A body sliding on a smooth inclined plane requires 4 s to reach the bottom, starting from rest at the top. How much time does it take to cover one-fourth the distance starting from rest at the top?

When a body slides on an inclined plane, component of weight along the plane produces an acceleration.

$a=\frac{mg\sin \theta }{m}=g\sin \theta =\text{ Constant }$

If s is the length of the inclined plane, then

$\begin{array}{l}s=0+\frac{1}{2}a{t}^2=\frac{1}{2}g\sin \theta \times t^2\\ \frac{s^{\mathrm{\prime }}}{s}=\frac{t^{\mathrm{\prime }2}}{t^2}\text{ or }\frac{s}{s^{\mathrm{\prime }}}=\frac{t^2}{t^{\mathrm{\prime }2}}\\ \text{ Given }t=4\mathrm{s}\text{ and }{s}^{\mathrm{\prime }}=\frac{s}{4}\\ t^{\mathrm{\prime }}=t\sqrt{\frac{s^{\mathrm{\prime }}}{s}}=4\sqrt{\frac{s}{4s}}=\frac{4}{2}=2\mathrm{s}\end{array}$