STRAIGHT LINE MOTION. 1 Equations of Straight Line Motion (Constant Acceleration)

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1 1 CHAPTER. STRAIGHT LINE MOTION (CONSTANT ACCELERATION) 1 INSTITIÚID TEICNEOLAÍOCHTA CHEATHARLACH INSTITUTE OF TECHNOLOGY CARLOW STRAIGHT LINE MOTION By definition, mechanics is the study of bodies at rest and in motion and the effect of forces on them. The subject of mechanics is subdivided into statics, which specifically focuses on bodies at rest, and dynamics, which focuses on bodies in motion. Because of its relevance to game development we are primarily concerned with the subject of dynamics which itself is divided into two specific areas, namely, kinematics, which focuses on the motion of bodies without regard to the forces that act on the body, and kinetics, which consider both the motion of bodies and the forces that act on or otherwise affect bodies in motion. 1 Equations of Straight Line Motion (Constant Acceleration) We define the following quantities which are needed to describe the motion of a body: Distance is the length of a given path. The unit of distance is the metre (m). Displacement defines the position of one point relative to another point: displacement includes both the distance between two points and the direction of the first point from the second point. Speed is the rate at which a moving body covers its path, no account being taken of the direction of motion. The unit of distance is the metre and the unit of time is the second, hence the unit of speed is the metre per second (ms 1 ). Velocity v is the rate of change of position with time when the direction of motion is specified. Velocity v is thus a vector quantity; its magnitude is its speed. Velocity is expressed in metre per second (ms 1 ). Acceleration a is the rate of change of velocity with respect to time, expressed in metres per second per second (ms ). The rate of decrease of velocity with time, i.e., negative acceleration is called deceleration.

2 CHAPTER. STRAIGHT LINE MOTION (CONSTANT ACCELERATION) Consider a particle moving in a straight line with a constant acceleration a and has an initial velocity u and a final velocity v after an interval of time t. V elocity C v v u u θ B A t D T ime In a velocity-time graph the acceleration is represented by the gradient of BC Therefore Hence a = v u t v = u + at Let the increase in displacement (position of one point relative to another) in time t be denoted by s. This is represented by the area of ABCD. Therefore Eliminating v from and yields s = ut + 1 t(v u) = ut + 1 vt 1 ut = 1 ut + 1 vt s = 1 (u + u + at)t = 1 (ut + at ) = 1 (u + v)t Therefore s = ut + 1 at

3 3 CHAPTER. STRAIGHT LINE MOTION (CONSTANT ACCELERATION) 3 Eliminating u from and yields s = 1 ((v at) + v)t = 1 (vt at ) Therefore s = vt 1 at Finally, eliminating t from and yields s = 1 (u + v)(v u) a = 1 a (uv u + v uv) Therefore v = u + as In summary, we have that when a particle moves in a straight line its displacement, velocity and acceleration can have one of only two possible directions. Motion in a straight line with constant acceleration occur frequently enough to justify obtaining general equations which can be applied to particular problems. ** Let P be a particle moving in a straight line. In general for motion in a straight line we have v = u + at s = ut + 1 at s = vt 1 at v = u + as where v is velocity, u initial velocity, s distance, a acceleration and t time.

4 4 CHAPTER. STRAIGHT LINE MOTION (CONSTANT ACCELERATION) 4 Example A particle is moving along a straight line with a constant acceleration (deceleration) of 3ms. If initially it has a velocity of 10ms 1, to find the time when the velocity is zero we have u = 10, v = 0, a = 3 We require t so using the appropriate equation of motion we have v = u + at 0 = 10 + ( 3)t t = seconds. Example A train traveling along a straight line with a constant acceleration is observed to travel consecutive distances of 1km in times of 30 seconds and 60 seconds respectively. To find the initial velocity of the train we have V elocity u T ime At s = 1000m, t = 30 seconds. Also at s = 000m, t = 90 seconds. Substituting this information into the equation of motion s = ut + 1 at yields 1000 = 30u + 450a 000 = 90u a Solving for u yields u = 38 9 ms 1.

