CHAPTER 2
Handout:
ACCELERATION
Acceleration of an object is defined as the rate of change of velocity per unit time.
SI unit of acceleration is ms-2
Where, change in velocity = final velocity – initial velocity
=v–u
velocity is measured in meters per second (m/s) and time in seconds (s)
a = v-u
t

If the velocity of an object increases, the object is undergoing acceleration. Hence, if the
velocity of an object decreases,
it is undergoing deceleration.

If the velocity of the object is constant, the acceleration is zero.
DECELERATION
Deceleration can be defined as the rate at which a moving object slows down, per unit time.
Deceleration also is known as negative acceleration. Hence it is denoted by – a.
SI unit of deceleration is ms-2
Where, change in velocity = final velocity – initial velocity
=v–u
Page 1 of 6
a = v-u
t
When in calculation the value of acceleration will be negative here but in the final answer, the
value of deceleration will itself be a positive number.
Examples:

A bus starts from rest and achieves a velocity of 20m/s in 10 seconds while moving to
the right. Calculate its acceleration.
Ans:
Given,
initial velocity, u = 0 m/s
final velocity, v = 20m/s
time, t = 10s
a = v-u
t
a = 20-0
10
a = 2m/s2
The acceleration of the bus is 2 m/s2 to the right.

A car travelling westwards at 30 ms−1 suddenly comes to a halt in 5 s. Find its
deceleration.
Ans:
Given,
initial velocity, u = 30 m/s
final velocity, v = 0m/s
time, t = 5s
a = v-u
t
a = 0 - 30
5
a = -6 m/s2
The deceleration of the car is -6 m/s2
Page 2 of 6
Page 3 of 6
VELOCITY - TIME GRAPH
•
The gradient of a distance-time graph gives you speed/ velocity.
•
The gradient of a velocity-time graph gives you acceleration.
•
The area under the velocity-time graph will give you the displacement travelled by the
said body.
Calculating Area under the graph
Page 4 of 6
Distance travelled = 1/2bh + lw
or,
Distance travelled = area of trapezium
SOLVED PROBLEM:
Question 01:
From the figure given of a travelling vehicle,
a) Explain what is happening between:
i. t= 0 and t = 2s.
ii. t = 2 and t = 5s.
b) Calculate:
i. acceleration of the vehicle
ii. the deceleration
iii. t = 5 and t = 8s.
iii. total distance travelled
Solution 01:
a) i. Between t= 0 and t = 2s, the vehicle is moving with constant acceleration.
ii. Between t = 2 and t = 5s, the vehicle is travelling at uniform velocity 3 m/s.
iii. Between t = 5 and t = 8s, the vehicle is undergoing deceleration.
b) i.
a = (v-u)/t
Page 5 of 6
=
(3-0) m/s
2s
= 1.5 m/s2
The acceleration of the vehicle is 1.5 m/s2
b) ii. a = (v-u)/t
a=
(0-3) m/s
3s
a = -1 m/s2
The deceleration of the vehicle is 1 m/s2
iii. Total distance travelled = Area under the graph = Area of the trapezium
= ½ x sum of parallel sides x height
= ½ x [8+(5-2)] x 3 m
= ½ x 11 x 3 m
= 16.5 m
The total distance travelled is 16.5 m.
Page 6 of 6