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

According to the question, we can say that, the energy of the machine is delivering a constant power which is further used up in moving a body (from rest) along a horizontal straight line.

$\therefore$ Energy supply = Pt

where, P = power

and t = time taken.

$\begin{array}{l}\Rightarrow \frac{1}{2}{\mathrm{mv}}^2=\mathrm{Pt}\\ \Rightarrow {\mathrm{v}}^2\propto \mathrm{t}\Rightarrow \mathrm{v}\propto \sqrt{\mathrm{t}}\\ \mathrm{or} \frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{c}\sqrt{\mathrm{t}} ...\left(\mathrm{i}\right)\end{array}$

where, c is any constant.

Now, integrating Eq. (i), we get

$\begin{array}{l}{\int }_0^{\mathrm{s}} \mathrm{ds}={\int }_0^{\mathrm{t}} \mathrm{c}\sqrt{\mathrm{t}}\mathrm{dt}\\ \Rightarrow \mathrm{s}=\mathrm{c}{\int }_0^{\mathrm{t}} {\mathrm{t}}^{\frac{1}{2}}\cdot \mathrm{dt}\\ \Rightarrow \mathrm{s}=\frac{2\mathrm{c}}{3}{\mathrm{t}}^{\frac{3}{2}}\\ \Rightarrow \mathrm{s}\propto {\mathrm{t}}^{3/2}\end{array}$

So, the distance moved by the body in time t is proportional to ${\mathrm{t}}^{\frac{3}{2}}$.