Starting from rest at the top of an inclined | vedclass

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(C) For a body starting from rest $(u = 0)$ and moving with constant acceleration $(a)$ along an inclined plane, the distance $(s)$ covered in time $(

Vedclass · Accepted Answer

$2$ seconds

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(C) For a body starting from rest $(u = 0)$ and moving with constant acceleration $(a)$ along an inclined plane, the distance $(s)$ covered in time $(t)$ is given by the equation of motion: $s = ut + \frac{1}{2}at^2$.Since $u = 0$, the equation simplifies to $s = \frac{1}{2}at^2$, which implies $s \propto t^2$.Let the total distance be $S$ and the total time taken be $T = 4$ seconds.We are asked to find the time $(t')$ taken to cover one-fourth of the distance, i.e., $s' = \frac{S}{4}$.Using the proportionality $s \propto t^2$, we have the ratio: $\frac{s'}{S} = \frac{(t')^2}{T^2}$.Substituting the values: $\frac{S/4}{S} = \frac{(t')^2}{4^2}$.$\frac{1}{4} = \frac{(t')^2}{16}$.$(t')^2 = \frac{16}{4} = 4$.$t' = \sqrt{4} = 2$ seconds.Therefore, the body covers one-fourth of the distance in $2$ seconds.

Starting from rest at the top of an inclined plane, a body reaches the bottom of the inclined plane in $4$ seconds. In what time does the body cover one-fourth of the distance starting from rest at the top?

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