Question

A point moves rectilinearly with deceleration which depends on the velocity $v$ of the particle as $a=k\sqrt{v}$, where $k$ is a positive constant. At the initial moment the velocity of the point is equal to $v_0$. What distance will it cover before it stops ?

• A. $\frac{3v_0^{\frac{3}{2}}}{2k}$
• B. $\frac{v_0^{\frac{1}{2}}}{2k}$
• C. $\frac{v_0^{\frac{3}{2}}}{3k}$
• D. $\frac{2v_0^{\frac{3}{2}}}{3k}$

Answer 1

The correct option is D $\frac{2v_0^{\frac{3}{2}}}{3k}$

In the problem it is clear that acceleration here is not constant and it depends on velocity. It means, we should use calculus as
$v\frac{dv}{dx}=-k\sqrt{v}$
$\sqrt{v}dv=-kdx$
On integrating $\underset{v_0}{\overset{v}{\int }}\sqrt{v}dv=-\underset{0}{\overset{x}{\int }}kdx$
$\frac{2}{3}(v_0^{\frac{3}{2}}-v^{\frac{3}{2}})=kx.....\left(i\right)$
Put $v=0$ in equ.$\left(i\right)$, we get
$x=\frac{2v_0^{\frac{3}{2}}}{3k}$
Hence, maximum displacement in the positive direction from origin $=\frac{2v_0^{\frac{3}{2}}}{3k}$