5 5 CHAPTER. STRAIGHT LINE MOTION (CONSTANT ACCELERATION) 5 Example A particle starts from a point O with a velocity of ms 1 and travels along a straight line with a constant acceleration of ms. Two seconds later a second particle starts from rest at O and travels along the same line with an acceleration of 6ms. We wish to find how far from O the second particle overtakes the first. V elocity 0 T ime When the second particle overtakes the first they will both have the same displacement from 0. Let displacement = d metres. 1 st particle takes T seconds to reach this point. nd particle takes (T-) seconds to reach this point. For the 1 st particle For the nd particle s = ut + 1 at d = T + T s = ut + 1 at d = 3(T ) Eliminating d from both equations yields T + T = 3(T ) T 7T + 6 = 0 (T 6)(T 1) = 0

6 6 CHAPTER. STRAIGHT LINE MOTION (CONSTANT ACCELERATION) 6 Therefore T = 6, since T = 1 is before the second particle starts. Finally, at T = 6 we have d = 48. The second particle overtakes the first 48m from 0. Example A particle moving in a straight line with a constant velocity of 5ms 1, at one instant and 4 seconds later it has a velocity of 15ms 1. Find the acceleration and the distance covered by the particle in 4 seconds. v = 5ms 1 v = 15ms 1 t = 0 t = 4 Let u = 5 ms 1, t = 4 seconds, v = 15 ms 1. v = u + at 15 = 5 + 4a a =.5 ms Also v = u + as 5 = 5 + 5s s = 40 m Exercise At the same instant two children, who are standing 4m apart, begin to cycle directly towards each other. James starts from rest at a point A, riding with a constant acceleration of ms and William rides with a constant speed of ms 1. Find how long it is before they meet. [Solution: They meet after 4 seconds]. Exercise A particle A starts from rest at a point O and moves on a straight line with constant acceleration ms. At the same instant another particle B, 1m behind O, is moving with velocity 5ms 1 and has constant acceleration of 3ms. How far from O are the particles when B overtakes A? [Solution: B overtakes A at a distance 4m from O].

7 7 CHAPTER. STRAIGHT LINE MOTION (CONSTANT ACCELERATION) 7 Exercise The driver of a truck travelling at 30ms 1, sees an obstacle 50m ahead. He decelerates at a constant rate and comes to a halt just in time. Find the deceleration and the breaking time. 7 50m 1.1 Vertical Motion under Gravity Before the time of Galileo it was thought that if two objects of different masses were dropped the heavier object would fall faster than the light one. In a famous series of experiments Galileo showed that this was not true. (He allegedly dropped objects from the top of the leaning tower of Pisa and timed their descent from the Cathedral clock opposite). The result that Galileo observed are that if air resistance is ignored all bodies (whatever their mass) have the same constant acceleration towards the center of the earth when they are moving under the action of their weight only. This acceleration is denoted by the letter g, and a good approximation to its value is 9.81 ms. When a body is thrown vertically upwards or is dropped it will move in a straight line. The only force acting on it will be its weight causing a constant acceleration g along the line, so the equations for motion in a straight line with constant acceleration will apply. In some problems it is convenient to take the downward direction as positive, in which case the acceleration is +g, but in other problems it is convenient to take the upward direction as positive, in which case the acceleration is g. Example A stone is thrown vertically downwards from the top of a tower and hits the ground 10 seconds later with a speed of 51 ms 1. Find the height of the tower. Taking the downward direction as positive i.e., g = ms

8 8 CHAPTER. STRAIGHT LINE MOTION (CONSTANT ACCELERATION) 8 s Now v = 51ms 1, t = 10 seconds and a = 9 81ms. The tower is 19 5 m high. s = vt 1 at = 51(10) 1 (9 81)(10) s = 19 5 m Example A ball is thrown vertically upwards from a point 0 5m above ground level with a speed of 7ms 1. Find the height above this point reached by the ball and the speed with which it hits the ground. Taking the upward direction as positive i.e., g = 9 81 ms 7ms 1 0.5m s = 0 Ground Level

