and planes carry people from one 3.3 Average velocity and average place to the other. The earth rotates once every twenty-four speed 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 speed within its local group of galaxies. 3.5 Acceleration Motion is change in position of an object with time. How 3.6 Kinematic equations for does the position change with time ? In this chapter, we shall uniformly accelerated motion learn how to describe motion. For this, we develop the 3.7 Relative velocity concepts of velocity and acceleration. We shall confine Summary ourselves to the study of motion of objects along a straight Points to ponder line, also known as rectilinear motion. For the case of Exercises rectilinear motion with uniform acceleration, a set of simple Additional exercises equations can be obtained. Finally, to understand the relative Appendix 3.1 nature of motion, we introduce the concept of relative velocity. In our discussions, we shall treat the objects in motion as point objects. This approximation is valid so far as the size 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 40 PHYSICS a rectangular coordinate system consisting of with the path of the car’s 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 O to P. Then the distance moved by the car is respect to this coordinate system. To measure OP = +360 m. This distance is called the path time, we position a clock in this system. This length traversed by the car. In the second coordinate system along with a clock constitutes case, the car moves from O to P and then moves a frame of reference. 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 — 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 reference depends upon the situation. For Displacement example, for describing motion in one dimension, It is useful to define another quantity we need only one axis. To describe motion in displacement as the change in position. Let two/three dimensions, we need a set of two/ x1 and x2 be the positions of an object at time t1 three axes. and t2. Then its displacement, denoted by ∆x, in Description of an event depends on the frame time ∆t = (t2 - t1), is given by the difference of reference chosen for the description. For between the final and initial positions : example, when you say that a car is moving on ∆x = x2 – x1 a road, you are describing the car with respect (We use the Greek letter delta (∆) to denote a to a frame of reference attached to you or to the change in a quantity.) ground. But with respect to a frame of reference If x2 > x1, ∆x is positive; and if x2 < x1, ∆x is attached with a person sitting in the car, the negative. car is at rest. Displacement has both magnitude and direction. Such quantities are represented by To describe motion along a straight line, we vectors. You will read about vectors in the next can choose an axis, say X-axis, so that it chapter. Presently, we are dealing with motion coincides with the path of the object. We then along a straight line (also called rectilinear measure the position of the object with reference motion) only. In one-dimensional motion, there to a conveniently chosen origin, say O, as shown are only two directions (backward and forward, in Fig. 3.1. Positions to the right of O are taken upward and downward) in which an object can as positive and to the left of O, as negative. move, and these two directions can easily be Following this convention, the position specified by + and – signs. For example, coordinates of point P and Q in Fig. 3.1 are +360 displacement of the car in moving from O to P is : m and +240 m. Similarly, the position coordinate ∆x = x2 – x1 = (+360 m) – 0 m = +360 m of point R is –120 m. The displacement has a magnitude of 360 m and Path length is directed in the positive x direction as indicated Consider the motion of a car along a straight by the + sign. Similarly, the displacement of the line. We choose the x-axis such that it coincides car from P to Q is 240 m – 360 m = – 120 m. The Fig. 3.1 x-axis, origin and positions of a car at different times. 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 of the car from O to P and back to Q. In this time and we have an x-t graph. Let us first case, the path length = (+360 m) + (+120 m) = + consider the simple case in which an object is 480 m. However, the displacement = (+240 m) – stationary, e.g. a car standing still at x = 40 m. (0 m) = + 240 m. Thus, the magnitude of The position-time graph is a straight line parallel displacement (240 m) is not equal to the path to the time axis, as shown in Fig. 3.2(a). length (480 m). If an object moving along the straight line The magnitude of the displacement for a covers equal distances in equal intervals of course of motion may be zero but the time, it is said to be in uniform motion along a corresponding path length is not zero. For straight line. Fig. 3.2(b) shows the position-time example, if the car starts from O, goes to P and graph of such a motion. Fig. 3.2 Position-time graph of (a) stationary object, and (b) an object in uniform motion. # x (m) t (s) ž Fig. 3.3 Position-time graph of a car. 