CHAPTER 2: Kinematics of Linear Motion (5 hours) SF017 CHAPTER 2: Kinematics of Linear Motion (5 hours)

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 (1 hour) SF017 Learning Outcomes : 2.1 Linear Motion (1 hour) 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

From the equation of instantaneous velocity, SF017 From the equation of instantaneous velocity, Therefore Simulation 2.1 Simulation 2.2 Simulation 2.3

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 (1 hour) SF017 Learning Outcome : 2.3 Freely falling bodies (1 hour) 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 Simulation 2.4 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

Learning Outcomes : 2.4 Projectile motion (2 hours) SF017 Learning Outcomes : 2.4 Projectile motion (2 hours) At the end of this chapter, students should be able to: Describe and use equations for projectile, Calculate: time of flight, maximum height, range and maximum range, instantaneous position and velocity.

2.4. Projectile motion y v v1 v1y 1 v2x v1x 2 sy=H v2y u v2 uy  x SF017 2.4. Projectile motion A projectile motion consists of two components: vertical component (y-comp.) motion under constant acceleration, ay= g horizontal component (x-comp.) motion with constant velocity thus ax= 0 The path followed by a projectile is called trajectory is shown in Figure 2.9. y x Figure 2.9 B A P Q C v1 v v1y sy=H 1 v2x t1 v1x v2 2 v2y t2 u uy Simulation 2.5  ux sx= R

but the y-component of the initial velocity is given by SF017 From Figure 2.9, The x-component of velocity along AC (horizontal) at any point is constant, The y-component (vertical) of velocity varies from one point to another point along AC. but the y-component of the initial velocity is given by

Velocity Point P Point Q SF017 Table 2.2 shows the x and y-components, magnitude and direction of velocities at points P and Q. Velocity Point P Point Q x-comp. y-comp. magnitude direction Table 2.2

SF017 2.4.1 Maximum height, H The ball reaches the highest point at point B at velocity, v where x-component of the velocity, y-component of the velocity, y-component of the displacement, Use

2.4.2 Time taken to reach maximum height, t’ SF017 2.4.2 Time taken to reach maximum height, t’ At maximum height, H Time, t = t’ and vy= 0 Use 2.4.3 Flight time, t (from point A to point C)

2.4.4 Horizontal range, R and value of R maximum SF017 2.4.4 Horizontal range, R and value of R maximum Since the x-component for velocity along AC is constant hence From the displacement formula with uniform velocity, thus the x-component of displacement along AC is and

From the trigonometry identity, thus SF017 From the trigonometry identity, thus The value of R maximum when  = 45 and sin 2 = 1 therefore Simulation 2.6

2.4.5 Horizontal projectile SF017 2.4.5 Horizontal projectile Figure 2.10 shows a ball bearing rolling off the end of a table with an initial velocity, u in the horizontal direction. Horizontal component along path AB. Vertical component along path AB. h x A B Figure 2.10 Simulation 2.7

= Time taken for the ball to reach the floor (point B), t SF017 Time taken for the ball to reach the floor (point B), t By using the equation of freely falling bodies, Horizontal displacement, x Use condition below : Figure 2.11 The time taken for the ball free fall to point A The time taken for the ball to reach point B = (Refer to Figure 2.11)

SF017 Since the x-component of velocity along AB is constant, thus the horizontal displacement, x Note : In solving any calculation problem about projectile motion, the air resistance is negligible. and

v1x Example 2.9 : u  = 60.0 x v1 v1y v2x v2y v2 P O SF017 Example 2.9 : Figure 2.12 x O u  = 60.0 y R H v2y v1x v1y v2x Q v1 P v2 Figure 2.12 shows a ball thrown by superman with an initial speed, u = 200 m s-1 and makes an angle,  = 60.0 to the horizontal. Determine a. the position of the ball, and the magnitude and direction of its velocity, when t = 2.0 s.

