Question

The retardation experienced by a moving motor boat after its engine is cut off, is given by $\frac{dv}{dt}=-8v^2$. If the magnitude of velocity at cut off is $v_0=3\text{m/s}$, the magnitude of the velocity $1\text{sec}$ after the cut off is (in $\text{m/s}$)

• A. $\frac{3}{25}$
• B. $\frac{24}{25}$
• C. $\frac{1}{8}$
• D. $\frac{3}{28}$

Answer 1

The correct option is A $\frac{3}{25}$

It is given that, the retardation is $\frac{dv}{dt}=-8v^2$

$\Rightarrow \frac{dv}{v^2}=-8dt$

Integrating on both sides we get,
${\int }_{v_0}^v\frac{dv}{v^2}=-{\int }_0^{t}8dt$,
here the limits are initial velocity which is the velocity at cut off $v_0=3\text{m/s}$ and the final velocity be $v$ $(v_0\to v)$, the limit for time be $0\to t$

$\Rightarrow {\begin{array}{c}-\frac{1}{v}\end{array}|}_{v_0}^v=-8t$

$-\frac{1}{v}+\frac{1}{v_0}=-8t$
$\Rightarrow \frac{1}{v}-\frac{1}{v_0}=8t$
$\frac{1}{v}=\frac{1}{v_0}+8t$
$\Rightarrow v=\frac{v_0}{1+8v_{0}t}$
$v=\frac{3}{1+8\times 3\times 1}=\frac{3}{25}\text{m/s}$

Hence option A is the correct answer