Question

A car (treat it as particle) of mass $m$ is accelerating on a level smooth road under the action of single force $F$. The power delivered to car is constant equal to $p$. If the velocity of the car at an instant is $v$, than after travelling how much distance velocity becomes double?

• A. $\frac{7m{v}^3}{3P}$
• B. $\frac{4m{v}^2}{3P}$
• C. $\frac{m{v}^4}{P}$
• D. $\frac{18m{v}^3}{7P}$

Answer 1

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

$power=force\times velocity$

$p=F\times v$

$p=mv\frac{dv}{ds}\times v$

$\underset{v}{\overset{2v}{\int }}m{v}^{2}dv=\underset{0}{\overset{s}{\int }}pds$

$\frac{m}{3}[8v^3-v^3]=s$

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

If the velocity of the car at an instant is v, than after travelling $s=\frac{7m{v}^3}{3p}$ much distance velocity becomes double.