The velocity of a particle moving in a positive direction of the $x$-axis varies as $v$ = $\alpha$$\sqrt{x}$, where $\alpha$ is a positive constant. Assuming that at the moment $t$ = 0, the particle was located at the point $x$= 0:

$$\begin{gathered} v=\alpha \sqrt{x} \\ \frac{dv}{dx}=\alpha \frac{1}{2}{x}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{\alpha }{2\sqrt{x}} \\ Thus acceleration, a=v\frac{dv}{dx}=\alpha \sqrt{x}\times \frac{\alpha }{2\sqrt{x}}=\frac{{\alpha }^2}{2} \\ Also \frac{dv}{dt}=\frac{{\alpha }^2}{2} \\ or v={\int }_0^t\frac{{\alpha }^2}{2}dt=\frac{{\alpha }^{2}t}{2} \\ Mean velocity, \overline{v}=\frac{{\int }_0^t\frac{{\alpha }^{2}t}{2}dt}{t}=\frac{{\alpha }^2{t}^2}{4t}=\frac{{\alpha }^{2}t}{4} \\ s=\frac{1}{2}a{t}^2=\frac{1}{2}\left[\frac{{\alpha }^2}{2}\right]t^2 \\ \therefore t=2\frac{\sqrt{s}}{\alpha } \\ Now, \overline{v}=\frac{{\alpha }^2\times 2\frac{\sqrt{s}}{\alpha }}{4}=\frac{\alpha \sqrt{s}}{2} \end{gathered}$$