The retardation experienced by a moving motor boat, after its engine is cut-off, is given by $\frac{d v}{d t} = k v^{3},$ where k is a constant. If $v_{0}$ is the magnitude of the velocity at cut-off, the magnitude of the velocity at time t after the cut-off is

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$\frac{v_{0}}{2}$

$v_{0} e^{- k t}$

$v_{0}$

$\frac{v_{0}}{\sqrt{2 v_{0}^{2} k t + 1}}$

answer is D.

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Detailed Solution

As, $\frac{d v}{d t} = - k v^{3}$

$\Rightarrow \int_{v_{0}}^{v} \frac{d v}{v^{3}} = - k \int_{0}^{t} d t$

$\Rightarrow - \frac{1}{k} \int_{v_{0}}^{v} v^{- 3} d v = t$

or $- \frac{1}{k} {| \frac{v^{- 3 + 1}}{- 3 + 1} |}_{v_{0}}^{v} = t$

$\frac{1}{2 k} | \frac{1}{v^{2}} - \frac{1}{v_{0}^{2}} | = t$

Or $\frac{1}{v^{2}} - \frac{1}{v_{0}^{2}} = 2 k t$

or $\frac{1}{v^{2}} = \frac{1}{v_{0}^{2}} + 2 k t$

or $\frac{1}{v^{2}} = \frac{1 + 2 v_{0}^{2} k t}{v_{0}^{2}}$

$o r v = \frac{v_{0}}{\sqrt{2 v_{0}^{2} k t + 1}}$

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