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(B) The distance $S$ covered by an object starting from rest under constant acceleration $a$ is given by $S = \frac{1}{2}at^2$.Since $a$ is constant,w

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$2$

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(B) The distance $S$ covered by an object starting from rest under constant acceleration $a$ is given by $S = \frac{1}{2}at^2$.Since $a$ is constant,we have $S \propto t^2$,which implies $t \propto \sqrt{S}$.Let $t_1 = 4 \, s$ be the time taken to cover the total distance $S_1 = S$.We want to find the time $t_2$ taken to cover the distance $S_2 = \frac{S}{4}$.Using the proportionality $t \propto \sqrt{S}$,we get $\frac{t_2}{t_1} = \sqrt{\frac{S_2}{S_1}}$.Substituting the values: $\frac{t_2}{4} = \sqrt{\frac{S/4}{S}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore,$t_2 = 4 	imes \frac{1}{2} = 2 \, s$.

$A$ body takes $4 \, s$ to reach the bottom of an inclined plane from the top. How much time (in $s$) will it take to cover one-fourth of the distance?

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