Engineering Mechanics

Chapter 18 : Motion Under Variable Acceleration „„„„„ 397

Similarly, in the next 4 sec the mean acceleration

9.6 8.4 9m/s 2

2

+

==

∴ Increase in velocity during these 4 sec
= 9 × 4 = 36 m/s
and velocity at the end of these 4 sec
= 39.2 + 36 = 75.2 m/s
∴Average velocity in these 4 sec
39.2 75.2
57.2 m/s
2

+

==.

and distance traversed in these 4 sec
= 57.2 × 4 = 228.8 m
∴ Distance traversed up to the end of these 4 sec
= 78.4 + 228.8 = 307.2 m
Similarly, calculate the mean accelerations, velocities and distances traversed by the car at the
end of each 4 sec and prepare the table as shown in the following table.

ta aav δt increase vav δs

in vv s

12

2

aa+

(t 2 – t 1 )(aav × t)^12

2

vv+

(vav × δt)

010 0 0

9.8 4 39.2 19.6 78.4

4 9.6 39.2 78.4

9.0 4 36.0 57.2 228.8

8 8.4 75.2 307.2

7.4 4 29.6 90.0 360.0

12 6.4 104.8 667.2

5.0 4 20.0 114.8 459.2

16 3.6 124.8 1126.4

1.8 4 7.2 128.4 513.6

20 6 132.0 1640.0

In the above table is given the speed of car and distance traversed at the end of each
4 seconds. Ans.

Example 18.11. A car starts from rest, and moves along a straight line. The distance
covered (s) in seconds (t) from the start, were observed to be as under :

t 0 51015202530

s 0 20 100 230 330 380 400

Calculate the velocity and acceleration of the car after 10 and 20 seconds from start.

Solution. In the first 5 seconds, the distance covered = 20 m

∴ Average velocity after 2.5 seconds of start (or in all 5 seconds from 0 to 5)

= 20/5 = 4 m/s