When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance.
Suppose, the velocity of moving vehicle is $v_{0}$. After applying brakes retardation : $-a$
Distance covered : $d_{\mathrm{s}}$ (stopping distance)
Final velocity : $v=0$
In equation $v^{2}-v_{0}^{2}=2 a x$
$\therefore 0-v_{0}^{2}=2(-a)\left(d_{\mathrm{s}}\right)\left(\because v=0, a=-a, s=d_{\mathrm{s}}\right)$
$\therefore v_{0}^{2}=2 a d_{\mathrm{s}}$
$\therefore d_{\mathrm{s}}=\frac{v_{0}^{2}}{2 a}$
Here, $a$ is value of retardation and it is constant.
Thus, stopping distance is proportional to square of the initial velocity.
$\therefore d_{\mathrm{s}} \propto v_{0}^{2}$
Stopping distance, $d_{\mathrm{s}} \propto v_{0}^{2}$
$\therefore \frac{(d \mathrm{~s})_{2}}{(d \mathrm{~s})_{1}}=\frac{\left(v_{0}\right)_{2}^{2}}{\left(v_{0}\right)_{1}^{2}}=(2)^{2}=4$
$\therefore$ Stopping distance becomes  $4$ times.