(^396) A Textbook of Engineering Mechanics
and now substituting initial velocity (u) = 6.78 m/s, acceleration (a) = 2.6 m/s^2 and distance trav-
elled (s) = 10 m in the equation of motion i.e. v^2 = u^2 + 2 as,
v^2 = (6.78)^2 + (2 × 2.6 × 10) = 98 or v = 9.9 m/s
Similarly, calculate the mean accelerations and velocities of the automobile at the end of each
10 metres and prepare the table as shown below.
sa aav δs
12
2
aa+
s 1 – s 2 u^2 + 2as v
0 2.2
2.3 10 46 6.78
10 2.4
2.6 10 98 9.9
20 2.8
2.4 10 146 12.08
30 2.0
1.8 10 182 13.49
40 1.6
1.3 10 218 14.76
50 1.0
From the above table, we find that the average velocity of the automobile between 40 and 50
or 45 m of its start is 14.76 m/s Ans.
Example 18.10. A car, starting from rest has an acceleration a in m/s^2 after t second from its
start as given in the following table:
t 0 4 8 12 16 20
a 10 9.6 8.4 0.4 3.6 0
Find the speed of the car at the end of each 4 sec interval and the distance traversed.
Solution. In the first 4 sec the mean acceleration
10 9.6 9.8 m/s 2
2
- =
∴ Increase in velocity during these 4 sec
= 9.8 × 4 = 39.2 m/s
and velocity at the end of 4 sec
= 0 + 39.2 = 39.2 m/s
∴ Average velocity in the first 4 sec
039.2
19.6 m/s
2
==
and distance traversed in the first 4 sec
= 19.6 × 4 = 78.4 m
∴ Distance traversed up to the end of 4 sec.
= 0 + 78.4 = 78.4 m