Topic 2 :

Kinematics of linear motion

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2.0 Kinematics of Linear Motion

Kinematics

Kinematics is the study of how things move

linear motion

The motion of an object along a straight path is called a linear motion.

Overview:

2.1 Linear Motion

a) Define

i. instantaneous velocity, average velocity and uniform velocity

ii.instantaneous acceleration, average acceleration and uniform

acceleration

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b) Compare the following quantities

i. instantaneous velocity, average velocity and uniform velocity

ii. instantaneous acceleration, average acceleration and uniform

acceleration

c) Sketch displacement-time, velocity-time and acceleration-times graphs.

4. Interpret displacement-time, velocity-time and acceleration-times graphs.

e) Determine the distance travelled, displacement, velocity and

acceleration from appropriate graphs.

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2.1. Linear motion (1-D)

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Example 1 :

Distance travelled = 200m�Displacement =120 m, in the direction of Northeast

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Example 2

An airplane flies 600 km north and then 400 km to the east. What is the distance and displacement travelled by this airplane?

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Total distance traveled d = 600 + 400 = 1000 km

Displacement, s = = 721.11 km

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Speed, v

• tells us how fast an object is moving.

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• scalar quantity

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• defined as the rate of total distance travelled

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Average speed is defined as a distance travelled divided by the elapsed time.

or

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SI unit : m s^–1

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Velocity, v

• tells us how fast object is moving & in which direction it is moving.
• vector quantity,

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• defined as change of displacement per unit time or the rate of change of displacement

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SI unit : m s^–1

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Velocity, v (vector quantity, unit→ms^-1 )

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Example 3:

An insect crawls along the edge of a rectangular swimming pool of length 27m & width 21m. If it crawls from corner A to corner B in 30 min.

(a) What is its average speed ?

(b) What is the magnitude of its average velocity ?

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Solution

Given : length L = 27m; width, W = 21m

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Total distance travel, d = 27 + 21 = 48 m

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Displacement, s =

(a)

(b)

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• An acceleration may be due to:

1) change in speed (magnitude),

2) change in direction or

3) change in both speed and direction.

• Velocity is vector quantity, 🡪 a change in velocity may thus involve either or both magnitude & direction.

Acceleration, a

• defined as the rate of change of velocity

• SI unit : m s^–2

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Acceleration, a (vector quantity, unit→ms^-2 )

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Deceleration :

object is slowing down

( direction of acceleration is opposite to the direction of the motion or velocity).

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Acceleration 🡪

Deceleration 🡪

Car in figure (a) & (d)

Car in figure (b) & (c)

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Example 4:

A car travels in a straight line along a road. Its distance, s is given as a function of time t by the equation:

(a) Calculate the average velocity of the car for the time interval t = 0 s and t = 10 s.

(b) Calculate the instantaneous velocity of the car at t = 5 s.

(c) Calculate the instantaneous acceleration of the car at t = 5 s.

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Solution

Given :

(a) At t₁= 0 s, s₁ = 0 m

At t₂=10 s, s₂ = 2.4(10)² – 0.12 (10)³ = 120 m

Average velocity,

(b) Instantaneous velocity,

At t = 5s,

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(c) Instantaneous acceleration,

At t = 5s,

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a car is moving with equal distance in same time interval

• No change in velocity
• No acceleration
• Nothing is change

Uniform motion

Non uniform motion

a car is moving with different distance in same time interval

• Change in velocity
• uniform acceleration.

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Graphical Methods

Displacement – time (s-t) graphs

Instantaneous velocity = gradient of (s-t) graph

We can determine velocity at point A by drawing a tangent line at A

Gradient at B is zero which indicates that velocity = 0 at that instant.

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Graphs for linear motion ( s – t )

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Velocity versus time graph (v-t)

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• At point A, the gradient is positive – a > 0(speeding up)
• At point B, the gradient is zero – a= 0
• At point C, the gradient is negative – a < 0(slowing down)

Uniform velocity

Uniform acceleration

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Graphs for linear motion ( v – t )

• Area under graph=Displacement

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Graphs for linear motion ( a – t )

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Physical meaning of displacement-time, velocity-time and

acceleration-time.

Area under the graph

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A toy train moves slowly along a straight track according to the displacement, s against time, t graph in Figure 2.1.

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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 = 11 s.

e. Determine the distance travelled by the toy train.

Figure 2.1

Example 5

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Solution :

a. 0 to 10 s : The train at rest.

10 to 14 s : The train moves in positive direction at a

constant velocity of 1.50 cm s⁻¹.

b.

Solution :�

c.

d.

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Solution :

e. The distance travelled by the train, d is given by

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A velocity-time (v-t) graph in Figure 2.2 shows the motion of a lift.

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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.

Example 6

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Solution :

1. 0 to 5 s : Lift moves upward from rest with a constant acceleration

of 0.4 m s⁻².

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 0.2 m s⁻².

15 to 20 s : Lift moving with constant velocity of 4 m s⁻¹.

20 to 25 s : Lift decelerates at a constant rate of 0.8 m s⁻².

25 to 30 s : Lift at rest or stationary.

30 to 35 s : Lift moves downward with a constant acceleration of 0.8 m s⁻².

35 to 40 s : Lift moving downward with constant velocity of 4 m s⁻¹.

40 to 50 s : Lift decelerates at a constant rate of 0.4 m s⁻² and comes to rest.

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Solution :

b.

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Solution :

c. i.

A₁

A₂

A₃

A₄

A₅

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Solution :

c. ii.

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d.

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1. Can you accelerate a body without speeding up or slowing down? Is it possible?

2. A car is traveling at 30 km h^–1 to the north. Then it turns to the west without changing its speed. Is the car accelerating?

