A body sliding on a smooth inclined plane req | vedclass

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(B) For a body starting from rest $(u = 0)$ and sliding down a smooth inclined plane with constant acceleration $(a)$,the distance covered $(S)$ in ti

Vedclass · Accepted Answer

$2$

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(B) For a body starting from rest $(u = 0)$ and sliding down a smooth inclined plane with constant acceleration $(a)$,the distance covered $(S)$ in time $(t)$ is given by the equation of motion:$S = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}at^2 = \frac{1}{2}at^2$.This implies that $S \propto t^2$,or $t \propto \sqrt{S}$.Given that the total time to cover the full distance $(S)$ is $t_1 = 4 \, s$.We want to find the time $(t_2)$ to cover one-fourth of the distance $(S_2 = \frac{S}{4})$.Using the proportionality $t \propto \sqrt{S}$,we have:$\frac{t_2}{t_1} = \sqrt{\frac{S_2}{S_1}} = \sqrt{\frac{S/4}{S}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore,$t_2 = \frac{1}{2} 	imes t_1 = \frac{1}{2} 	imes 4 \, s = 2 \, s$.

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

Similar Questions

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$A$ 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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