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Step-by-Step Solution
Step 1: Express Power in terms of Force and Velocity
For an automobile of mass $m$ supplied with a constant power $P$, we know that:
$$
P = F \cdot v
$$
where $F$ is the net force and $v$ is the velocity of the automobile.
Step 2: Relate Force to Mass and Acceleration
Since $F = m \frac{dv}{dt}$ and $v = \frac{dx}{dt}$, it is often useful to use the chain rule approach so that:
$$
F = m \frac{dv}{dt} = m \, v \frac{dv}{dx}.
$$
Substituting $F$ into $P = F \cdot v$ gives:
$$
P = m \, v \left( v \frac{dv}{dx} \right) = m \, \frac{v^2 \, dv}{dx}.
$$
Step 3: Perform the First Integration to Relate Velocity and Position
Rearrange and integrate from $x = 0$ to $x$ and $v = 0$ to $v$:
$$
\int_0^x \frac{P}{m} \, dx = \int_0^v v^2 \, dv.
$$
This gives:
$$
\frac{P x}{m} = \frac{v^3}{3}.
$$
Hence,
$$
v = \left(\frac{3 P x}{m}\right)^{\frac{1}{3}}.
$$
Step 4: Express Velocity as the Time Derivative of Position
Since $v = \frac{dx}{dt}$, we have:
$$
\frac{dx}{dt} = \left(\frac{3 P x}{m}\right)^{\frac{1}{3}}.
$$
We can separate variables:
$$
\frac{dx}{x^{\frac{1}{3}}} = \left(\frac{3P}{m}\right)^{\frac{1}{3}} dt.
$$
Step 5: Perform the Second Integration to Obtain Position as a Function of Time
Integrate both sides from $t=0$ to $t$ and $x=0$ to $x$:
$$
\int_0^x x^{-\frac{1}{3}} \, dx = \left(\frac{3P}{m}\right)^{\frac{1}{3}} \int_0^t dt.
$$
The left-hand side integral is:
$$
\int x^{-\frac{1}{3}} \, dx = \int x^{\frac{-1}{3}} \, dx
= \frac{x^{\frac{2}{3}}}{\frac{2}{3}} = \frac{3}{2} x^{\frac{2}{3}}.
$$
The right-hand side integral is simply:
$$
\left(\frac{3P}{m}\right)^{\frac{1}{3}} \cdot t.
$$
So equating both,
$$
\frac{3}{2} x^{\frac{2}{3}} = \left(\frac{3P}{m}\right)^{\frac{1}{3}} t.
$$
Solving for $x$, we get:
$$
x^{\frac{2}{3}} = \frac{2}{3} \left(\frac{3P}{m}\right)^{\frac{1}{3}} t.
$$
Hence,
$$
x = \left[ \frac{2}{3} \left(\frac{3P}{m}\right)^{\frac{1}{3}} t \right]^{\frac{3}{2}}.
$$
Step 6: Simplify the Expression
On simplifying, we ultimately find:
$$
x = \left(\frac{8P}{9m}\right)^{\frac{1}{2}} \, t^{\frac{3}{2}}.
$$
This matches the correct answer provided, which is:
$$
\boxed{x = \left(\frac{8P}{9m}\right)^{\frac{1}{2}} \, t^{\frac{3}{2}}.}
$$
