(^398) „„„„„ A Textbook of Engineering Mechanics
Similarly, in the next 5 seconds, the distance covered
= 100 – 20 = 80 m
∴ Average velocity after 7.5 seconds of start (or in all 5 seconds from 5 to 10)
= 80/5 = 16 m/s
and rate of increase in average velocity (or acceleration ) in 5 seconds (from 2.5 to 7.5 seconds)
16 – 4 2.4 m/s 2
5

Similarly, calculate the distances covered, average velocities and accelerations in each 5 sec-
onds and prepare the table as shown below.
ts δt δs
s
v
t
δ

δ
v 2 – v 1
21
21

vv

a

tt

=

(t 2 – t 1 )(s 2 – s 1 )

00

5204

5 20 12 12/5 = 2.4

58016

10 100 10 10/5 = 2.0

5 130 26

15 230 – 6 – 6/5 = – 1.2

5 100 20

20 330 – 10 – 10/5 = – 2.0

55010

25 380 – 6 – 6/5 = – 1.2

5204

30 400

Velocity of the car after 10 and 20 seconds from start

From the average velocity column, we find that average velocity of the car after 7.5 seconds

= 16 m/s

and average velocity of the car after 12.5 seconds

= 26 m/s

∴ Average velocity of the car after 10 seconds from start

16 26

21 m/s

2

+

== Ans.

Similarly, velocity of the car after 20 seconds from start

20 10

15 m/s

2

+

== Ans.

Acceleration of the car after 10 and 20 seconds from start

From the acceleration (i.e. last) column, we find that acceleration of the car after 10 and 20

seconds from start is 2.0 m/s^2 and – 20 m/s^2 respectively. Ans.