A body sliding on a smooth inclined plane requires $4 \mathrm{~s}$ to reach the botom, starting from rest at the top. How much time does it take to cover one fourth the distance starting from rest at the top?

Question

Aicharya · Accepted Answer

When a body slides on an inclined plane, component of weight along the plane produces an acceleration.
$a=\frac{m g \sin \hat{\theta}}{m}=g \sin \theta=$ constant. If $s$ be the length of the inclined plane, then
$\begin{array}{l}
\qquad s=0+\frac{1}{2} a t^{2}=\frac{1}{2} g \sin \theta \times t^{2} \\
\frac{s^{\prime}}{s}=\frac{t^{2}}{t^{2}} \text { or } \frac{s}{s^{\prime}}=\frac{t^{2}}{t^{\prime 2}} \\
\text { Given, } t=4 s \text { and } s^{\prime}=\frac{s}{4} \\
\qquad t^{\prime}=1 \sqrt{\frac{s^{\prime}}{s}}=4 \sqrt{\frac{s}{4 s}}=\frac{4}{2}=2 \mathrm{~s}
\end{array}$

A body sliding on a smooth inclined plane requires \(4 \mathrm{~s}\) to reach the botom, starting from rest…

Solution