An automobile of mass 'm' accelerates starting from origin and initially at rest,

Question:

An automobile of mass 'm' accelerates starting from origin and initially at rest, while the engine supplies constant power P. The position is given as a function of time by : 

  1. $\left(\frac{9 P}{8 m}\right)^{\frac{1}{2}} t^{\frac{3}{2}}$

  2. $\left(\frac{8 \mathrm{P}}{9 m}\right)^{\frac{1}{2}} \mathrm{t}^{\frac{2}{3}}$

  3. $\left(\frac{9 m}{8 P}\right)^{\frac{1}{2}} t^{\frac{3}{2}}$

  4. $\left(\frac{8 \mathrm{P}}{9 m}\right)^{\frac{1}{2}} \mathrm{t}^{\frac{3}{2}}$


Correct Option: , 4

Solution:

P = const.

$\mathrm{P}=\mathrm{Fv}=\frac{\mathrm{mv}^{2} \mathrm{dv}}{\mathrm{dx}}$

$\int_{0}^{x} \frac{P}{m} d x=\int_{0}^{v} v^{2} d v$

$\frac{\mathrm{Px}}{\mathrm{m}}=\frac{\mathrm{v}^{3}}{3}$

$\left(\frac{3 \mathrm{Px}}{\mathrm{m}}\right)^{1 / 3}=v=\frac{\mathrm{dx}}{\mathrm{dt}}$

$\left(\frac{3 \mathrm{P}}{\mathrm{m}}\right)^{1 / 3} \int_{0}^{\mathrm{t}} \mathrm{dt}=\int_{0}^{\mathrm{x}} \mathrm{x}^{-1 / 3} \mathrm{dx}$

$\Rightarrow x=\left(\frac{8 P}{9 m}\right)^{1 / 2} t^{3 / 2}$ 

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