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 travelling 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

The correct option is A $\frac{7m{v}^3}{3P}$

Power delivered to the car=$P=Fv=constant$
$⟹P=mv\frac{dv}{ds}v$

$=m{v}^2\frac{dv}{ds}$
$⟹{\int }_0^{s}ds={\frac{m}{P}}_v^{2v}\int v^{2}dv$

$⟹s=\frac{7m{v}^3}{3P}$