1. Given: The body moves in one direction under a constant power source.

2. Use the definition of power: $P = \frac{dW}{dt}$ Since work done changes kinetic energy, $P = \frac{d}{dt}\left(\frac{1}{2}mv^2\right)$

3. Since $P$ is constant, $P = m v \frac{dv}{dt}$

4. But $v = \frac{dx}{dt}$. Also, another direct relation for power is $P = Fv$ and with $F = m\frac{dv}{dt}$, $P = m v \frac{dv}{dt}$ Rearranging, $v\,dv = \frac{P}{m}dt$

5. Integrate both sides: $\int v\,dv = \int \frac{P}{m}dt$ $\frac{v^2}{2} = \frac{P}{m}t + C$

6. Assuming the body starts from rest at $t=0$, we get $C=0$. Hence, $v^2 \propto t$ $v \propto t^{1/2}$

7. Now displacement is $x = \int v\,dt \propto \int t^{1/2} dt$ $x \propto t^{3/2}$

8. Therefore, displacement is proportional to $\boxed{t^{3/2}}$

9. Option check:

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

So the correct option is B.