CHAPTER 2: Kinematics of Linear Motion SF017 CHAPTER 2: Kinematics of Linear Motion

2.0 Kinematics of Linear motion SF017 2.0 Kinematics of Linear motion is defined as the studies of motion of an objects without considering the effects that produce the motion. There are two types of motion: Linear or straight line motion (1-D) with constant (uniform) velocity with constant (uniform) acceleration, e.g. free fall motion Projectile motion (2-D) x-component (horizontal) y-component (vertical)

Learning Outcomes : 2.1 Linear Motion SF017 Learning Outcomes : 2.1 Linear Motion At the end of this chapter, students should be able to: Define and distinguish between Distance and displacement Speed and velocity Instantaneous velocity, average velocity and uniform velocity Instantaneous acceleration, average acceleration and uniform acceleration, Sketch graphs of displacement-time, velocity-time and acceleration-time. Determine the distance travelled, displacement, velocity and acceleration from appropriate graphs.

2.1. Linear motion (1-D) 2.1.1. Distance, d scalar quantity. SF017 2.1. Linear motion (1-D) 2.1.1. Distance, d scalar quantity. is defined as the length of actual path between two points. For example : The length of the path from P to Q is 25 cm. P Q

2.1.2 Displacement, Example 2.1 : vector quantity. SF017 2.1.2 Displacement, vector quantity. is defined as the distance between initial point and final point in a straight line. The S.I. unit of displacement is metre (m). Example 2.1 : An object P moves 30 m to the east after that 15 m to the south and finally moves 40 m to west. Determine the displacement of P relative to the original position. Solution : N E W S O P  30 m 15 m 10 m

2.1.3 Speed, v The magnitude of the displacement is given by SF017 The magnitude of the displacement is given by and its direction is 2.1.3 Speed, v is defined as the rate of change of distance. scalar quantity. Equation:

2.1.4 Velocity, is a vector quantity. SF017 2.1.4 Velocity, is a vector quantity. The S.I. unit for velocity is m s-1. Average velocity, vav is defined as the rate of change of displacement. Equation: Its direction is in the same direction of the change in displacement.

Instantaneous velocity, v SF017 Instantaneous velocity, v is defined as the instantaneous rate of change of displacement. Equation: An object moves in a uniform velocity when and the instantaneous velocity equals to the average velocity at any time.

Gradient of s-t graph = velocity SF017 Therefore s t s1 t1 The gradient of the tangent to the curve at point Q = the instantaneous velocity at time, t = t1 Q Gradient of s-t graph = velocity

2.1.5 Acceleration, vector quantity. SF017 2.1.5 Acceleration, vector quantity. The S.I. unit for acceleration is m s-2. Average acceleration, aav is defined as the rate of change of velocity. Equation: Its direction is in the same direction of motion. The acceleration of an object is uniform when the magnitude of velocity changes at a constant rate and along fixed direction.

Instantaneous acceleration, a SF017 Instantaneous acceleration, a is defined as the instantaneous rate of change of velocity. Equation: An object moves in a uniform acceleration when and the instantaneous acceleration equals to the average acceleration at any time.

Gradient of v-t graph = acceleration SF017 Deceleration, a is a negative acceleration. The object is slowing down meaning the speed of the object decreases with time. Therefore v t Q v1 t1 The gradient of the tangent to the curve at point Q = the instantaneous acceleration at time, t = t1 Gradient of v-t graph = acceleration

2.1.6 Graphical methods s s t t s t SF017 2.1.6 Graphical methods Displacement against time graph (s-t) s t s t Gradient increases with time Gradient = constant s t (a) Uniform velocity (b) The velocity increases with time (c) Q Gradient at point R is negative. R P The direction of velocity is changing. Gradient at point Q is zero. The velocity is zero.

