A man is moving with 36 kmph. The time of reaction is 0.9 seconds. On seeing an obstacle in the path, he applies brakes and decelerates at 5 m/s2, the total distance covered before he stops is:
19 m
Given
u = 10 m/s
Distance covered within reaction time, d = u x t = 10 x 0.9 = 9 m
a = -5 m/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 = (10) (2) + (1/2) (-5)(2)(2)
s = 20 + (-10) = 10
Total displacement = 9 + 10 = 19