Question

The breaks applied to a car produce a negative acceleration of 6m/s 2 the car takes 2 seconds to stop after applying brakes, calculate the distance it travels during this time.

Answer 1

Retardation = 6 m/s^2
t = 2 sec
v = 0
Since retardation, 
(u-v)/t = 6
u/2 = 6
u = 12 m/s^2
We know, 
S= ut - (1/2)at^2
= 12*2 - (1/2)*6*2*2
= 24-12
= 12 m (Answer)