b. the time taken for the ball reaches the maximum height, H and SF017 b. the time taken for the ball reaches the maximum height, H and calculate the value of H. c. the horizontal range, R d. the magnitude and direction of its velocity when the ball reaches the ground (point P). e. the position of the ball, and the magnitude and direction of its velocity at point Q if the ball was hit from a flat-topped hill with the time at point Q is 45.0 s. (Given g = 9.81 m s-2) Solution : The component of Initial velocity :

a. i. position of the ball when t = 2.0 s , Horizontal component : SF017 Solution : a. i. position of the ball when t = 2.0 s , Horizontal component : Vertical component : therefore the position of the ball is (200 m, 326 m)

a. ii. magnitude and direction of ball’s velocity at t = 2.0 s , SF017 Solution : a. ii. magnitude and direction of ball’s velocity at t = 2.0 s , Horizontal component : Vertical component : Magnitude, Direction, from positive x-axis anticlockwise

b. i. At the maximum height, H : SF017 Solution : b. i. At the maximum height, H : Thus the time taken to reach maximum height is given by ii. Apply

c. Flight time = 2(the time taken to reach the maximum height) SF017 Solution : c. Flight time = 2(the time taken to reach the maximum height) Hence the horizontal range, R is d. When the ball reaches point P thus The velocity of the ball at point P, Horizontal component: Vertical component:

therefore the direction of ball’s velocity is SF017 Solution : Magnitude, Direction, therefore the direction of ball’s velocity is e. The time taken from point O to Q is 45.0 s. i. position of the ball when t = 45.0 s, Horizontal component : from positive x-axis anticlockwise

therefore the position of the ball is (4500 m, 2148 m) SF017 Solution : Vertical component : therefore the position of the ball is (4500 m, 2148 m) e. ii. magnitude and direction of ball’s velocity at t = 45.0 s , Horizontal component :

therefore the direction of ball’s velocity is SF017 Solution : Magnitude, Direction, therefore the direction of ball’s velocity is from positive x-axis anticlockwise

SF017 Example 2.10 : A transport plane travelling at a constant velocity of 50 m s1 at an altitude of 300 m releases a parcel when directly above a point X on level ground. Calculate a. the flight time of the parcel, b. the velocity of impact of the parcel, c. the distance from X to the point of impact. (Given g = 9.81 m s-2) Solution : 300 m d X

The parcel’s velocity = plane’s velocity thus SF017 Solution : The parcel’s velocity = plane’s velocity thus a. The vertical displacement is given by Thus the flight time of the parcel is and

b. The components of velocity of impact of the parcel : SF017 Solution : b. The components of velocity of impact of the parcel : Horizontal component : Vertical component : Magnitude, Direction, therefore the direction of parcel’s velocity is from positive x-axis anticlockwise

c. Let the distance from X to the point of impact is d. SF017 Solution : c. Let the distance from X to the point of impact is d. Thus the distance, d is given by

Exercise 2.4 : Use gravitational acceleration, g = 9.81 m s2 SF017 Exercise 2.4 : Use gravitational acceleration, g = 9.81 m s2 A basketball player who is 2.00 m tall is standing on the floor 10.0 m from the basket, as in Figure 2.13. If he shoots the ball at a 40.0 angle above the horizontal, at what initial speed must he throw so that it goes through the hoop without striking the backboard? The basket height is 3.05 m. ANS. : 10.7 m s1 Figure 2.13

SF017 Exercise 2.4 : An apple is thrown at an angle of 30 above the horizontal from the top of a building 20 m high. Its initial speed is 40 m s1. Calculate a. the time taken for the apple to strikes the ground, b. the distance from the foot of the building will it strikes the ground, c. the maximum height reached by the apple from the ground. ANS. : 4.90 s; 170 m; 40.4 m A stone is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m s1 at 40 above the horizontal. How far above or below its original level will the stone strike the opposite wall? ANS. : 10.3 m below the original level.

SF017 THE END… Next Chapter… CHAPTER 3 : Momentum and Impulse