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(C) Given that the power $P$ delivered by the machine is constant.The work done by the machine in time $t$ is $W = P \cdot t$.According to the work-en

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$t^{3/2}$

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(C) Given that the power $P$ delivered by the machine is constant.The work done by the machine in time $t$ is $W = P \cdot t$.According to the work-energy theorem,the work done is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2} m v^2 - 0 = \frac{1}{2} m v^2$.Equating the two,we get $\frac{1}{2} m v^2 = P \cdot t$.This implies $v^2 \propto t$,or $v \propto t^{1/2}$.Since velocity $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = k \cdot t^{1/2}$ (where $k$ is a constant).Integrating with respect to time $t$ to find the distance $x$:$x = \int k \cdot t^{1/2} dt = k \cdot \frac{t^{1/2 + 1}}{1/2 + 1} = k \cdot \frac{t^{3/2}}{3/2}$.Therefore,$x \propto t^{3/2}$.

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

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