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A car of mass ‘m’ accelerates on a level smooth road when a constant power ‘p’ is delivered to it. If the velocity of the car at any instant is ‘v’, then its velocity will be doubled after travelling a distance

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$\frac{7 m v^{3}}{3 p}$

$\frac{4 m v^{3}}{3 p}$

$\frac{m v^{3}}{p}$

$\frac{18 m v^{3}}{7 p}$

answer is A.

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Detailed Solution

$P = F . V$

$= m . \frac{d v}{d s} . v^{2}$

$\int d s = \frac{m}{p} \int_{v}^{2 v} v^{2} d v$

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