Question

A car of mass $m$ is accelerated on a plane road under the influence of a force in such a way that its velocity becomes $v_2$ from $v_1$ in a distance $s$. If a constant power $P$ is generated by the car engine, then the velocity $v_2$ will be

• A. ${(\frac{Ps}{2m}+v_1^2)}^{\frac{1}{2}}$
• B. ${(\frac{Ps}{2m}+v_1^2)}^{\frac{1}{3}}$
• C. ${(\frac{3Ps}{m}+v_1^3)}^{\frac{1}{2}}$
• D. ${(\frac{3Ps}{m}+v_1^3)}^{\frac{1}{3}}$

Answer 1

The correct option is D ${(\frac{3Ps}{m}+v_1^3)}^{\frac{1}{3}}$

We know that,
$P=Fv=m\left(\frac{dv}{dt}\right)v$

$\frac{P}{m}.dt=v.dv$

$v^{2}dv=\frac{P}{m}vdt=\frac{P}{m}ds,{\int }_{v_1}^{v_2}{v}^{2}dv=\frac{P}{m}{\int }_0^{x}ds$

${\left[\frac{v^3}{3}\right]}_{v_1}^{v_2}=\frac{P}{m}s,v_2^3-v_1^3=\frac{3Ps}{m},v_2={[\frac{3Ps}{m}+v_1^3]}^{\frac{1}{3}}$