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 portion of the x-t graph between t = 0 s and t = 8 and picks up speed till t = 10 s and thereafter s is blown up and shown in Fig. 3.4. As seen moves with uniform speed till t = 18 s. Then the from the plot, the average velocity of the car brakes are applied and the car stops at between time t = 5 s and t = 7 s is : t = 20 s and x = 296 m. The position-time graph x 2 − x1 ( 27.4 − 10.0 ) m for this case is shown in Fig. 3.3. We shall refer v= = = 8.7 m s –1 t 2 − t1 (7 − 5 ) s to this graph in our discussion in the following sections. Geometrically, this is the slope of the straight line P1P2 connecting the initial position P1 to 3.3 AVERAGE VELOCITY AND AVERAGE the final position P2 as shown in Fig. 3.4. SPEED The average velocity can be positive or negative When an object is in motion, its position depending upon the sign of the displacement. It changes with time. But how fast is the position is zero if the displacement is zero. Fig. 3.5 shows changing with time and in what direction? To the x-t graphs for an object, moving with positive describe this, we define the quantity average velocity (Fig. 3.5a), moving with negative velocity velocity. Average velocity is defined as the (Fig. 3.5b) and at rest (Fig. 3.5c). change in position or displacement (∆x) divided by the time intervals (∆t), in which the displacement occurs : x 2 − x1 ∆x v= = (3.1) t 2 − t1 ∆t where x2 and x1 are the positions of the object at time t2and t1, respectively. Here the bar over Fig. 3.5 Position-time graph for an object (a) moving the symbol for velocity is a standard notation with positive velocity, (b) moving with negative velocity, and (c) at rest. used to indicate an average quantity. The SI unit for velocity is m/s or m s–1, although km h–1 Average velocity as defined above involves is used in many everyday applications. only the displacement of the object. We have seen Like displacement, average velocity is also a earlier that the magnitude of displacement may vector quantity. But as explained earlier, for be different from the actual path length. To motion in a straight line, the directional aspect describe the rate of motion over the actual path, of the vector can be taken care of by + and – we introduce another quantity called average signs and we do not have to use the vector speed. 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 : Total path length Average speed = (3.2) Total time interval Average speed has obviously the same unit (m s–1) 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. Fig. 3.4 The average velocity is the slope of line P1P2. In that case, the magnitude of average velocity 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 t has been moving over a given time interval but Example 3.1 A car is moving along a does not tell us how fast it moves at different straight line, say OP in Fig. 3.1. It moves instants of time during that interval. For this, from O to P in 18 s and returns from P to Q we define instantaneous velocity or simply in 6.0 s. What are the average velocity velocity v at an instant t. and average speed of the car in going (a) The velocity at an instant is defined as the from O to P ? and (b) from O to P and back limit of the average velocity as the time interval to Q ? ∆t becomes infinitesimally small. In other words, Answer (a) ∆x v = lim (3.3a) Displacement ∆t → 0 ∆t Average velocity = Time interval dx (3.3b) = dt + 360 m v= = + 20 m s −1 lim where the symbol ∆t →0 stands for the operation 18 s of taking limit as ∆tg0 of the quantity on its Path length right. In the language of calculus, the quantity Average speed = Time interval on the right hand side of Eq. (3.3a) is the differential coefficient of x with respect to t and 360 m dx = = 20 m s−1 is denoted by (see Appendix 3.1). It is the 18 s dt rate of change of position with respect to time, Thus, in this case the average speed is equal to the magnitude of the average velocity. at that instant. (b) In this case, We can use Eq. (3.3a) for obtaining the value of velocity at an instant either graphically or Displacement +240 m Average velocity = = numerically. Suppose that we want to obtain Time interval (18 + 6.0) s graphically the value of velocity at time t = 4 s (point P) for the motion of the car represented =+10 m s-1 in Fig. 3.3. The figure has been redrawn in Fig. 3.6 choosing different scales to facilitate the Path length OP + PQ Average speed = = Time interval ∆t = (360 +120 ) m = 20 m s-1 24 s Thus, in this case the average speed is not equal to the magnitude of the average velocity. This happens because the motion here involves 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. t If the car in Example 3.1 moves from O to P and comes back to O in the same time interval, Fig. 3.6 Determining velocity from position-time average speed is 20 m/s but the average velocity graph. Velocity at t = 4 s is the slope of the is zero ! tangent to the graph at that instant. 