A car of mass m is accelerating on a level sm | vedclass

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Question

(A) The power delivered to the car is given by $P = Fv$.Since $F = ma = m \frac{dv}{dt} = m \frac{dv}{ds} \frac{ds}{dt} = mv \frac{dv}{ds}$,we can sub

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$\frac{7mv^3}{3P}$

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(A) The power delivered to the car is given by $P = Fv$.Since $F = ma = m \frac{dv}{dt} = m \frac{dv}{ds} \frac{ds}{dt} = mv \frac{dv}{ds}$,we can substitute this into the power equation:$P = (mv \frac{dv}{ds}) v = mv^2 \frac{dv}{ds}$.Rearranging the terms to integrate with respect to velocity and distance:$v^2 dv = \frac{P}{m} ds$.Integrating both sides from initial velocity $v$ to final velocity $2v$ and distance $0$ to $S$:$\int_{v}^{2v} v^2 dv = \int_{0}^{S} \frac{P}{m} ds$.Evaluating the integrals:$\left[ \frac{v^3}{3} \right]_{v}^{2v} = \frac{P}{m} S$.$\frac{(2v)^3 - v^3}{3} = \frac{PS}{m}$.$\frac{8v^3 - v^3}{3} = \frac{PS}{m}$.$\frac{7v^3}{3} = \frac{PS}{m}$.Solving for $S$:$S = \frac{7mv^3}{3P}$.

$A$ car of mass $m$ is accelerating on a level smooth road under the action of a single force $F$. The power delivered to the car is constant and equal to $P$. If the velocity of the car at an instant is $v$,then after travelling how much distance does its velocity become double?

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