Question

An automobile of mass ${}^{\prime }{m}^{\prime }$ accelerates starting from origin and initailly at rest, while the engine supplies constant power $P$ . The position is given as a function of time by :

• A. $\left[\sqrt{\left(\frac{9p}{8m}\right)}\right]t^{\frac{3}{2}}$
• B. $\left[\sqrt{\left(\frac{8p}{9m}\right)}\right]t^{\frac{2}{3}}$
• C. $\left[\sqrt{\left(\frac{9m}{8p}\right)}\right]t^{\frac{3}{2}}$
• D. $\left[\sqrt{\left(\frac{8p}{9m}\right)}\right]t^{\frac{3}{2}}$

Answer 1

The correct option is D $\left[\sqrt{\left(\frac{8p}{9m}\right)}\right]t^{\frac{3}{2}}$

Given, constant power is supplied. As we know
$P=Fv=mav=mv\left(v\frac{dv}{dx}\right)=\frac{m{v}^{2}dv}{dx}$
$\Rightarrow {\int }_0^x\frac{P}{m}dx={\int }_0^v{v}^{2}dv$
$\Rightarrow \frac{Px}{m}=\frac{v^3}{3}$
$\Rightarrow {\left(\frac{3Px}{m}\right)}^{\frac{1}{3}}=v$
Also, we have
$v=\frac{dx}{dt}$
$\Rightarrow {\left(\frac{3p}{m}\right)}^{\frac{1}{3}}{\int }_0^{t}dt={\int }_0^x{x}^{-\frac{1}{3}}dx$
$\Rightarrow x={\left(\frac{8p}{9m}\right)}^{\frac{1}{2}}{t}^{\frac{3}{2}}$
Or, $x=\left[\sqrt{\left(\frac{8p}{9m}\right)}\right]t^{\frac{3}{2}}$