44 PHYSICS calculation. Let us take ∆t = 2 s centred at instant for motion of the car shown in Fig. 3.3. t = 4 s. Then, by the definition of the average For this case, the variation of velocity with time velocity, the slope of line P1P2 ( Fig. 3.6) gives is found to be as shown in Fig. 3.7. the value of average velocity over the interval 3 s to 5 s. Now, we decrease the value of ∆t from 2 s to 1 s. Then line P1P2 becomes Q1Q2 and its slope gives the value of the average velocity over the interval 3.5 s to 4.5 s. In the limit ∆t → 0, the line P1P2 becomes tangent to the position- time curve at the point P and the velocity at t = 4 s is given by the slope of the tangent at that point. It is difficult to show this process graphically. But if we use numerical method to obtain the value of the velocity, the meaning of the limiting process becomes clear. For the graph shown in Fig. 3.6, x = 0.08 t3. Table 3.1 gives the value of ∆x/∆t calculated for ∆t 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= Fig. 3.7 Velocity–time graph corresponding to motion shown in Fig. 3.3.  ∆t   ∆t  The graphical method for the determination t −  and t 2 =  t +  and the fourth and  2   2  of the instantaneous velocity is always not a the fifth columns give the corresponding values convenient method. For this, we must carefully plot the position–time graph and calculate the 3 of x, i.e. x (t1) = 0.08 t1 and x (t2) = 0.08 t 23. The value of average velocity as ∆t becomes smaller sixth column lists the difference ∆x = x (t2) – x and smaller. It is easier to calculate the value (t1) and the last column gives the ratio of ∆x and of velocity at different instants if we have data ∆t, i.e. the average velocity corresponding to the of positions at different instants or exact value of ∆t listed in the first column. expression for the position as a function of time. We see from Table 3.1 that as we decrease Then, we calculate ∆x/∆t from the data for the value of ∆t from 2.0 s to 0.010 s, the value of decreasing the value of ∆t and find the limiting the average velocity approaches the limiting value as we have done in Table 3.1 or use value 3.84 m s–1 which is the value of velocity at differential calculus for the given expression and dx dx t = 4.0 s, i.e. the value of at t = 4.0 s. In this calculate at different instants as done in dt dt manner, we can calculate velocity at each the following example. ∆x Table 3.1 Limiting value of at t = 4 s ∆t MOTION IN A STRAIGHT LINE 45 t This led to the concept of acceleration as the rate Example 3.2 The position of an object of change of velocity with time. moving along x-axis is given by x = a + bt2 where a = 8.5 m, b = 2.5 m s–2 and t is The average acceleration a over a time measured in seconds. What is its velocity at interval is defined as the change of velocity t = 0 s and t = 2.0 s. What is the average divided by the time interval : velocity between t = 2.0 s and t = 4.0 s ? a = v 2 – v1 = ∆v t 2 – t1 ∆t (3.4) Answer In notation of differential calculus, the velocity is where v2 and v1 are the instantaneous velocities dx d or simply velocities at time t2 and t1. It is the v= dt = dt ( a + bt ) = 2b t = 5.0 t m s 2 -1 average change of velocity per unit time. The SI unit of acceleration is m s–2. At t = 0 s, v = 0 m s–1 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 ) − x (2.0 ) connecting the points corresponding to (v2, t2) Average velocity = 4.0 − 2.0 and (v 1 , t 1 ). The average acceleration a + 16b – a – 4b for velocity-time graph shown in Fig. 3.7 for = = 6.0 × b 2.0 different time intervals 0 s - 10 s, 10 s – 18 s, = 6.0 × 2.5 = 15 m s-1 t and 18 s – 20 s are : From Fig. 3.7, we note that during the period 24 – 0) m s –1 t =10 s to 18 s the velocity is constant. Between 0 s - 10 s a=( = 2.4 m s –2 period t =18 s to t = 20 s, it is uniformly (10 – 0) s decreasing and during the period t = 0 s to t 24 – 24) m s –1 = 10 s, it is increasing. Note that for uniform 10 s - 18 s a=( = 0 m s –2 motion, velocity is the same as the average (18 – 10) s velocity at all instants. 