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

We begin by recalling the definition of mechanical power. When a force $$F$$ moves an object with instantaneous speed $$v$$, the power delivered is

$$P = F\,v.$$

For the automobile of mass $$m$$, the only horizontal force producing acceleration is the net engine force. Newton’s second law gives

$$F = m\,a = m\,\frac{dv}{dt}.$$

Substituting this value of $$F$$ in the power expression, we get

$$P = \left(m\frac{dv}{dt}\right)v.$$

So

$$m\,v\,\frac{dv}{dt} = P.$$

To separate the variables, multiply both sides by $$dt$$:

$$m\,v\,dv = P\,dt.$$

Now integrate each side. At $$t = 0$$ the car is at rest, so the lower limit for velocity is $$0$$.

$$\int_{0}^{v} m\,v\,dv = \int_{0}^{t} P\,dt.$$

Carry out the integrations:

$$m\left[\frac{v^{2}}{2}\right]_{0}^{v} = P\,[t]_{0}^{t}.$$

So

$$\frac{m\,v^{2}}{2} = P\,t.$$

Solving for $$v^{2}$$ gives

$$v^{2} = \frac{2P}{m}\,t.$$

Taking the positive square root (speed is non-negative), we obtain the velocity as a function of time:

$$v = \sqrt{\frac{2P}{m}}\;t^{1/2}.$$

Next, we use the relationship between velocity and position, namely

$$v = \frac{dx}{dt}.$$

Substituting the expression for $$v$$ just found, we have

$$\frac{dx}{dt} = \sqrt{\frac{2P}{m}}\;t^{1/2}.$$

Separate the variables:

$$dx = \sqrt{\frac{2P}{m}}\;t^{1/2}\,dt.$$

Integrate again, with the initial condition that the car starts from the origin $$x = 0$$ at $$t = 0$$:

$$\int_{0}^{x} dx = \sqrt{\frac{2P}{m}}\int_{0}^{t} t^{1/2}\,dt.$$

The left integral simply yields $$x$$. For the right side, recall the power rule of integration, $$\int t^{n}\,dt = \frac{t^{\,n+1}}{n+1}$$. Here $$n = \tfrac{1}{2}$$, so

$$\int t^{1/2}\,dt = \frac{t^{3/2}}{3/2} = \frac{2}{3}t^{3/2}.$$

Therefore,

$$x = \sqrt{\frac{2P}{m}}\;\left(\frac{2}{3}\,t^{3/2}\right).$$

Simplify the coefficient. The numerical factor becomes $$\frac{2}{3}$$, and under the radical we can combine constants:

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

Thus the position as a function of time is finally

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

This matches Option D.

Hence, the correct answer is Option 4.