Given:

• Mass of automobile $= m$
• Initial velocity $u=0$
• Constant power supplied by engine $= P$
• Starts from origin

Use power relation

Power is the rate of change of kinetic energy: $P = \frac{d}{dt}\left(\frac{1}{2}mv^2\right)$

Since $P$ is constant, $\frac{1}{2}mv^2 = Pt + C$

At $t=0$, $v=0$, so $C=0$. Hence, $\frac{1}{2}mv^2 = Pt$

Therefore, $v^2 = \frac{2Pt}{m}$ $v = \sqrt{\frac{2Pt}{m}}$

Relate velocity to position

Since $v = \frac{dx}{dt}$ we get $\frac{dx}{dt} = \sqrt{\frac{2P}{m}}\, t^{1/2}$

Integrating, $x = \sqrt{\frac{2P}{m}} \int t^{1/2} dt$ $x = \sqrt{\frac{2P}{m}} \cdot \frac{2}{3} t^{3/2} + C$

Since $x=0$ at $t=0$, $C=0$.

Thus, $x = \frac{2}{3}\sqrt{\frac{2P}{m}}\, t^{3/2}$

Simplify coefficient

$\frac{2}{3}\sqrt{\frac{2P}{m}} = \sqrt{\frac{4}{9}}\sqrt{\frac{2P}{m}} = \sqrt{\frac{8P}{9m}}$

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

Match with options

This matches Option D.

Comparison with stored answer

Stored correct answer: D

My derived answer: D

Hence, they agree.