Question

The deceleration experienced by a moving motorboat 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. $\frac{v_0}{\sqrt{(2v_0^{2}kt+1)}}$
• B. $v_0{e}^{-kt}$
• C. $\frac{v_0}{2}$
• D. $v_0$

Answer 1

Answer: (A)

$\frac{v_0}{\sqrt{(2v_0^{2}kt+1)}}$

Given, $\frac{dv}{dt}=-k{v}^3\Rightarrow \frac{dv}{v^3}=-kdt$
On integration we get, $\frac{-1}{2v^2}=-kt+cAtt=0v=v_o$
$\therefore \frac{-1}{2v_o^2}=c$
Putting this in the integrated equation we get,
$\frac{-1}{2v^2}=-kt-\frac{1}{2v_o^2}\Rightarrow \frac{1}{2v_o^2}+kt=\frac{1}{2v^2}\Rightarrow v=\frac{v_o}{\sqrt{1+2v_o^{2}kt}}$