Question

An automobile of mass m accelerates, starting from rest, while the engine supplies constant power P, its position varies with respect to time as?

• A.
• B.
• C.
• D. independent of 't'

Answer 1

The correct option is B

As Fv = P = Constant

i.e. m$\frac{dv}{dt}$v = P $[AsF=\frac{mdv}{dt}]$ or $\int vdv$= $\int \frac{p}{m}dt$

By intergrating both sides we get $\frac{v^2}{2}$ =$\frac{P}{m}$t + $C_1$

As initially the body is at rest i.e. v=0 at t=0 , so $C_1$=0 $\therefore$ v= $(\frac{2Pt}{m}{)}^{\frac{1}{2}}$

Position

For the above expression v = $(\frac{2Pt}{m}{)}^{\frac{1}{2}}$ or $\frac{ds}{dt}$ = $(\frac{2Pt}{m}{)}^{\frac{1}{2}}$ = $[Asv=\frac{ds}{dt}]$
$\frac{ds}{dt}={\left(\frac{2pt}{m}\right)}^{\frac{1}{2}}\int ds=\int {\left(\frac{2pt}{m}\right)}^{\frac{1}{2}}dtS={\left(\frac{2pt}{m}\right)}^{\frac{1}{2}}\int t^{\frac{1}{2}}dt={\left(\frac{2p}{m}\right)}^{\frac{1}{2}}\frac{t^{\frac{1}{2}+1}}{\frac{3}{2}}={\left(\frac{2p}{m}\right)}^{\frac{1}{2}}\frac{2}{3}{t}^{\frac{3}{2}}=\sqrt{\frac{8p}{9m}{t}^3}$