Question

The declaration experienced by a moving motor boat, after its engine is cut-off is given by $\frac{dv}{dt}=-k{v}^3$ where k is constant If $v_0$ is the magnitude of the velocity at cut-off, the magnitude of the velocity at a time t after the cut off is

• A. $\frac{v_0}{2}$
• B. $v_0$
• C. $v_0{e}^{-ti}$
• D. $\frac{v_0}{\sqrt{(2v_0^{2}kt+1)}}$

Answer 1

The correct option is D $\frac{v_0}{\sqrt{(2v_0^{2}kt+1)}}$

Given$,$

$\frac{dv}{dt}=-k{v}^3$

$\frac{dv}{v^3}=-kdt$

$\underset{v}{\overset{v_0}{\int }}\frac{dv}{v^3}=-k\underset{0}{\overset{t}{\int }}dt$

$\Rightarrow \left[\frac{1}{2{v_0}^2}-\frac{1}{2v^2}\right]=-kt$

$\Rightarrow \frac{1}{2}{v_0}^2+kt=\frac{1}{2v^2}$

$\Rightarrow \left(1+2{v_0}^{2}kt\right)=\frac{2{v_0}^2}{2v^2}=\frac{{v_0}^2}{v^2}$

$\Rightarrow v=\frac{v_0}{\sqrt{2{v_0}^{2}kt+1}}$

Hence,

option $\left(D\right)$ is correct answer.