0 – 24 ) m s –1 Instantaneous speed or simply speed is the 18 s - 20 s a=( = – 12 m s –2 magnitude of velocity. For example, a velocity of (20 – 18) s + 24.0 m s–1 and a velocity of – 24.0 m s–1 — 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, a (m s–2) 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 during its course of motion. How to describe this change? Should it be described as the rate of Fig. 3.8 Acceleration as a function of time for motion change in velocity with distance or with time ? represented in Fig. 3.3. This was a problem even in Galileo’s time. It was Instantaneous acceleration is defined in the same first thought that this change could be described way as the instantaneous velocity : by the rate of change of velocity with distance. But, through his studies of motion of freely falling ∆ v dv a = lim = (3.5) objects and motion of objects on an inclined ∆t dt ∆t → 0 plane, Galileo concluded that the rate of change The acceleration at an instant is the slope of of velocity with time is a constant of motion for the tangent to the v–t curve at that instant. For all objects in free fall. On the other hand, the the v–t curve shown in Fig. 3.7, we can obtain change in velocity with distance is not constant acceleration at every instant of time. The – it decreases with the increasing distance of fall. resulting a – t curve is shown in Fig. 3.8. We see 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 –12 m s –2 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 (c) An object is moving in negative direction acceleration over that period. 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 (d) An object is moving in positive direction change in speed (magnitude), a change in till time t1, and then turns back with the direction or changes in both. Like velocity, same negative acceleration, for example acceleration can also be positive, negative or the motion of a car from point O to point zero. Position-time graphs for motion with Q in Fig. 3.1 till time t1 with decreasing positive, negative and zero acceleration are speed and turning back and moving with shown in Figs. 3.9 (a), (b) and (c), respectively. the same negative acceleration. Note that the graph curves upward for positive An interesting feature of a velocity-time graph acceleration; downward for negative for any moving object is that the area under the acceleration and it is a straight line for zero curve represents the displacement over a acceleration. As an exercise, identify in Fig. 3.3, given time interval. A general proof of this the regions of the curve that correspond to these three cases. 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 vo at t = 0 and v at time t, we have v − v0 a= or, v = v0 + a t (3.6) t −0 Fig. 3.9 Position-time graph for motion with (a) positive acceleration; (b) negative Fig. 3.10 Velocity–time graph for motions with acceleration, and (c) zero acceleration. constant acceleration. (a) Motion in positive Let us see how velocity-time graph looks like direction with positive acceleration, for some simple cases. Fig. 3.10 shows velocity- (b) Motion in positive direction with negative acceleration, (c) Motion in negative time graph for motion with constant acceleration direction with negative acceleration, for the following cases : (d) Motion of an object with negative (a) An object is moving in a positive direction acceleration that changes direction at time with a positive acceleration, for example t1. Between times 0 to t1, its moves in the motion of the car in Fig. 3.3 between positive x - direction and between t1 and t = 0 s and t = 10 s. t2 it moves in the opposite direction. MOTION IN A STRAIGHT LINE 47 statement requires use of calculus. We can, 1 however, see that it is true for the simple case of = 2 (v –v0 ) t + v0t an object moving with constant velocity u. Its velocity-time graph is as shown in Fig. 3.11. Fig. 3.11 Area under v–t curve equals displacement of the object over a given time interval. The v-t curve is a straight line parallel to the time axis and the area under it between t = 0 and t = T is the area of the rectangle of height u Fig. 3.12 Area under v-t curve for an object with and base T. Therefore, area = u × T = uT which uniform acceleration. is the displacement in this time interval. How As explained in the previous section, the area come in this case an area is equal to a distance? under v-t curve represents the displacement. Think! Note the dimensions of quantities on Therefore, the displacement x of the object is : the two coordinate axes, and you will arrive at the answer. 1 x = 2 (v –v 0 ) t + v 0 t (3.7) Note that the x-t, v-t, and a-t graphs shown in several figures in this chapter have sharp But v − v0 = a t kinks at some points implying that the 1 functions are not differentiable at these Therefore, x = a t 2 + v 0t 2 points. In any realistic situation, the 1 functions will be differentiable at all points or, x = v0t + at 2 (3.8) and the graphs will be smooth. 