3. How would you draw a displacement time graph for a stationary object?

4. What would the gradient of a distance time graph represent?

5. What does the area a speed-time graph represent?

Test your understanding

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Check your answers.

1. Yes, it is possible.

2. Yes, the car is accelerating.

3.

4. Speed

5. Distance

a) Apply equations of motion with uniform acceleration:

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2.2 Uniformly accelerated motion

b) Apply equations of motion for free fall:

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(Experiment 3: Free fall motion)

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Uniformly accelerated motion

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where v : final velocity

u : initial velocity

a : uniform (constant) acceleration

t : time

s : displacement

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Equations of motion

Sign Convention

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Note:

• Acceleration : +ve, Deceleration: -ve
• Special case : uniform velocity,

u = v ⇒ a = 0 , s = ut

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Images that are equal spaced indicated that the object is moving with constant speed or velocity.

-Velocity constant

-No acceleration ( a=0)

A racing car is running on track

A cyclist is starting a race

An increasing distance between the images shows that the object is speeding up

-change in velocity

-uniform acceleration

- initial velocity u = 0

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A car is speeding up after a green signal

A car is stopping at a red light

A car is slowing down for a red light

Uniformly accelerated motion.

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A plane lands on a runway at velocity 50 m s⁻¹ and decelerates at constant rate. The plane travels 1.0 km before stops. Calculate

a. the deceleration of the plane.

b. the time taken for the plane to stop.

Solution :

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a. Use

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Hence the deceleration of the plane is 1.25 m s⁻².

Example 7

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Solution :

b. By using the equation of linear motion,

OR

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A particle moves along horizontal line according to the equation

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Where s is displacement in meters and t is time in seconds.

At time, t =2 s, determine

a. the displacement of the particle,

b. Its velocity, and

c. Its acceleration.

Solution :

a. t =2 s ;

Example 8

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Solution :

b. Instantaneous velocity at t = 2 s,

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Use

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Thus

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Solution :

c. Instantaneous acceleration at t = 2 s,

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Use

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Hence

Uniformly accelerated motion

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where v : final velocity

u : initial velocity

g : 9.81ms^-2

t : time interval

s : vertical displacement

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Equations of motion (free fall)

Free falling body(θ=90º)

• Free fall motion is the vertical motion of a body at constant acceleration, g under the influence of gravitational field without air resistance.
• In the earth’s gravitational field (9.81 ms^-2), the constant acceleration
• known as acceleration due to gravity or free-fall acceleration or gravitational acceleration. (a=-g)
• the value is g = 9.81 m s⁻²
• the direction is towards the centre of the earth (downward). Negative sign indicates that it is always directed downward.

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• Note:
• In solving any problem involving free falling body or free fall motion, the assumption made is to ignore the air resistance.

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Free fall �downward at u=0ms^-1(drop)

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Case 1

An apple is drop from a certain height to the ground.

Case 2

An apple is thrown downwards from a certain height to the ground.

Free fall �downward at u=0ms^-1(drop)

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Example

An apple is drop from a certain height to the ground.

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Example

An apple is thrown downwards from a certain height to the ground.

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s=+(max height),

v=0 (stop to move

downwards)

s=0,(initial)

u=+(upward)

t= 0

s=0(same level

with initial),

v=- (downward)

,increasing

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s=-(below initial

level),

v=- (downward,

max value)

t=max, flight time

s=+(above initial

level)

v=+ (upward)

,decreasing

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s=0(same level with initial),

v=- (downward)

t= (max, flight time)

s=0,(initial)

u=+(upward)

t=0

s= +(max height),

v=0 (stop to move

downwards)

s=+(above initial level)

v=- (downward)

, increasing

s=+(above initial

level),

v=+ (upward)

,decreasing

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Example

An apple is thrown upwards from the ground with initial velocity equal to 20 ms^-1 and fall back to the ground.

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Example

An apple is thrown upwards from certain height with initial velocity equal to 20 ms^-1 and fall back to the ground.

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• Sign convention:

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• Table 2.1 shows the equations of linear motion and free falling body.

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Table 2.1

From the sign convention thus,

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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 displacement is

s = −150 m

Hence

Example 9

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Solution :

b. The book’s velocity is given by

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Therefore, the book’s velocity is 54.2 m s⁻¹ downwards.

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OR

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A book is thrown downwards 150 m from the ground at 20 ms^-1. 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 displacement is

s = −150 m

Hence

Example 10

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Solution :

b. The book’s velocity is given by

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Therefore, the book’s velocity is 57.81 m s⁻¹ downwards.

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OR

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A ball is thrown straight upwards with an initial velocity of 15 m s⁻¹ from the roof of a building. The building is 40 m high. Calculate

a. the maximum height of the ball from the ground.

b. the time taken for the ball returns to the original level.

c. the time taken for the ball strikes the ground if it misses the edge of the roof.

d. the velocity of the ball just before strike the ground.

(Given g = 9.81 m s⁻²)

Example 11

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Solution :

a. At point B (highest point), v = 0 and s = h thus

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Therefore, the maximum height H from the ground is

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b. From point A to C, s = 0 m thus

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Solution :

c. From point A to D, s= −40 m thus

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By using

OR

Time don’t have negative value.

a

b

c

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Solution :

d. Time taken from A to D is t = 4.77 s thus

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From A to D, s = −40 m

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Therefore, the ball’s velocity at D is 31.8 m s⁻¹ downwards.

OR

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1. 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

2. A falling stone takes 0.30 s to travel past a window 2.2 m tall as shown in Figure 1.

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From what height above the top of the windows did the stone fall?

ANS. : 1.75 m

Test your understanding

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