# To define concepts such as distance, displacement, speed, velocity, and acceleration.

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1 Chapter 7 Kinematics of a particle Overview In kinematics we are concerned with describing a particle s motion without analysing what causes or changes that motion (forces). In this chapter we look at particles moving in a straight line. The objectives here are as follows: To define concepts such as distance, displacement, speed, velocity, and acceleration. To study motion in a straight line with constant acceleration mathematically. We will use metric units: metres for distance and seconds for time. Before defining the fairly basic concepts mentioned above, it is helpful to define what we mean by a particle. A particle is a mathematical abstraction: it is a physical object concentrated at a single point - that is, an object with no extension in space. The kinematic laws written down in this chapter hold for such abstract entities; the laws apply equally to extended objects, as we shall demonstrate in later chapters. This generalization requires extra concepts and mathematical tools. To avoid this extra layer of complication, we shall for the present deal only with point particles. 7.1 Distance, Displacement, Speed, Velocity, Acceleration Our first definition is that speed. This is easy for the case of an object that is moving steadily: Definition 7.1 The speed of an object moving steadily is speed = distance travelled. time taken 76

2 7.1. Distance, Displacement, Speed, Velocity, Acceleration 77 If the object is not moving steadily, that is, if the speed in the definition is not constant, we can still look at the average speed, which is defined in all cases as average speed = distance travelled. time taken Speed is therefore measured in metres per second (m/s). Example: A cyclist travels on a straight road. The first mile takes one hour. The second mile takes two hours (bad cyclist!). Draw the cyclist s speed on a distance-time graph. Hence, compute the cyclist s average speed. Solution: 1 hour = 3600 s 1 mile = 8 km = 1600 m 5 Distance-time grpah: distance (*1000m) O 3.6 A time(*1000s) 7.2 B 1.6 Figure 7.1: Cyclist s speed from O to A = Slope of OA = = 0.44 = 0.44 m/s Cyclist s speed from A to B = Slope of BC = = 0.22 = 0.22 m/s

3 78 Chapter 7. Kinematics of a particle Cyclist s average speed from O to B = = = 0.30 m/s = Slope of AC In the previous example, it can be seen that the slope of the distance-time graph is the cyclist s speed. The gradient changes is piecewise-constant and hence, the cyclist s speed is too at each point in time the cyclist s speed is constant and the speed changes abruptly at one point. In contrast, objects whose speed varies smoothly will have a curves distance-time graph, as in the following example. Example: A car starts from rest at a traffic light and moves away a distance d after time t, according to the following table: t d d What is the average speed of the car, between 2s and 4s? Here, time is expressed in seconds and distance in metre. Solution: A distance-time graph is shown in Figure distance(metres) time(secs) Figure 7.2: Distance-time graph

4 7.1. Distance, Displacement, Speed, Velocity, Acceleration 79 Reading off from the graph (or from the table), the verage speed from 2 sec to 4 sec is 24 2 = 12 m/s The instantaneous speed If we want to find the speed at a particular time we have to take the average over smaller and smaller intervals. The slope of these chords approaches the slope of the tangent (Figure 7.1.1): tangent We then have the following very important result / definition: Definition 7.2 The instantaneous speed at a particular value of t is given by the slope of the tangent of the distance-time graph at the point where t has that value. Equivalently, using what we have already learned from the Calculus review, we can write where v = ds dt, v is the instantaneous speed (function of time) s is the distance travelled at time t Velocity Recall the notion of displacement as the signed distance from the origin. This gives rise to related notion of instantaneous velocity

5 80 Chapter 7. Kinematics of a particle Definition 7.3 The instantaneous velocity is the rate of change of the displacement with respect to time. Equivalently, it is the speed with directional information. Example: A particle moves at constant speed but its displacement-time graph has the following characteristics: Time (s) Displacement from Origin Draw the displacement-time graph. Solution Figure displacement(m) time(s) Particle travels 4 m in 2 seconds. Speed is 2 m/s, and is constant! But Slope of graph is ±2 Velocity alternates between +2 and 2. A change in the velocity is called acceleration.

6 7.1. Distance, Displacement, Speed, Velocity, Acceleration 81 Definition 7.4 The instantaneous acceleration is the rate of change of velocity with respect to time: where v is the velocity. Also, If a > 0 then the velocity is increasing speeding up. a = dv dt, (7.1) If a < 0 then the velocity is decreasing slowing down. Sometimes a negative acceleration is called deceleration. Units of acceleration are therefore units of velocity divided by units of time. We write this as m/s 2 and we say this is metres per second per second or metres per second squared. The slope of a velocity-time plot tells us the acceleration. Generally, the acceleration will depend on time. However, there is a very nice mathematical theory of kinematics for the case of constant acceleration, which we study below. We first of all have a look at some examples. Example: A racing car accelerates rapidly from the grid. Starting from a stationary state, it reaches 100 kilometres per hour in just 4 seconds. What is the average acceleration over this period? Solution: The instantaneous acceleration is given by Equation (7.1). The average acceleration is got by replacing the infinitesimal amounts dv and dv with finite increments: Average Acceleration over a period t = v t, and in the present example, v = 100 kph and t = 4 sec. Hence, Average acceleration = change in velocity time taken = 100 kph 4 sec 100 kph = Average acceleration = Compare this to the acceleration of gravity, 9.81 m/s 2! m/s 28 m/s 28 m/s 4 sec = 7 m/s2

7 82 Chapter 7. Kinematics of a particle 20 speed(m/s) A B 5 O I II III E F C time(secs) Example: A particle moves in the positive direction and has the velocity-time curve shown in Figure Compute the (constant) acceleration in each of the phases I, II, and III. Solution: Phase I: The speed increases from 0 to 12 m/s in 40s. The acceleration = = 0.3 m/s2. The slope of OA = = 0.3. Phase II: The speed increases form 12m/s to 17m/s. The acceleration = 5 20 = 0.25 m/s2. The slope of AB = 5 20 = Phase III: The speed decreases from 17 m/s to 0 m/s in 32 s. The deceleration = = 0.53 m/s 2.

