What is stopping distance for a vehicle? What will be the stopping distance if the initial velocity is doubled?

(N/A) When brakes are applied to a moving vehicle,the distance it travels before coming to a complete stop is called the stopping distance.
Suppose the initial velocity of the moving vehicle is $v_{0}$. After applying brakes,the retardation is $-a$.
Let the distance covered be $d_{s}$ (stopping distance) and the final velocity be $v = 0$.
Using the kinematic equation $v^{2} - v_{0}^{2} = 2as$:
$0 - v_{0}^{2} = 2(-a)(d_{s})$
$v_{0}^{2} = 2ad_{s}$
$d_{s} = \frac{v_{0}^{2}}{2a}$
Here,$a$ is the magnitude of retardation,which is constant.
Thus,the stopping distance is proportional to the square of the initial velocity: $d_{s} \propto v_{0}^{2}$.
If the initial velocity is doubled,i.e.,$(v_{0})_{2} = 2(v_{0})_{1}$,then:
$\frac{(d_{s})_{2}}{(d_{s})_{1}} = \frac{(v_{0})_{2}^{2}}{(v_{0})_{1}^{2}} = (2)^{2} = 4$.
Therefore,the stopping distance becomes $4$ times the original distance.