Question

An automobile of mass $M$ accelerate from rest with a constant power $P$ generated by the engine. Assume that $80\%$ of power generated by the engine accelerates the automobile. The distance covered by automobile in time $t$ is given by

• A. $\sqrt{\frac{8}{5}\frac{P{t}^3}{M}}$
• B. $\sqrt{\frac{8}{15}\frac{P{t}^3}{M}}$
• C. $\sqrt{\frac{32}{45}\frac{P{t}^3}{M}}$
• D. $\sqrt{\frac{4}{5}\frac{P{t}^3}{M}}$

Answer 1

The correct option is C $\sqrt{\frac{32}{45}\frac{P{t}^3}{M}}$

Let the mass of the body be $M$ and $P$ be the constant power supplied

Now , $P=F.v$ , where $F$ is the force applied by the source

Hence ,

$P=M\times a\times v$ , where $a$ is the acceleration of the body .

But $a=\frac{dv}{dt}$

Hence ,

$P=M\times \frac{dv}{dt}\times v$

i.e, $\frac{8P}{10}=F.v=M\frac{dv}{dt}v$

Or

$P.dt=Mv.dv$

i.e, $\frac{8P}{10}dt=Mv.dv$

Integrating both sides from $0$ to $t$ and velocity from $0$ to $v$ , we get

$Pt=\frac{M{v}^2}{2}$

i.e, $\frac{4P}{5}t=\frac{M{v}^2}{2}$

$v=\sqrt{\frac{8Pt}{5M}}$

$\frac{dx}{dt}=\sqrt{\frac{8Pt}{5M}}$

$x=\sqrt{\frac{8Pt}{5M}}dt$

$x=\sqrt{\frac{8P}{5M}}\left(\frac{2t^{3/2}}{3}\right)$

$x=\frac{2}{3}\sqrt{\frac{8P}{5M}}\times t^{3/2}$

Hence , $x=\sqrt{\frac{32P{t}^3}{45M}}$ $\underset{–}{\text{is the distance covered by automobile in time t}}$.