8 7.2. Mathematical description of kinematics with constant acceleration Mathematical description of kinematics with constant acceleration velocity v B u A t v - u O t time Consider a particle launched from the origin (zero initial displacement) with initial velocity v. At a final time t, the particle has a velocity v (Figure 7.2). The particle experiences a constant acceleration a. We now work out a mathematical description between the final displacement s, and the quantities a, u, and v. Our starting-point is the definition of the instantaneous acceleration in calculus notation: We integrate this expression: t 0 a = dv dt. t a dt = dv 0 dt dt, = v(t) v(0), = v u. However, the acceleration is constant and therefore comes outside the integral, to give t a dt = at, 0 hence v = u + at (7.2) We can also make use of the expression and integrate as before to obtain v = ds dt s = t 0 vt.

9 84 Chapter 7. Kinematics of a particle Use the expression (7.2) in the integrand here: s = t 0 (u + at) dt. Hence, s = ut at2. (7.3) It is sometimes useful to work with expressions that do not involve t. We can eliminate t between Equations (7.2) and (7.3) by taking Equation (7.2) and writing Substitute into Equation (7.3): t = v u a. s = ut at2, ( ) v u = u a ( ) 2 v u, a = uv u v2 uv v2. a Hence, as = 1 2 ( v 2 u 2). Tidy up to obtain v 2 = u 2 + 2as. (7.4)

10 7.2. Mathematical description of kinematics with constant acceleration 85 Example: A particle starts from a point A with velocity 3 m/s and moves with a constant acceleration of 1 2 m/s2 along a straight line AB. It reaches B with a ve- 3m/s 1 2 m/s2 5m/s locity of 5 m/s. Find: A s B (a) the displacement from A to B. (b) the time taken from A to B. Solution: For motion in a straight line with constant acceleration we have five quantities: u, v, s, a, t. Each of the equations contains four of the five. Which of the equations should we use? It depends on the problem. In the present context we anticipate that different formulas should be for (a) and (b) above. In particular, we know: u = 3 v = 5 a = 1 2, so we should focus either on eliminating s or t. (a) s =? Avoid formula with t in it: v 2 u 2 = 2as 25 9 = s s = 16 metres (b) t =? Avoid formula with s in it: v = u + at 5 = t t = 4 seconds Example: The driver of a train begins the approach to the station by applying the brakes to produce a steady deceleration of 0.2 m/s 2 and bring the train to rest at the platform in 1 min 30 secs.

11 86 Chapter 7. Kinematics of a particle Find: (a) the speed when the brakes were applied, (b) the distance travelled. Solution: We know: t = 90, v = 0, and a = 0.2. (a) Initial speed u =? v = u + at. 0 = u u = 2 90 = 18 m/s. 10 (b) Distance travelled s =? v 2 u 2 = 2as. 0 (18) 2 = 2( 0.2)s. s = 810 m. Example: A world-calss sprinter accelerates with constant acceleration to his maximum speed in 4.0m/s. he then maintains this speed for the remainder of a 100-m race, finishing with a total time of 9.1s. (a) What is the runner s average acceleration during the first 4.0s? (b) What is his average acceleration during the last 5.1s? (c) What is his average accleration for the entire race? (d) Explain why your answer to part (c) is not the average of the answers to parts (a) and (b). Solution: It is helpful to view a velocity-versus-time plot, as in the sketch in Figure 7.3.

12 7.2. Mathematical description of kinematics with constant acceleration 87 Figure 7.3: The strategy for Part (a) is to find the displacements s 0 and s 1 in Phases I and II in terms of the acceleration a. In Phase I we have s 0 = 1 2 at2 0, where t 0 = 4s is the duration of Phase I. In Phase II we have s 1 = v 0 (T t 0 ), where v 0 is the unknown final velocity, T = 9.1s and T t 0 = 5.1 is the duration of Phase II. The velocity v 0 and the acceleration a are connected via a = v 0 /t = v 0 = at 0. Combine these results to get s 0 = 1 2 at2 0, s 1 = at 0 (T t 0 ). But hence hence Fill in the numbers: s 0 + s 1 = 100m, 1 2 at2 0 + at 0 (T t 0 ) = 100m, at 0 ( T 1 2 t 0) = 100m. a (4s) (7.1s) = 100m,

13 88 Chapter 7. Kinematics of a particle hence a = (100/28.4) = 3.5 m/s 2, correct to two significant figures. For part (b) the velocity is constant and hence the acceleration is zero. For part (c) the average acceleration over the whole race is a av = v final v initial T = v 0 T = a(t 0/T ). Filling int he numbers, this is a av = (3.52) (4/9.1) = 1.5m/s 2, correct to two significant figures. For part (d), a simple average of the two accelerations gives ( )/2 = 1.75m/s 2, which is the wrong answer. The reason this is the wrong answer is because the average should be weighted by how much time is spent in each phase. The corrected weighted average would be ( ) ( ) t0 T t0 a av = a + 0 = a(t 0 /T ), T T in agreement with the average computed in part (c).

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