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:

When a constant power $$P$$ is delivered to a body of mass $$m$$ starting from rest, the work-energy theorem tells us that all power goes into kinetic energy: $$P = \frac{d}{dt}\!\left(\frac{1}{2}mv^2\right) = mv\frac{dv}{dt}$$.

Rearranging: $$mv\,dv = P\,dt$$. Integrating the left side from $$v=0$$ to $$v$$ and the right side from $$0$$ to $$t$$: $$\int_0^v mv'\,dv' = \int_0^t P\,dt'$$, which gives $$\frac{1}{2}mv^2 = Pt$$.

So the velocity at time $$t$$ is $$v = \sqrt{\frac{2P}{m}} \cdot t^{1/2}$$, i.e., $$v \propto t^{1/2}$$.

The distance $$x$$ is found by integrating velocity: $$x = \int_0^t v\,dt' = \int_0^t \sqrt{\frac{2P}{m}}\,(t')^{1/2}\,dt' = \sqrt{\frac{2P}{m}} \cdot \frac{(t')^{3/2}}{3/2}\Bigg|_0^t = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3}\,t^{3/2}$$.

Therefore $$x \propto t^{3/2}$$.