Area under the v-t graph = displacement SF017 Velocity versus time graph (v-t) The gradient at point A is positive – a > 0(speeding up) The gradient at point B is zero – a= 0 The gradient at point C is negative – a < 0(slowing down) t1 t2 v t (a) t2 t1 v t (b) t1 t2 v t (c) B Uniform acceleration C Uniform velocity A Area under the v-t graph = displacement

SF017 Example 2.2 : A toy train moves slowly along a straight track according to the displacement, s against time, t graph in Figure 2.1. a. Explain qualitatively the motion of the toy train. b. Sketch a velocity (cm s-1) against time (s) graph. c. Determine the average velocity for the whole journey. d. Calculate the instantaneous velocity at t = 12 s. e. Determine the distance travelled by the toy train. 2 4 6 8 10 12 14 t (s) s (cm) Figure 2.1

v (cm s1) t (s) Solution : SF017 Solution : a. 0 to 6 s : The train moves at a constant velocity of 6 to 10 s : The train stops. 10 to 14 s : The train moves in the same direction at a constant velocity of b. 2 4 6 8 10 12 14 t (s) 0.68 1.50 v (cm s1)

e. The distance travelled by the toy train is 10 cm. SF017 Solution : c. d. e. The distance travelled by the toy train is 10 cm.

SF017 Example 2.3 : A velocity-time (v-t) graph in Figure 2.2 shows the motion of a lift. a. Describe qualitatively the motion of the lift. b. Sketch a graph of acceleration (m s2) against time (s). c. Determine the total distance travelled by the lift and its displacement. d. Calculate the average acceleration between 20 s to 40 s. 5 10 15 20 25 30 35 t (s) -4 -2 2 4 v (m s1) Figure 2.2 40 45 50

15 to 20 s : Lift moving with constant velocity of SF017 Solution : a. 0 to 5 s : Lift moves upward from rest with a constant acceleration of 5 to 15 s : The velocity of the lift increases from 2 m s1 to 4 m s1 but the acceleration decreasing to 15 to 20 s : Lift moving with constant velocity of 20 to 25 s : Lift decelerates at a constant rate of 25 to 30 s : Lift at rest or stationary. 30 to 35 s : Lift moves downward with a constant acceleration of 35 to 40 s : Lift moving downward with constant velocity of 40 to 50 s : Lift decelerates at a constant rate of and comes to rest.

a (m s2) t (s) Solution : b. 5 10 15 20 25 30 35 40 45 50 -0.4 -0.2 SF017 Solution : b. t (s) 5 10 15 20 25 30 35 40 45 50 -0.4 -0.2 0.2 0.6 a (m s2) -0.6 -0.8 0.8 0.4

v (m s1) t (s) Solution : c. i. 5 10 15 20 25 30 35 -4 -2 2 4 40 45 SF017 Solution : c. i. 5 10 15 20 25 30 35 t (s) -4 -2 2 4 v (m s1) 40 45 50 A2 A3 A1 A4 A5

SF017 Solution : c. ii. d.

SF017 Exercise 2.1 : Figure 2.3 shows a velocity versus time graph for an object constrained to move along a line. The positive direction is to the right. a. Describe the motion of the object in 10 s. b. Sketch a graph of acceleration (m s-2) against time (s) for the whole journey. c. Calculate the displacement of the object in 10 s. ANS. : 6 m Figure 2.3

SF017 Exercise 2.1 : A train pulls out of a station and accelerates steadily for 20 s until its velocity reaches 8 m s1. It then travels at a constant velocity for 100 s, then it decelerates steadily to rest in a further time of 30 s. a. Sketch a velocity-time graph for the journey. b. Calculate the acceleration and the distance travelled in each part of the journey. c. Calculate the average velocity for the journey. Physics For Advanced Level, 4th edition, Jim Breithaupt, Nelson Thornes, pg.15, no. 1.11 ANS. : 0.4 m s2,0 m s2,-0.267 m s2, 80 m, 800 m, 120 m; 6.67 m s1.

