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(C) Given, Power $(P) =$ constant.Kinetic Energy $(KE) = \frac{1}{2} m v^{2}$.We know that, $P = \frac{d(KE)}{dt} = \frac{d}{dt} (\frac{1}{2} m v^{2})

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$s=a t^{3 / 2}, a$ is a constant

Vedclass · Answer

(C) Given, Power $(P) =$ constant.Kinetic Energy $(KE) = \frac{1}{2} m v^{2}$.We know that, $P = \frac{d(KE)}{dt} = \frac{d}{dt} (\frac{1}{2} m v^{2}) = m v \frac{dv}{dt}$.Since $P$ is constant, $m v \frac{dv}{dt} = P$.Integrating both sides with respect to time $t$:$\int m v dv = \int P dt \Rightarrow \frac{1}{2} m v^{2} = P t$ (assuming initial velocity is $0$ at $t=0$).$v^{2} = \frac{2 P}{m} t \Rightarrow v = \sqrt{\frac{2 P}{m}} t^{1/2}$.Since $v = \frac{ds}{dt}$, we have $\frac{ds}{dt} = \sqrt{\frac{2 P}{m}} t^{1/2}$.Integrating with respect to $t$:$s = \int \sqrt{\frac{2 P}{m}} t^{1/2} dt = \sqrt{\frac{2 P}{m}} \cdot \frac{t^{3/2}}{3/2} = \frac{2}{3} \sqrt{\frac{2 P}{m}} t^{3/2}$.Thus, $s \propto t^{3/2}$, which means $s = a t^{3/2}$ where $a$ is a constant.

$A$ body starts from rest, under the action of an engine working at a constant power and moves along a straight line. The displacement $s$ is given as a function of time $t$ as:

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