Question

The retardation experienced by moving motorboat after its engine is cut off at the instant $t=0$ is given by $a=-\alpha v^3$ where $\alpha$ is a positive constant with appropriate dimensions. If velocity at $t=0$ is $v_0$ , then velocity as a function of time is given as

• A. $v_0{e}^{-\alpha t}$
• B. $\frac{v_0}{\sqrt{1+2\alpha v_0^{2}t}}$
• C. $\frac{v_0}{\sqrt{1+\alpha v_0^{2}t}}$
• D. $v_0{e}^{-\alpha t/2}$

Answer 1

The correct option is B $\frac{v_0}{\sqrt{1+2\alpha v_0^{2}t}}$

Given $a=-\alpha v^3$

$a=\frac{dv}{dt}$

$-\frac{dv}{dt}=\alpha t$

Integrating both sides we get,

$\frac{1}{2v^2}+c=0\alpha t$

At $t=0,v=v_0$

$c=-\frac{1}{2v_0^2}$

At $t=t$,

$\frac{1}{2v^2}-\frac{1}{2v_0^2}=\alpha t$

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

$v=\sqrt{\frac{v_0^2}{2v_0^2\alpha t+1}}$

$v=\frac{v_0}{\sqrt{2v_0^2\alpha t+1}}$