When the speed of a car is $u$, the minimum distance over which it can be stopped is $s$. If the speed becomes $nu$, what will be the minimum distance over which it can be stopped during same time? Assume the acceleration is constant and same in both cases.

The correct option is D $n^{2}s$
For 1st case:
Using 3rd equation of kinematics,
$v^2-u^2=2as$
Here, $v=0$ and $a=-ve$ and displacement = s
$0-u^2=-2as$
$\Rightarrow a=\frac{u^2}{2s}$ .......(1)

In 2nd case:
$u\to nu$
$v\to 0$
and $a=$ constant

Let displacement = s'

$0^2-(nu{)}^2=-2a{s}^{\prime }$

Use 'a' from eq(1)
$n^2{u}^2=2\left(\frac{u^2}{2s}\right)s^{\prime }$
$\Rightarrow s^{\prime }=n^{2}s$