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(C) Given that power $P$ is constant. We know that $P = Fv = mav = m \left( \frac{dv}{dt} \right) v$. Integrating this expression: $\frac{P}{m} dt = v

Vedclass · Accepted Answer

$t^{3/2}$

Vedclass · Answer

(C) Given that power $P$ is constant. We know that $P = Fv = mav = m \left( \frac{dv}{dt} \right) v$. Integrating this expression: $\frac{P}{m} dt = v dv$. Integrating both sides: $\int \frac{P}{m} dt = \int v dv \implies \frac{P}{m} t = \frac{v^2}{2}$. Thus,$v^2 = \frac{2P}{m} t$,which gives $v = \sqrt{\frac{2P}{m}} t^{1/2}$. Since $v = \frac{ds}{dt}$,we have $ds = \sqrt{\frac{2P}{m}} t^{1/2} dt$. Integrating with respect to $t$: $s = \int \sqrt{\frac{2P}{m}} t^{1/2} dt = \sqrt{\frac{2P}{m}} \left( \frac{t^{3/2}}{3/2} \right) = \sqrt{\frac{2P}{m}} \left( \frac{2}{3} t^{3/2} \right)$. Therefore,$s \propto t^{3/2}$.

$A$ body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time $t$ is proportional to

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