Question

A car moving at a certain speed stops on applying brakes within 16 m. If the speed of the car is doubled, maintaining the same retardation. Then at what distance does it stop? Also calculate the percentage change in this distance.

Answer 1

Given v $=$ 0, u $=$ initial velocity
retardation $=$ a, s $=$ 16 m
$\therefore a=\frac{-u^2}{2s}$
$u_1=2u;a_1=a;s_1=?$
$s=\frac{-4u^2}{2a}$
From (1)
$s_1=\frac{-4u^2}{2(-u^2)}\left(2s\right)$
$s_1=4\times 16=64m$
% change $=\frac{s_1-s}{s}\times 100$
$=\frac{4s-s}{s}\times 100=300$%