2 Equation (3.7) can also be written as What this means physically is that v + v0 acceleration and velocity cannot change x= t = vt (3.9a) values abruptly at an instant. Changes are 2 always continuous. where, 3.6 KINEMATIC EQU ATIONS FOR UNIFORMLY ACCELERATED MOTION v + v0 v= (constant acceleration only) 2 For uniformly accelerated motion, we can derive (3.9b) some simple equations that relate displacement (x), time taken (t), initial velocity (v 0), final Equations (3.9a) and (3.9b) mean that the object velocity (v) and acceleration (a). Equation (3.6) has undergone displacement x with an average already obtained gives a relation between final velocity equal to the arithmetic average of the and initial velocities v and v0 of an object moving initial and final velocities. with uniform acceleration a : From Eq. (3.6), t = (v – v0)/a. Substituting this in Eq. (3.9a), we get v = v0 + at (3.6) This relation is graphically represented in Fig. 3.12. v + v0 v − v0 v 2 − v 02 x =vt = = The area under this curve is : 2 a 2a Area between instants 0 and t = Area of triangle ABC + Area of rectangle OACD v 2 = v02 + 2ax (3.10) 48 PHYSICS This equation can also be obtained by t substituting the value of t from Eq. (3.6) into = 0 (v 0 + at ) dt Eq. (3.8). Thus, we have obtained three important equations : 1 x – x 0 = v0 t + a t2 v = v0 + at 2 1 1 x = x 0 + v0 t + a t2 x = v 0t + at 2 2 2 We can write v 2 = v02 + 2ax (3.11a) dv dv dx dv a= = =v 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 v x v dv = 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 – v02 = a (x – x 0 ) = 0 as non-zero, say x0. Then Eqs. (3.11a) are 2 modified (replacing x by x – x0 ) to : v 2 = v02 + 2a ( x – x 0 ) v = v0 + at The advantage of this method is that it can be 1 2 used for motion with non-uniform acceleration x = x 0 + v 0t + at (3.11b) also. 2 Now, we shall use these equations to some v 2 = v 02 + 2a ( x − x 0 ) (3.11c) important cases. t t Example 3.4 A ball is thrown vertically Example 3.3 Obtain equations of motion t upwards with a velocity of 20 m s–1 from for constant acceleration using method of the top of a multistorey building. The calculus. height of the point from where the ball is thrown is 25.0 m from the ground. (a) How Answer By definition high will the ball rise ? and (b) how long will it be before the ball hits the ground? dv Take g = 10 m s–2. a = dt dv = a dt Answer (a) Let us take the y-axis in the Integrating both sides v t vertically upward direction with zero at the dv = a dt ground, as shown in Fig. 3.13. v0 0 Now vo = + 20 m s–1, t =a dt (a is constant) a = – g = –10 m s–2, 0 v = 0 m s–1 v – v 0 = at If the ball rises to height y from the point of v = v 0 + at launch, then using the equation ( v 2 = v02 + 2 a y – y 0 ) dx we get Further, v= dt 0 = (20)2 + 2(–10)(y – y0) dx = v dt Solving, we get, (y – y0) = 20 m. Integrating both sides x t (b) We can solve this part of the problem in two dx = v dt x0 0 ways. Note carefully the methods used. MOTION IN A STRAIGHT LINE 49 0 = 25 +20 t + (½) (-10) t2 Or, 5t2 – 20t – 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. t t Example 3.5 Free-fall : Discuss the motion of an object under free fall. Neglect air resistance. Answer An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by g. If air resistance is neglected, the object is Fig. 3.13 said to be in free fall. If the height through which the object falls is small compared to the FIRST METHOD : In the first method, we split earth’s radius, g can be taken to be constant, the path in two parts : the upward motion (A to equal to 9.8 m s–2. Free fall is thus a case of B) and the downward motion (B to C) and motion with uniform acceleration. calculate the corresponding time taken t1 and We assume that the motion is in y-direction, t2. Since the velocity at B is zero, we have : more correctly in –y-direction because we v = vo + at choose upward direction as positive. Since the 0 = 20 – 10t1 acceleration due to gravity is always downward, Or, t1 = 2 s it is in the negative direction and we have This is the time in going from A to B. From B, or a = – g = – 9.8 m s–2 the point of the maximum height, the ball falls The object is released from rest at y = 0. Therefore, freely under the acceleration due to gravity. The v0 = 0 and the equations of motion become: ball is moving in negative y direction. We use equation v= 0–gt = –9.8 t m s–1 y = 0 – ½ g t2 = –4.9 t 2 m 1 2 y = y0 + v 0t + at v2 = 0 – 2 g y = –19.6 y m2 s–2 2 These equations give the velocity and the We have, y0 = 45 m, y = 0, v0 = 0, a = – g = –10 m s–2 distance travelled as a function of time and also 0 = 45 + (½) (–10) t22 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 1 2 y = y0 + v 0t + at 2 Now y0 = 25 m y=0m vo = 20 m s-1, a = –10m s–2, t = ? (a) 50 PHYSICS traversed during successive intervals of time. Since initial velocity is zero, we have 1 2 y=− gt 2 Using this equation, we can calculate the position of the object after different time intervals, 