Question

A vehicle of mass M is accelerated on a horizontal frictionless road under a force changing its velocity from u to v in distance S. A constant power P is given by the engine of the vehicle, then $v=$

• A. ${(u^3+\frac{2PS}{M})}^{1/3}$
• B. ${(\frac{PS}{M}+u^3)}^{1/2}$
• C. ${(\frac{PS}{M}-u^2)}^{1/3}$
• D. ${(\frac{3PS}{M}+u^3)}^{1/3}$

Answer 1

The correct option is D ${(\frac{3PS}{M}+u^3)}^{1/3}$

Using $P=Fv=M\left(\frac{dv}{dt}\right)v$
i.e. $v^{2}dv=\frac{P}{M}vdt=\frac{P}{M}dS$
Integrating ${\int }_u^v{v}^{2}dv=\frac{P}{M}{\int }_0^{S}dS$
$v^3-u^3=\frac{3PS}{M}$
or $v={(\frac{3PS}{M}+u^3)}^{1/3}$