Question

A man is moving with 36 kmph. On seeing an obstacle in the path, he applies brakes and decelerates at 5 m/s2, if the time of reaction is 0.9 seconds. Then the total distance covered before he stops is:

• A. 19 m
• B. 16 m
• C. 18m
• D. 17 m

Answer 1

The correct option is A

19 m

u =36 kmph = 10 m/s

Distance covered within reaction time, d = u x t = 10 x 0.9 = 9m

a = -5m/s² , v = 0 m/s,

we know that, a = (v - u) / t.

So, t = (v - u) / a = (-10)/(-5) = 2s

We also know that, the distance is given as

s = ut + (1/2) at²

s = (10m/s) (2s) + (1/2) (-5m/s)(2)(2)

s = 20 m + (-10m) = 10m

Total displacement = 9m + 10 m = 19m