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(C) Power is defined as the rate of doing work,given by $P = Fv$.Since $F = ma$ and $a = \frac{dv}{dt}$,we have $P = m \left( \frac{dv}{dt} \right) v

Vedclass · Accepted Answer

$t^{3/2}$

Vedclass · Answer

(C) Power is defined as the rate of doing work,given by $P = Fv$.Since $F = ma$ and $a = \frac{dv}{dt}$,we have $P = m \left( \frac{dv}{dt} \right) v = k$ (where $k$ is a constant).Rearranging the terms,we get $v dv = \frac{k}{m} dt$.Integrating both sides,we obtain $\frac{v^2}{2} = \frac{k}{m} t$,which implies $v = \sqrt{\frac{2k}{m}} t^{1/2}$.Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \sqrt{\frac{2k}{m}} t^{1/2}$.Integrating with respect to time $t$,we get $x = \int \sqrt{\frac{2k}{m}} t^{1/2} dt = \sqrt{\frac{2k}{m}} \left( \frac{t^{3/2}}{3/2} \right) = \frac{2}{3} \sqrt{\frac{2k}{m}} t^{3/2}$.Thus,the displacement $x$ is proportional to $t^{3/2}$.

$A$ body is moving uni-directionally under the influence of a source of constant power. Its displacement in time $t$ is proportional to

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