The work done by the engine with constant power $P$ over a time $t$ changes the kinetic energy of the car. The work-energy theorem states $W = \Delta K$. The work done is $W = \int P dt$. Since the car starts from rest, its kinetic energy is $K = \frac{1}{2}mv^2$.

From the definition of power $P = Fv$ and Newton's second law $F=ma=m\frac{dv}{dt}$, we have:

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

Separating variables and integrating from rest ($v=0$ at $t=0$):

$$\int_0^t P d\tau = \int_0^{v(t)} mv' dv'$$ $$Pt = \frac{1}{2}m v(t)^2$$

Solving for velocity as a function of time:

$$v(t) = \sqrt{\frac{2P}{m}} t^{1/2}$$

The distance $d$ is found by integrating the velocity from $t=0$ to $t=T$:

$$d = \int_0^T v(t) dt = \int_0^T \sqrt{\frac{2P}{m}} t^{1/2} dt = \sqrt{\frac{2P}{m}} \left[\frac{t^{3/2}}{3/2}\right]_0^T = \frac{2}{3}\sqrt{\frac{2P}{m}}T^{3/2}$$

Since the track distance $d$ and the car's mass $m$ are constant, we can establish a relationship between power $P$ and time $T$:

$$d = \text{constant} \implies \sqrt{P}T^{3/2} = \text{constant}$$

Squaring both sides gives a simpler relation:

$$PT^3 = \text{constant}$$

To find the relationship between the differential changes $dP$ and $dT$, we take the total differential of this expression:

$$d(PT^3) = 0$$ $$T^3 dP + P(3T^2 dT) = 0$$

Solving for $dT$:

$$3PT^2 dT = -T^3 dP$$ $$dT = -\frac{T^3 dP}{3PT^2} = -\frac{T}{3P} dP$$

The negative sign indicates that an increase in power ($dP > 0$) results in a decrease in time ($dT < 0$).