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Q.

A car of mass $M$ is accelerating on a level smooth road under the action of a single force $F$ acting along the direction of motion. The power delivered to the car is constant and equal to $P$ . The velocity of the car at an instant is $V$ . After travelling a distance of $\frac{7 M V^{3}}{3 P},$ the velocity becomes $K V$ then $K =$

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a

8

b

4

c

2

d

1

answer is C.

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Detailed Solution

$\begin{array}{l} P = F V = M \frac{d V}{d t} V = M V^{2} \frac{d V}{d s} \\ P d s = M V^{2} d V \\ P s = \frac{M}{3} [V_{1}^{3} - V^{3}] \\ P [\frac{7 M V^{3}}{3 P}] = \frac{M}{3} [V_{1}^{3} - V^{3}] \\ 7 M V^{3} = M [V_{1}^{3} - V^{3}] \Rightarrow V_{1}^{3} = 8 V^{3} \\ \Rightarrow V_{1} = 2 V \end{array}$

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