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  $x=0:$

 $v=\alpha \sqrt{x}$
 $\frac{dv}{dx}=\alpha \frac{1}{2}{x}^{-1/2}=\frac{\alpha }{2\sqrt{x}}.$
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  $v=\underset{0}{\overset{t}{\int }}\frac{{\alpha }^2}{2}dt=\frac{{\alpha }^{2}t}{2}.$
Mean velocity,
                           $\overline{v}=\frac{{\displaystyle \underset{0}{\overset{t}{\int }}\frac{{\alpha }^{2}t}{2}dt}}{t}=\frac{{\alpha }^2{t}^2}{4t}=\frac{{\alpha }^{2}t}{4}$
                           $s=\frac{1}{2}a{t}^2=\frac{1}{2}\left(\frac{{\alpha }^2}{2}\right)t^2$
                             $\therefore t=2\frac{\sqrt{s}}{\alpha }$
Now  $\overline{v}=\frac{{\alpha }^2\times \frac{2\sqrt{s}}{\alpha }}{4}=\frac{\alpha \sqrt{s}}{2}.$