Question

If the speed of a motor car is doubled, how much more distance will it cover before stopping under the same retarding force?

Answer 1

Stopping Distance:

$s=\frac{v^2-u^2}{2a}$

$s=\frac{0-u^2}{2\left(\frac{-F}{m}\right)}$

$s=\frac{m{u}^2}{2F}$

As m and F are constant

$s\propto u^2$

$\frac{s_2}{s_1}={\left(\frac{u_2}{u_1}\right)}^2$

$\frac{s_2}{s_1}={\left(\frac{2u}{u}\right)}^2$

$s_2=4s_1$

Stopping distance increases to 4times