A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as $\alpha = a\sqrt{v}$ , where a is a positive constant. At the initial moment the velocity of the point is equal to v₀. What distance will it transverse before it stops? What will it take to cover the distance.?

Given 

$$\begin{gathered} -\frac{dv}{dt}=a\sqrt{v} \\ \text{ or }\frac{dv}{\sqrt{v}}=adt \\ \text{ or } {\int }_{v_0}^v{v}^{-1/2}dv={\int }_0^t(-a)dt \\ {\left|\frac{v^{1/2}}{1/2}\right|}_{v_0}^v=-at \\ \sqrt{v}-\sqrt{v_0}=\frac{-at}{2}\text{ or }dt \frac{dv}{\sqrt{v}}=adt \\ \text{ or } {\int }_{v_0}^v{v}^{-1/2}dv={\int }_0^t(-a)dt \\ {\left|\frac{v^{1/2}}{1/2}\right|}_{v_0}^v=-at \\ \sqrt{v}-\sqrt{v_0}=\frac{-at}{2}\sqrt{v}=\sqrt{v_0}-\frac{at}{2}Forstop,vtobezeroi.e.,0=\sqrt{v_0}-\frac{at}{2} \\ \therefore t=\frac{2\sqrt{v_0}}{a}Againfromequation\left(i\right),wehave\frac{ds}{dt}={\left(\sqrt{v_0}-\frac{at}{2}\right)}^2 \\ ds={\left(\sqrt{v_0}-\frac{at}{2}\right)}^{2}dt \end{gathered}$$

or On integrating, we get

$$\begin{gathered} \text{ At }t=\frac{2\sqrt{v_0}}{a},s=\frac{{\left(\sqrt{v_0}-\frac{at}{2}\right)}^3}{3\times \left(\frac{-a}{2}\right)}=-\frac{2}{3a}{\left(\sqrt{v_0}-\frac{at}{2}\right)}^3 \\ t=\frac{2\sqrt{v_0}}{a}, \\ s=\frac{-2}{3a}{\left(\sqrt{v_0}-\frac{a2\sqrt{v_0}}{a}\right)}^3 \\ =\left(\frac{2}{3a}\right)v_0^{3/2} \end{gathered}$$