0, τ , 2 τ , 3 τ … which are given in second column of Table 3.2. If we take (–1/ 2) g τ 2 as y0 — the position coordinate after first time interval , then third column gives (b) the positions in the unit of yo. The fourth column gives the distances traversed in successive s. We find that the distances are in the simple ratio 1: 3: 5: 7: 9: 11… 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. t t 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 an important factor for road safety and (c) depends on the initial velocity (v0) and the braking capacity, or deceleration, –a that Fig. 3.14 Motion of an object under free fall. (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 t in terms of vo and a. t Example 3.6 Galileo’s law of odd Answer Let the distance travelled by the vehicle numbers : “The 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…...].” Prove it. 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 τ and find out the distances the square of the initial velocity. Doubling the Table 3.2 MOTION IN A STRAIGHT LINE 51 initial velocity increases the stopping distance by a factor of 4 (for the same deceleration). 1 2 d=− gtr For the car of a particular make, the braking 2 distance was found to be 10 m, 20 m, 34 m and 2d 50 m corresponding to velocities of 11, 15, 20 Or, tr = s and 25 m/s which are nearly consistent with g the above formula. Given d = 21.0 cm and g = 9.8 m s–2the reaction time is Stopping distance is an important factor 2×0.21 considered in setting speed limits, for example, tr = s ≅ 0.2 s. t 9.8 in school zones. t t 3.7 RELATIVE VELOCITY Example 3.8 Reaction time : When a situation demands our immediate You must be familiar with the experience of action, it takes some time before we travelling in a train and being overtaken by really respond. Reaction time is the another train moving in the same direction as time a person takes to observe, think you are. While that train must be travelling faster and act. For example, if a person is than you to be able to pass you, it does seem driving and suddenly a boy appears on slower to you than it would be to someone the road, then the time elapsed before standing on the ground and watching both the he slams the brakes of the car is the trains. In case both the trains have the same reaction time. Reaction time depends velocity with respect to the ground, then to you on complexity of the situation and on the other train would seem to be not moving at an individual. all. To understand such observations, we now You can measure your reaction time introduce the concept of relative velocity. by a simple experiment. Take a ruler Consider two objects A and B moving and ask your friend to drop it vertically uniformly with average velocities vA and vB in through the gap between your thumb one dimension, say along x-axis. (Unless and forefinger (Fig. 3.15). After you otherwise specified, the velocities mentioned in catch it, find the distance d travelled this chapter are measured with reference to the by the ruler. In a particular case, d was ground). If xA (0) and xB (0) are positions of objects found to be 21.0 cm. Estimate reaction A and B, respectively at time t = 0, their positions time. xA (t) and xB (t) at time t are given by: xA (t ) = xA (0) + vA t (3.12a) xB (t) = xB (0) + vB t (3.12b) Then, the displacement from object A to object B is given by xBA(t) = xB (t) – xA (t) = [ xB (0) – xA (0) ] + (vB – vA) t. (3.13) Equation (3.13) is easily interpreted. It tells us that as seen from object A, object B has a velocity vB – vA because the displacement from A to B changes steadily by the amount vB – vA in each unit of time. We say that the velocity of Fig. 3.15 Measuring the reaction time. object B relative to object A is vB – vA : vBA = vB – vA (3.14a) Answer The ruler drops under free fall. Therefore, vo = 0, and a = – g = –9.8 m s–2. The Similarly, velocity of object A relative to object B distance travelled d and the reaction time tr are is: related by vAB = vA – vB (3.14b) 52 PHYSICS Fig. 3.16 Position-time graphs of two objects with equal velocities. This shows: vBA = – vAB (3.14c) Fig. 3.17 Position-time graphs of two objects with unequal velocities, showing the time of meeting. Now we consider some special cases : (a) If vB = vA, vB – vA = 0. Then, from Eq. (3.13), xB (t) – xA (t) = xB (0) – xA (0). Therefore, the two objects stay at a constant distance (xB (0) – xA (0)) apart, and their position–time graphs are straight lines parallel to each other as shown in Fig. 3.16. The relative velocity vAB or vBA is zero in this case. (b) If vA > vB, vB – vA is negative. One graph is steeper than the other and they meet at a common point. For example, suppose vA = 20 m s-1 t(s) and xA (0) = 10 m; and vB = 10 m s-1, xB (0) = 40 m; then the time at which they meet is t = 3 s (Fig. 