1. Given: A machine delivers constant power $P$ to a body initially at rest, moving along a horizontal straight line.

2. Use the definition of power: $P = \frac{dW}{dt} = \frac{d}{dt}\left(\frac{1}{2}mv^2\right)$ Since the motion is horizontal and no change in potential energy is involved, the work done goes into kinetic energy.

3. Therefore, $P = \frac{d}{dt}\left(\frac{1}{2}mv^2\right) = mv\frac{dv}{dt}$

4. Rearranging, $v\,dv = \frac{P}{m}\,dt$

5. Integrate both sides. Initially, the body is at rest, so at $t=0$, $v=0$: $\int_0^v v\,dv = \frac{P}{m}\int_0^t dt$ $\frac{v^2}{2} = \frac{P}{m}t$ $v = \sqrt{\frac{2Pt}{m}}$

6. Now find displacement $x$ using $\frac{dx}{dt} = v = \sqrt{\frac{2P}{m}}\, t^{1/2}$

7. Integrate again: $x = \int_0^t \sqrt{\frac{2P}{m}}\, t^{1/2} dt$ $x = \sqrt{\frac{2P}{m}}\cdot \frac{2}{3} t^{3/2}$

8. Hence, $x \propto t^{3/2}$

9. Option check:

   • A: $t^{3/2}$ ✅
   • B: $t^{1/2}$ ❌
   • C: $t^{1/4}$ ❌
   • D: $t^{3/4}$ ❌

Therefore, the correct answer is A.