Question

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

• A. $1s$
• B. $2s$
• C. $3s$
• D. $4s$

Answer 1

Answer: (B)

$2s$

Body at rest starts sliding down.
Given: $t=4s$
Here, $u=0\Rightarrow a=g\sin \theta$
Distance covered, $s=0\times 4+\frac{1}{2}\left(g\sin \theta \right){\left(4\right)}^2$ ......(i)

In second case:
Body rolls up from rest.
Hence, $u=0\Rightarrow a=-g\sin \theta$
Here, distance covered
$\frac{s}{4}=0\times t-\frac{1}{2}\left(g\sin \theta \right)\left(t^2\right)$ .....(ii)

Dividing equation (i) and (ii), we get:
$\frac{s}{\frac{s}{4}}=\frac{\frac{1}{2}\left(g\sin \theta \right){\left(4\right)}^2}{-\frac{1}{2}\left(g\sin \theta \right){\left(t\right)}^2}\Rightarrow t=2s$