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(C) Given that power $P$ is constant.We know that $P = F \cdot v = m \cdot a \cdot v = m \cdot \frac{dv}{dt} \cdot v$.Rearranging the terms, we get $v

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$t^{\frac{3}{2}}$

Vedclass · Answer

(C) Given that power $P$ is constant.We know that $P = F \cdot v = m \cdot a \cdot v = m \cdot \frac{dv}{dt} \cdot v$.Rearranging the terms, we get $v \cdot dv = \frac{P}{m} \cdot dt$.Integrating both sides, $\int v \cdot dv = \int \frac{P}{m} \cdot dt$, which gives $\frac{v^2}{2} = \frac{P}{m} \cdot t$.Thus, $v = \sqrt{\frac{2P}{m}} \cdot t^{\frac{1}{2}}$.Since $v = \frac{dx}{dt}$, we have $\frac{dx}{dt} = \sqrt{\frac{2P}{m}} \cdot t^{\frac{1}{2}}$.Integrating with respect to time $t$, $x = \int \sqrt{\frac{2P}{m}} \cdot t^{\frac{1}{2}} \cdot dt = \sqrt{\frac{2P}{m}} \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}$.Therefore, $x = \frac{2}{3} \sqrt{\frac{2P}{m}} \cdot t^{\frac{3}{2}}$.Since $P$ and $m$ are constants, $x \propto t^{\frac{3}{2}}$.

$A$ box 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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