3.17). At this instant they are both at a Fig. 3.18 Position-time graphs of two objects with position xA (t) = xB (t) = 70 m. Thus, object A velocities in opposite directions, showing the time of meeting. overtakes object B at this time. In this case,vBA = 10 m s–1 – 20 m s–1 = – 10 m s–1 = – vAB. t Example 3.9 Two parallel rail tracks run (c) Suppose vA and vB are of opposite signs. For north-south. Train A moves north with a example, if in the above example object A is speed of 54 km h–1, and train B moves south moving with 20 m s–1 starting at xA(0) = 10 m with a speed of 90 km h–1. What is the and object B is moving with – 10 m s–1 starting (a) velocity of B with respect to A ?, at xB (0) = 40 m, the two objects meet at t = 1 s (b) velocity of ground with respect to B ?, (Fig. 3.18). The velocity of B relative to A, and vBA = [–10 – (20)] m s–1 = –30 m s–1 = – vAB. In this (c) velocity of a monkey running on the case, the magnitude of vBA or vAB ( = 30 m s–1) is roof of the train A against its motion greater than the magnitude of velocity of A or (with a velocity of 18 km h–1 with that of B. If the objects under consideration are respect to the train A) as observed by a man standing on the ground ? 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 Answer Choose the positive direction of x-axis represent instantaneous velocities. to be from south to north. Then, MOTION IN A STRAIGHT LINE 53 vA = + 54 km h–1 = 15 m s–1 B = 0 – vB = 25 m s–1. vB = – 90 km h–1 = – 25 m s–1 In (c), let the velocity of the monkey with respect Relative velocity of B with respect to A = vB – vA= to ground be vM. Relative velocity of the monkey – 40 m s–1 , i.e. the train B appears to A to move with respect to A, with a speed of 40 m s–1 from north to south. vMA = vM – vA = –18 km h–1 = –5 ms–1. Therefore, Relative velocity of ground with respect to vM = (15 – 5) m s–1 = 10 m s–1. t 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 : ∆x = x2 – 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 : ∆x v= ∆t 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 ∆t becomes infinitesimally small : ∆x dx v = lim v = lim = ∆t → 0 ∆t → 0 ∆t dt 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 : ∆v a= ∆t 9. Instantaneous acceleration is defined as the limit of the average acceleration as the time interval ∆t goes to zero : ∆ v dv a = lim a = lim = ∆t → 0 ∆t → 0 ∆t dt 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 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 1 2 x = v0 t + at 2 2 2 v = v0 + 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 – x0). 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’s 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). 56 PHYSICS Fig. 3.19 3.3 A woman starts from her home at 9.00 am, walks with a speed of 5 km h–1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h–1. Choose suitable scales and plot the x-t graph of her motion. 3.4 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, 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 and otherwise how long the drunkard takes to fall in a pit 13 m away from the start. 3.5 A jet airplane travelling at the speed of 500 km h–1 ejects its products of combustion at the speed of 1500 km h–1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ? 3.6 A car moving along a straight highway with speed of 126 km h–1 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 ? 3.7 Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h–1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s–2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them ? 3.8 On a two-lane road, car A is travelling with a speed of 36 km h–1. Two cars B and C approach car A in opposite directions with a speed of 54 km h–1 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 ? 3.9 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–1 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? 3.10 A player throws a ball upwards with an initial speed of 29.4 m s–1. (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’s hands ? (Take g = 9.8 m s–2 and neglect air resistance). MOTION IN A STRAIGHT LINE 57 3.11 Read each statement below carefully and state with reasons and examples, if it is true or false ; A particle in one-dimensional motion (a) with zero speed at an instant may have non-zero acceleration at that instant (b) with zero speed may have non-zero velocity, (c) with constant speed must have zero acceleration, (d) with positive value of acceleration must be speeding up. 3.12 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. 3.13 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]. 3.14 A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h–1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1. 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 !] 3.15 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 ? 