Question

A point moves in a straight line with retardation $d$ which depends on velocity of particle v as $d=C\sqrt{v}$, where $C$ is a positive constant. If initial velocity of the particle is $v_0$ the total distance traveled by the particle before stopping is :

• A. $\frac{1}{3C}{V}_0^{1/2}$
• B. $\frac{2}{3C}{V}_0^{3/2}$
• C. $\frac{2}{C}{V}_0^{3/2}$
• D. $\frac{V_0}{C}$

Answer 1

The correct option is B $\frac{2}{3C}{V}_0^{3/2}$

$d=c\sqrt{v}\Rightarrow -\frac{vdv}{dx}=c\sqrt{v}\Rightarrow -\sqrt{v}dv=cdx\Rightarrow {\int }_{vo}^0-\sqrt{v}dv=c{\int }_0^{x}dx\Rightarrow cx=\frac{2}{3}{v}_o^{3/2}\therefore x=\frac{2}{3c}{v}_o^{3/2}$