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A particle starting from rest. Its acceleration at an instant $t$ is $a = \frac{10}{v + 1} {ms}^{- 2}$ , where $v$ is instantaneous velocity. The distance travelled after which the particle attain a velocity of $15 {ms}^{- 1}$ is

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$123.75 m$

$375 m$

$200 m$

$1237.5 m$

answer is A.

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Detailed Solution

$∵ a = \frac{v d v}{d x}$ or $v d v = a d x$

or $v d v = \frac{10}{v + 1} d x$
or $\int_{0}^{15} v (v + 1) d v = 10 \int_{0}^{x} d x$
or ${[\frac{v^{3}}{3} + \frac{v^{2}}{2}]}_{0}^{15} = 10 x$
or $\frac{1}{10} [\frac{15^{3}}{3} + \frac{15^{2}}{2}] = x$
$∴ x = 123.75 m$

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