3.16 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 58 PHYSICS 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 30 km h–1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h–1. If the muzzle speed of the bullet is 150 m s–1, with what speed does the bullet hit the thief’s car ? (Note: Fig. 3.21 Obtain that speed which is relevant for damaging the thief’s car). 3.19 Suggest a suitable physical situation for each of the following graphs (Fig 3.22): Fig. 3.22 3.20 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, – 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 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 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 Fig. 3.25 and D ? Additional Exercises 3.23 A three-wheeler starts from rest, accelerates uniformly with 1 m s–2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1,2,3….) versus n. What do you expect this plot to be during accelerated motion : a straight line or a parabola ? 3.24 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–1. 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 ? 3.25 On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed –1 9 km h (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–1. 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 ? Fig. 3.26 3.26 Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 m s–1 and 30 m s–1. 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–2. Give the equations for the linear and curved parts of the plot. 60 PHYSICS Fig. 3.27 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 What is the average speed of the particle over the intervals in (a) and (b) ? 3.28 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 t1 to t2: (a) x(t2 ) = x(t1) + v (t1) (t2 – t1) +(½) a (t2 – t1)2 (b) v(t2 ) = v(t1) + a (t2 – t1) (c) vaverage = (x(t2) – x(t1))/(t2 – t1) (d) aaverage = (v(t2) – v(t1))/(t2 – t1) (e) x(t2 ) = x(t1) + vaverage (t2 – t1) + (½) aaverage (t2 – t1)2 (f) x(t2 ) – x(t1) = area under the v-t curve bounded by the t-axis and the dotted line shown. MOTION IN A STRAIGHT LINE 61 APPENDIX 3.1 : ELEMENTS OF CALCULUS Differential Calculus Using the concept of ‘differential coefficient’ or ‘derivative’, 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 + ∆x, y + ∆y). The slope of the line joining P and Q is given by: ∆y (y + ∆y ) − y tanθ = = (2) ∆x ∆x Suppose now that the point Q moves along the curve towards P. In this process, ∆y and ∆x ∆y decrease and approach zero; though their ratio will not necessarily vanish. What happens ∆x to the line PQ as ∆y→ 0, ∆x→ 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 θ approaches the slope of the tangent at P, denoted by m: ∆y (y + ∆y ) − y m = lim = lim (3) ∆x ∆x → 0 ∆x → 0 ∆x The limit of the ratio ∆y/∆x as ∆x 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 + ∆y = f (x + ∆x), we can write the definition of the derivative as: dy df ( x ) ∆y  f ( x + ∆x ) – f ( x )  = = lim = lim   dx dx ∆x →0 ∆x ∆x → 0  ∆x  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. 62 PHYSICS d (a u ) du du du dx =a ; =. dx dx dt dx dt d(uv) dv du d (u / v ) 1 du dv =u +v ; = 2 –u dx dx dx dx v dx dx du du dx = dv dv dx d d (sin x ) = cos x ; dx (cos x ) = – sin x dx d d (tan x ) = sec2 x ; (cot x ) = – cos ec 2 x dx dx d d (sec x) = tan x sec x ; (cosec 2 x ) = – cot x co sec x dx dx d du d 1 (u )n = n u n –1 ; ( ln u) = dx dx du u d u (e ) = eu du In terms of derivatives, instantaneous velocity and acceleration are defined as ∆x dx v = lim = ∆t → 0 ∆t dt Dv dv d2 x a = lim = = Dt Æ0 Dt 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 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 ∆Ai = F (x i )(x i – x i –1) = F (x i )∆x where ∆x 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→∞ ), 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: N N A= ∑ ∆A = ∑ F (x )∆x i =1 i i =1 i The limit of this sum as N→∞ is known as the integral of F(x) over x from a to b. It is given a special symbol as shown below: b A= ∫ F (x )dx a The integral sign ∫ 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. dg(x ) Suppose we have a function g (x) whose derivative is f (x), i.e. f ( x ) = dx The function g (x) is known as the indefinite integral of f (x) and is denoted as: g( 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 ∫ f (x ) dx = g(x ) b a ≡ g(b ) – g(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, 2 2 x3 8 1 7 ∫ 2 x dx = = – = 3 3 3 3 1 1 Clearly, to evaluate definite integrals, we need to know the corresponding indefinite integrals. Some common indefinite integrals are 64 PHYSICS x n +1 x n dx = (n ≠ –1) n +1 1 ( )dx = ln x ( x > 0) x sin x dx = – cos x cos x dx = sin x e x dx = e x This introduction to differential and integral calculus is not rigorous and is intended to convey to you the basic notions of calculus.