CHAPTER 2 KINEMATICS OF LINEAR MOTION

Chapter 2 : Kinematics of Linear Motion

2.1 Linear Motion
2.2 Uniformly accelerated

motion

Matriculation Physics 1

2.0 Introduction

Kinematics

Description of the motion of objects
without consideration of what
causes the motion.

2

2.1 Linear Motion

Learning Outcomes

At the end of this chapter, students should be able to:

• Define

i) Instantaneous velocity, average velocity and uniform
velocity

ii) Instantaneous acceleration, average acceleration and
uniform acceleration

• Compare the following quantities

i) instantaneous velocity, average velocity and uniform velocity

ii) instantaneous acceleration, average acceleration and
uniform acceleration

• Sketch displacement-time, velocity-time and graphs.

• Interpret displacement-time, velocity-time and graphs.

• Determine the distance travelled, displacement, velocity
and acceleration from appropriate graphs.

• Discuss the physical meaning of displacement-time, 3
velocity-time and acceleration-time graphs.

Linear motion – motion of an object along a
straight line path.

Distance, d
-- the total length of travel in moving from one

location to another.
-- scalar quantity.
-- always positive.

Displacement, s
-- straight line distance from the initial

position to the final position of an object.

-- Vector quantity

-- can be positive, negative or zero.

4

Initial final
Position Position

5

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

6

Example

An air plane flies 600 km north and then 400 km
to the east.

N 400 km E
Final

600 km N

Begin

Total distance traveled, d = 600 + 400 = 1000 km

Displacement, s = 6002  4002 = 721.11 km (magnitude)

7

Test your understanding

1. You walk from your house to friend’s house then
to the grocery shop. Calculate:

(i) the distance traveled.
(ii) the displacement.

2. An athlete runs four laps of a 400 m track. 8
What is the athlete’s total displacement?

Speed, v
-- Rate of change of distance

v  distance traveled d
time to travel that distance t

-- S.I. unit : m s–1 ; scalar quantity.

9

Velocity, v
-- tells us how fast object is moving & in which

direction it is moving.

-- is the rate of change of displacement.

  displacement  s
velocity, v
travel time t

-- vector quantity ; SI unit : m s–1

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Average velocity, vav

 changein displacement
vav 
time taken for the change

  s  s2  s1
v av
t t2  t1

Instantaneous velocity, v
-- velocity at a specified position or instant of

time along the path of motion.

-- commonly referred as `velocity at point A `
or ` velocity at time t `.

limv  s  ds
t0 t dt

11

-- equal to the gradient at any point on the
curve of a displacement – time (s – t) graph.

Slope BE = average velocity
Slope at C = Instantaneous velocity

t
s

12

Uniform velocity
-- motion with constant velocity
-- acceleration, a = 0 m s–2

The displacement increases by equal amounts in
equal times.

13

Example 1
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 speed ?
(b) What is the magnitude of its average velocity ?

14

Solution
Given : Length, L = 27m; Width, W = 21m

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

Displacement, s = 272  212  34.21m

(a) Speed, v  d  48  0.0267 m s 1

t 30(60)

(b) Averagevelocity,   s  34.21
vav 30(60)
t
 0.019 m s 1

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Acceleration, a

-- time rate of change of velocity.

acceleration  change in velocit y
time taken

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

-- An acceleration may due to:
1) change in speed (magnitude),
2) change in direction or
3) change in both speed and direction.

16

Quick Test
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?

Answer : YES !
Reason : there is a change in direction

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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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Car in figure (a) & (d)  accelerating
Car in figure (b) & (c)  decelerating

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Average acceleration
-- change in velocity divided by the time taken

to make the change.

 change in velocit y  v2  v1  v
a
time to make the change t 2  t1 t

-- vector quantity.
-- SI unit : m s–2

Instantaneous acceleration
-- acceleration at a particular instant of time.

lima  v  dv  d 2s
t dt dt 2
t  0

21

Uniform acceleration
-- the acceleration does not depend on time or
always constant.

a  dv  constant
dt

-- velocity changes at a constant rate.

A v - t plot is a straight line whose gradient is
equal to acceleration.

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• For linear motion, + and – signs is used to
indicate direction of motion, velocity &
acceleration.
• With horizontal direction we may take :

-- to the right ( + )
-- to the left ( – )
• With vertical direction we may take :
-- upward ( + )
-- downward ( – )

The sign convention you choose is
entirely up to you. It doesn’t matter as
long as you keep the same sign
convention for the entire calculation.

23

Example 2
A car travels in a straight line along a road. Its
distance, s is given as a function of time, t by the
equation :

s(t)  2.4t 2  0.12t 3

(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 : s(t)  2.4t 2  0.12t 3

(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,   s2  s1  120  0
vav
t2  t1 10  0

  12 m s 1
vav

(b) Instantaneous velocity, v  ds
dt

 d (2.4t 2  0.12t 3 )
dt

25

v  4.8t  0.36t 2

At t = 5s, v  4.8(5)  0.36(5)2

v  15 m s1

(c) Instantaneous acceleration,

a  dv  d (4.8t  0.36t 2 )
dt dt
a  4.80.72t

At t = 5s, a  4.8  0.72(5)

a  1.2 m s2

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Graphical Methods
Displacement – time (s - t) graphs

Instantaneous velocity v  ds = gradient of (s-t) graph
dt

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velocity – time (v - t) graphs

acceleration, a = dv = gradient of (v - t) graph
dt

Displacement of the object = shaded area
under the (v - t) graph

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Gradient of the slope = a Area under the slope = s

Acceleration, a = constant
value

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Example 3 ( PSPM Session 2008/09 )

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Figure 5 shows a graph of displacement x against
time t of an object moving along x-axis. Calculate
(a) Average velocity for the time interval, 1 s to 4 s.
(b) Average speed for the time interval, 1 s to 4 s.
(c) Instantaneous velocity at t = 2.5 s.
(d) Instantaneous acceleration at t = 5.5 s.

