A particle is moving unidirectional on a horizontal plane under the action of a constant power supplying energy source. The displacement (s) – time (t) graph that describes the motion of the particle is (graphs are drawn schematically and are not to scale):

Here Power is constant,
Hence, $P\mathit{=}Fv\mathit{=}mav\mathit{=}av\mathit{=}constant\mathit{.}$

$$\begin{gathered} av=k \\ v\frac{dv}{dx}\times v=k \\ {\int }_0^v{v}^{2}dv=k{\int }_0^{x}dx \\ \therefore v={\left(3k\right)}^{1/3}\times {\left(x\right)}^{1/3}\Rightarrow v=A{\left(x\right)}^{1/3} where A={\left(3k\right)}^{1/3}=constant ....\left(1\right) \end{gathered}$$

Also, $v=\frac{dx}{dt}$$\therefore \frac{dx}{dt}=A{\left(x\right)}^{1/3}\Rightarrow {\int }_0^d\frac{dx}{{\left(x\right)}^{1/3}}=A{\int }_0^{t}dt$

$$Simplifying further we get,

$\Rightarrow d=A{\left(t\right)}^{3/2} ...\left(2\right)$

From equations $\left(1\right) \& \left(2\right)$ its clear that the graph won't be straight line so first & third options cant be correct.
Now, from equation $\left(1\right)$ its clear that when $x$ is increased then the slope $\frac{ds}{dt}$ should also increase which is possible in second option. Hence option $2$ is correct.