9 9 CHAPTER. STRAIGHT LINE MOTION (CONSTANT ACCELERATION) 9 Now u = 7ms 1, v = 0ms 1 and a = 9 81ms. v = u + as 0 = 49 + ( 9 81)s s = 5 m Therefore the ball reaches a height of 5m above its initial position. When ball hits the ground it is 0 5m below its initial position. Given u = 7ms 1, s = 0 5m and a = 9 81ms v = u + as v = 49 + ( 9 81)( 0 5) v = ±7 66ms 1 Therefore the ball hits the ground with a speed of 7 66ms 1. Example A youth playing with a ball in a garden surrounded by a wall 5m high and kicks the ball vertically up from a height of 0 4m with a speed of 14ms 1. For how long is the ball above the height of the wall? Give your answer corrected to significant figures. Taking the upward direction as positive i.e., g = 9 81 ms 1m 0 4m 14ms 1 s = 0 Ground Level

10 10 CHAPTER. STRAIGHT LINE MOTION (CONSTANT ACCELERATION) 10 Now u = 14ms 1, s = 1m and a = 9 81ms. s = ut + 1 at 1 = 14t 4 9t 0 3 = t 0 7t Now we have 7t 0t + 3 = 0. Using the quadratic formula we get t = 0 ± = 698 or The ball is at the height of the top of the wall at two different times. Therefore it takes seconds to reach the top of the wall when going up, and returns to that height 698 seconds from the start. So the ball is above the wall for 5 seconds, correct to two significant figures. Example An object is thrown vertically upwards with a speed of 14ms 1. Two seconds later a second similar object is dropped from the same point. Find where the two objects meet. 14ms 1 s = 0 (initial position) d m The two objects will meet when they have the same displacement (d metres) from the starting point. If the time taken by the first object to reach this displacement is T seconds, the time taken by the second object is (T ) seconds. Taking the upward direction as positive i.e., g = 9 81 ms

11 11 CHAPTER. STRAIGHT LINE MOTION (CONSTANT ACCELERATION) 11 For the 1 st object For the nd object s = ut + 1 at d = 14T 4 9T s = ut + 1 at d = 4 9(T ) Eliminating d from both equations yields T 0 7T = 0 7(T ) 0.8T =.8 T = 3 5 seconds Finally, at T = 3 5 seconds d = 11m. Therefore the objects meet 11m below their initial position. Example A stone is dropped from the top of a building and at the same time a second stone is thrown vertically upwards from the bottom of the building with a speed of 0ms 1. They pass each other 3 seconds later. Find the height of the building. Second Object s d F irst Object d 1 t = 3 seconds s = 0 (initial position) Taking the upward direction as positive i.e., g = 9 81 ms

12 1 CHAPTER. STRAIGHT LINE MOTION (CONSTANT ACCELERATION) 1 We wish to determine d 1 and d at t = 3 seconds. Let height of building be denoted by s where s = d 1 + d For the 1 st object For the nd object s = ut + 1 at d 1 = 0(3) + 1 ( 9 81)(9) d 1 = Therefore the building is 60m high. s = ut + 1 at d = ( 9 81)(9) d = Exercise A missile is fired vertically upwards with speed 47ms 1. Find, to 3 significant figures, the time that elapses before the missile returns to the firing position. Exercise A particle is projected vertically upwards from ground level with a speed of 50ms 1. For how long will it be more than 70m above the ground? Exercise A particle A is fired directly upward with a speed of 40ms 1 and one second later another one, B, is projected directly upward from the same point with the same velocity. Find the time it takes for the particles to collide and the height above the point of projection at which they meet. [Solution: t = 3 5sec, s = 78 8m].