Learning Outcome : 2.2 Uniformly accelerated motion (1 hour) SF017 Learning Outcome : 2.2 Uniformly accelerated motion (1 hour) At the end of this chapter, students should be able to: Derive and apply equations of motion with uniform acceleration:

2.2. Uniformly accelerated motion SF017 2.2. Uniformly accelerated motion From the definition of average acceleration, uniform (constant) acceleration is given by where v : final velocity u : initial velocity a : uniform (constant) acceleration t : time (1)

SF017 From equation (1), the velocity-time graph is shown in Figure 2.4 : From the graph, The displacement after time, s = shaded area under the graph = the area of trapezium Hence, velocity v u time t Figure 2.4 (2)

By substituting eq. (1) into eq. (2) thus SF017 By substituting eq. (1) into eq. (2) thus From eq. (1), From eq. (2), (3) multiply (4)

Therefore the equations (2) and (3) can be written as SF017 Notes: equations (1) – (4) can be used if the motion in a straight line with constant acceleration. For a body moving at constant velocity, ( a = 0) the equations (1) and (4) become Therefore the equations (2) and (3) can be written as constant velocity

SF017 Example 2.4 : A plane on a runway accelerates from rest and must attain takeoff speed of 148 m s1 before reaching the end of the runway. The plane’s acceleration is uniform along the runway and of value 914 cm s2. Calculate a. the minimum length of the runway required by the plane to takeoff. b. the time taken for the plane cover the length in (a). Solution : a. Use

b. By using the equation of linear motion, SF017 Solution : b. By using the equation of linear motion, OR

SF017 Example 2.5 : A bus travelling steadily at 30 m s1 along a straight road passes a stationary car which, 5 s later, begins to move with a uniform acceleration of 2 m s2 in the same direction as the bus. Determine a. the time taken for the car to acquire the same velocity as the bus, b. the distance travelled by the car when it is level with the bus. Solution : a. Given Use

SF017 b. From the diagram, c b b b c Therefore

SF017 Example 2.6 : A particle moves along horizontal line according to the equation Where s is displacement in meters and t is time in seconds. At time, t = 3 s, determine a. the displacement of the particle, b. Its velocity, and c. Its acceleration. Solution : a. t =3 s ;

b. Instantaneous velocity at t = 3 s, Use SF017 Solution : b. Instantaneous velocity at t = 3 s, Use Thus

c. Instantaneous acceleration at t = 3 s, Use SF017 Solution : c. Instantaneous acceleration at t = 3 s, Use Hence

SF017 Exercise 2.2 : A speedboat moving at 30.0 m s-1 approaches stationary buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of -3.50 m s-2 by reducing the throttle. a. How long does it take the boat to reach the buoy? b. What is the velocity of the boat when it reaches the buoy? No. 23,pg. 51,Physics for scientists and engineers with modern physics, Serway & Jewett,6th edition. ANS. : 4.53 s; 14.1 m s1 An unmarked police car travelling a constant 95 km h-1 is passed by a speeder traveling 140 km h-1. Precisely 1.00 s after the speeder passes, the policemen steps on the accelerator; if the police car’s acceleration is 2.00 m s-2, how much time passes before the police car overtakes the speeder (assumed moving at constant speed)? No. 44, pg. 41,Physics for scientists and engineers with modern physics, Douglas C. Giancoli,3rd edition. ANS. : 14.4 s

SF017 Exercise 2.2 : A car traveling 90 km h-1 is 100 m behind a truck traveling 75 km h-1. Assuming both vehicles moving at constant velocity, calculate the time taken for the car to reach the truck. No. 15, pg. 39,Physics for scientists and engineers with modern physics, Douglas C. Giancoli,3rd edition. ANS. : 24 s A car driver, travelling in his car at a constant velocity of 8 m s-1, sees a dog walking across the road 30 m ahead. The driver’s reaction time is 0.2 s, and the brakes are capable of producing a deceleration of 1.2 m s-2. Calculate the distance from where the car stops to where the dog is crossing, assuming the driver reacts and brakes as quickly as possible. ANS. : 1.73 m

Learning Outcome : 2.3 Freely falling bodies SF017 Learning Outcome : 2.3 Freely falling bodies At the end of this chapter, students should be able to: Describe and use equations for freely falling bodies. For upward and downward motion, use a = g = 9.81 m s2

SF017 2.3 Freely falling bodies is defined as the vertical motion of a body at constant acceleration, g under gravitational field without air resistance. In the earth’s gravitational field, the constant acceleration known as acceleration due to gravity or free-fall acceleration or gravitational acceleration. the value is g = 9.81 m s2 the direction is towards the centre of the earth (downward). Note: In solving any problem involves freely falling bodies or free fall motion, the assumption made is ignore the air resistance.