Solution

(a) v  saverage  (3)  2  1.67 m s1
t 4 1

(b) v average x  2  4  3  3 m s1
t 4 1

31

(c)  dx  gradient at t  2.5 s
d tt 2.5 s
v|

 (3)  4
32

 7 m s1

(d)   dv  0 m s2
a
dt

32

Test your concept

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?

33

6. The distance – time graph in figure below
represents the motion of an ant in 7 seconds.
Describe its motion.

34

7. A body moves along the x-axis. Assume that a
positive sign represents a direction to the right.
The velocity, v of the body is related to time, t
through the equation

v  2  3t 2

where v and t are measured in m s-1 and s

respectively. t = 0 when x = 0.
Determine
(a) the displacement
(b) the acceleration
at the instant of time t = 1 s

(1 m, -6 ms-2)

35

22.2.2 ULniinfeoarmr Mlyoaticocnelerated motion

2.2.1 Uniform acceleration

Learning Outcomes

At the end of this chapter, students should be
able to:
• a) Apply equations of motion with uniform

acceleration:

a vu v  u  at
t
s  ut  1 at2
s  1 (u  v)t 2
2
v2  u2  2as

36

Kinematics Equation for uniform acceleration

Assume a car has uniform acceleration & consider
the motion between X and Y :

u = initial velocity ( velocity on passing X ) 37
v = final velocity ( velocity on passing Y )
a = acceleration
s = displacement ( in moving from X to Y )
t = time taken ( to move from X to Y )

velocity – time graph for the car

Acceleration, a = gradient of graph v – t

a  v  u  at  v u
t

38

rearranged to give: 39

v  u  a t … (1)

Distance traveled, s = area under the graph
= area of trapezium

s  1 (u  v) t … (2)
2

Substitute (1) into (2) :

s  1 (u  u  a t) t
2

s  u t  1 a t 2 … (3)
2

From (1) : v = u + at , get an expression for t :

t  vu
a

Substitute into (2) :

s  1 (v  u)(v  u )
2a

 (v  u)(v  u)
2a

2as  v2  u 2

v 2  u 2  2 a s … (4)

40

Example 4

The driver of a pickup truck going 100 km h–1
applies the brakes, giving the truck a uniform
deceleration of 6.50 m s–2 while it travels 20.0m.
(a) What is the speed of the truck in kilometers

per hour at the end of this distance ?
(b) How much time has elapsed ?

Solution

Given :
a = – 6.50 m s–2 ( deceleration )
s = 20.0 m
u = 100 km hour –1

 100(1000)m  27.78 m s1
60(60)s

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(a) Final velocity , v = ?

v2  u 2  2as
v2  (27.78)2  2(6.5)(20)
v2  511.7284
v  22.62 ms1

Convert to km h–1

22.62 km
v  1000

1 hour
60(60)

v  81.43 k m hour1

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(b) Assume : time elapsed, t

v  u  at

22.62  27.78  (6.5)t

 6.5t  5.16

t  5.16
6.5

t  0.794 s

43

Example 5

A park ranger driving on a back country road
suddenly sees a deer ‘frozen’ in his headlights.
The ranger, who is driving at 11.4 m s−1
immediately applies the brakes and slows with
an acceleration of 3.8 m s−2.
a) If the deer is 20.0 m from the ranger’s

vehicle when the brakes are applied, how
close does the ranger come to hitting the
deer?
b) How much time is needed for the ranger’s
vehicle to stop?

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Solution

s
Given: u = 11.4 m s−1 ; a = − 3.80 m s−2
• 1st find the distance traveled before stopped

45

From: v2  u 2  2as
(0)2  (11.4)2  2(3.8)(s)

s   129.96  17.1 m
 7.6

The distance between the stopped vehicle & deer:

x  2017.1 2.9 m

(b) Time needed to stop = ?

From: v  u  at

0  11.4  (3.8)t

t  3.0s

46

Example 6

A toy car moves with an acceleration of 2 m s–2
from rest for 2.0 s. It then moves with constant
velocity for another 3.0 s. It finally comes to rest
after another 1.0 s.

(a) Sketch a velocity-time graph to shown the
motion of the toy car.

(b) What is the velocity of the toy car after first
2 seconds ?

(c) Calculate the deceleration of the car.
(d) What is the total displacement of the car for

the whole journey.

(e) Sketch the acceleration-time graph for the
motion of toy car.

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Solution Deceleration
(a) v (m s–1) V decreases
from 4 m s–1 to
4 0 m s–1 in 1 s

t (s)

02 56

(b) Using : v  u  at

 0  2(2)
v  4 m s 1

48

(c) Using : v  u  at

a  vu  04
t1

a  4 m s 2

(d) Displacement, s = area under the graph

s  1 (3  6)(4)
2

s  18 m

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(e) Acceleration – time graph
a (m s–2)

2 t (s)
02
56
–4

50

Follow up exercise

1 The speed limit in a school zone is 30 km h–1. A

driver traveling at this speed sees a child run

onto the road 13 m ahead of his car. He applies

the brakes and the car decelerates at a uniform

rate of 8.0 m s–2. If the driver’s reaction time is

0.25 s, will the car stop before hitting the child ?

~ U Think!~

2 (a) Is it possible for an object moving at non zero

velocity has a zero acceleration? Explain.

(b) A car is capable of accelerating at 0.60 m s–2.

Calculate the time needed for this car to go from

a speed of 5.5 m s–1 to a speed of 8.0 m s–1.

(4.17 s) 51