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Applied Mechanics - Engineering

Q1:

Time required to stop a car moving with a velocity 20 m/sec within a distance of 40 m, is

A 2 sec

B 3 sec

C 4 sec

D 5 sec

E 6 sec

ANS:C - 4 sec

vi = 20m/sec
vf = 0m/sec
S = 40m
t = ?

Solution:
We know that,
S=vi * t + 1/2at^2.
But , Here , a=?

We know that,
2aS = vf^2 - vi^2.

2 * a * 40 = 0 - 20^2.
a = - 400/80 = -5m/sec^2.

Now putting values in eqn S = vi * t + 1/2at^2.
40m = 20 * t + (-5 * t^2)/2.
40m = 20t - 2.5t^2.
25t^2 - 200t + 400 = 0.
t^2 - 8t + 16 = 0.
(t-4)^2=0.
t - 4 = 0.
t = 4sec-> Ans