Linear motion Freely falling bodies SF017 Sign convention: Table 2.1 shows the equations of linear motion and freely falling bodies. + - From the sign convention thus, Linear motion Freely falling bodies Table 2.1

SF017 An example of freely falling body is the motion of a ball thrown vertically upwards with initial velocity, u as shown in Figure 2.5. Assuming air resistance is negligible, the acceleration of the ball, a = g when the ball moves upward and its velocity decreases to zero when the ball reaches the maximum height, H. velocity = 0 H u Figure 2.5 v

s v =0 H t t1 2t1 v u t t1 2t1 u a t t1 2t1 g SF017 t s H t1 2t1 v =0 The graphs in Figure 2.6 show the motion of the ball moves up and down. Derivation of equations At the maximum height or displacement, H where t = t1, its velocity, hence therefore the time taken for the ball reaches H, t v u u t1 2t1 t a g t1 2t1 Figure 2.6

To calculate the maximum height or displacement, H: use either SF017 To calculate the maximum height or displacement, H: use either maximum height, Another form of freely falling bodies expressions are Where s = H OR

SF017 Example 2.7 : A ball is thrown from the top of a building is given an initial velocity of 10.0 m s1 straight upward. The building is 30.0 m high and the ball just misses the edge of the roof on its way down, as shown in figure 2.7. Calculate a. the maximum height of the stone from point A. b. the time taken from point A to C. c. the time taken from point A to D. d. the velocity of the ball when it reaches point D. (Given g = 9.81 m s2) A B C D u =10.0 m s1 30.0 m Figure 2.7

SF017 Solution : a. At the maximum height, H, vy = 0 and u = uy = 10.0 m s1 thus b. From point A to C, the vertical displacement, sy= 0 m thus A B C D u 30.0 m

SF017 Solution : c. From point A to D, the vertical displacement, sy= 30.0 m thus By using A B C D u 30.0 m a b c Time don’t have negative value. OR

u Solution : d. Time taken from A to D is t = 3.69 s thus B SF017 Solution : d. Time taken from A to D is t = 3.69 s thus From A to D, sy = 30.0 m Therefore the ball’s velocity at D is A B C D u 30.0 m OR

SF017 Example 2.8 : A book is dropped 150 m from the ground. Determine a. the time taken for the book reaches the ground. b. the velocity of the book when it reaches the ground. (Given g = 9.81 m s-2) Solution : a. The vertical displacement is sy = 150 m Hence uy = 0 m s1 150 m

b. The book’s velocity is given by SF017 Solution : b. The book’s velocity is given by Therefore the book’s velocity is OR

SF017 Exercise 2.3 : A ball is thrown directly downward, with an initial speed of 8.00 m s1, from a height of 30.0 m. Calculate a. the time taken for the ball to strike the ground, b. the ball’s speed when it reaches the ground. ANS. : 1.79 s; 25.6 m s1 A falling stone takes 0.30 s to travel past a window 2.2 m tall as shown in Figure 2.8. From what height above the top of the windows did the stone fall? ANS. : 1.75 m Figure 2.8 to travel this distance took 0.30 s

SF017 Exercise 2.3 : A ball is thrown directly downward, with an initial speed of 8.00 m s1, from a height of 30.0 m. Calculate a. the time taken for the ball to strike the ground, b. the ball’s speed when it reaches the ground. ANS. : 1.79 s; 25.6 m s1 A falling stone takes 0.30 s to travel past a window 2.2 m tall as shown in Figure 2.8. From what height above the top of the windows did the stone fall? ANS. : 1.75 m Figure 2.8 to travel this distance took 0.30 s