Question

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 traveling how much distance it becomes double?

• A. $\frac{7m{v}^3}{3P}$
• B. $\frac{4m{v}^3}{3P}$
• C. $\frac{m{v}^3}{P}$
• D. $\frac{18m{v}^3}{7P}$

Answer 1

Answer: (A)

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

$P=Fv=constant$
$\therefore F=\frac{P}{v}$
$\therefore mv\left(\frac{dv}{ds}\right)=\frac{P}{v}$
$\therefore {\int }_0^{s}dS=\frac{m}{P}{\int }_v^{2v}{v}^{2}dv$
Solving we get, $S=\frac{7m{v}^3}{3P}$