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$\text{Hint: } \frac{d(K.E)}{dt} = \text{Power}(P)$ $\text{Step 1: Find the velocity variation as a function of time.}$ $\text{As the particle is moving with constant power i.e., } P = P_0$ $\text{The power of the particle is given by; } P = \frac{dK}{dt}$ $K = P_0t \Rightarrow \frac{1}{2}mv^2 = P_0t \Rightarrow v = \sqrt{\frac{2P_0t}{m}}$ $\Rightarrow v \propto \sqrt{t}$ $\text{Step 2: Find the correct graph for displacement-time } (s-t).$ $\Rightarrow \frac{ds}{dt} = kt^{\frac{1}{2}} \ldots (1) \quad [\text{where } k \left( k = \sqrt{\frac{2P_0}{m}} \right) \text{ is a constant}]$ $\text{Integrate both sides equation (1), we get;}$ $\Rightarrow \int ds = \int kt^{\frac{1}{2}} dt$ $\Rightarrow s = k't^{\frac{3}{2}} \quad \left[ k' = \frac{2}{3}k \right]$ $\Rightarrow s \propto t^{\frac{3}{2}}$ $\text{Therefore, the correct graph for displacement-time } (s-t) \text{ is shown in the figure below,}$ $\text{Hence, option (1) is the correct answer.}$