I don't think you initially defined $T$, so I'm not entirely sure what it originally used for. The overall idea, however, is that because the newt force is zero (and the resistive force stays the same, i.e. $-600$, then the new forward force has to be $T' = +600$.
Physics is often (e.g. besides high-energy) much more pragmatic than mathematics per se -- one expression of this is the mixed use of things like primes, or superscripts, etc. Similarly, negative signs often aren't tracked quite as closely (as they should be). Because the positive or negative are purely conventional, and it is conceptually clear that the resulting force (and velocity) need to be opposite the resistive force -- then you can just give them the appropriate sign. It is, of course, better to treat them accurately and consistently, but note that this is a common location of hand-waving in introductory level textbooks.
You're right that the reduction in velocity takes some time. Let me spell out in more words/equations what you seem to already understand!
You're demanding that the power, $P=F_{\text{engine}}v$ is constant. You also have some $F_{\text{friction}}$, the force due to rolling friction, and $F_{\text{gravity}}$, the force pulling the car backwards along its axis of motion due to gravity.
From free body diagrams, $F_{\text{friction}}=\mu g M \cos(\theta)$, where $\theta$ is the angle of the slope, $\mu$ is the coefficient of rolling friction, and $gM$ is the normal force due to gravity. The force of gravity along the direction of motion is $F_{\text{gravity}}=g M \sin(\theta)$.
If the car's velocity is in equilibrium, then we have:
\begin{align*}
0&=F_{\text{engine}}-F_{\text{friction}}-F_{\text{gravity}}\\
&=\frac{P}{v}-\mu g M\cos(\theta)-g M\sin(\theta)
\end{align*}
So
$$v=\frac{P}{g M(\mu\cos(\theta)+\sin(\theta))}$$
This $v$ is the equilibrium velocity. If $\theta=0$ this just gives the velocity needed to push a block at constant power. If $\theta=\pi/2$ this just gives the velocity needed to lift a block at constant power. For a low coefficient of rolling friction, pushing is easier than lifting, and the higher theta is the lower the velocity for constant power is.
I think you've grasped all this intuitively in fewer words and equations! Obviously if someone is going straight and starts going up a hill without stepping more on the gas pedal, they're going to slow down over time.
There are a lot of assumptions made, however. For example, the relation between $F_{\text{engine}}$ and the power $P$ is probably wrong. You would expect the force that an engine delivers to depend on the velocity, the load placed on it, and all sorts of other things. My point is: this is kind of a weird constraint to put on an engine (for example, it breaks down when the car starts from a standstill and $v=0$). If you're working for BMW then you need a better model, but it suffices to us. We have a description showing that the car has to slow down.
Unfortunately it's harder to figure out how the slow-down happens. There is in fact an unbalanced force, so we have to start plugging stuff into Newton's law $F=ma$. We have to worry about the equation
$$ma=\frac{P}{v}-\mu g M\cos(\theta)-g M\sin(\theta)$$
which is an ugly, ugly thing. It's so ugly because it's nonlinear. The $v$ on the bottom completely screws up any non-calculus based methods you've learned so far (You wind up with an ordinary differential equation of the form $y''(t)=A/y'(t)+B$, where $y''$ is acceleration and $y'$ is velocity). It's easier to resort to numerics to solve it.
Let's look at the following situation: We have a 1 metric ton car which is being pulled by 1000 watts of power (this is a bit over 1 horsepower). It goes on flat ground at constant velocity for a bit, then goes up a three degree slope. That is: $\mu=0.01$, $g=9.81 \text{m}/\text{s}^2$, $M=1000\text{kg}$, $P=1000\text{watts}$, $\theta=\pi/60$. The graph of velocity over time looks as follows:
It starts at about ten meters per second and travels along at constant velocity. When it hits the slope, it takes about twenty seconds to slow all the way down to a bit less than two meters per second.
The graph was generated with the following Mathematica code. (Note: HeavisideTheta[t] returns 0 if t<0 and 1 if t>0)
theta=Pi/60; mu=0.01; m=1000; g=9.81; power=1000; t0=-50;tend=50;
gravityForce[t_]=g m Sin[theta] HeavisideTheta[t];
rollingFriction[t_]=mu g m-(1-Cos[theta])mu g m HeavisideTheta[t];
engineForce[t_]=power/ x'[t];
soln=NDSolve[{m x''[t]==engineForce[t]-rollingFriction[t]-gravityForce[t],x[t0]==0,x'[t0]==power/(g m mu)},{x[t],x'[t]},{t,t0,tend}];
Plot[x'[t]/.soln,{t,t0,tend},PlotLabel->"Car Velocity over time",AxesLabel->{"t (s)","v (m/s)"}]
Best Answer
The power required to sustain $v$ against friction is:
$$ P_0 = F_0 v = 1000N \times 12 \frac{m}{s} = 12 kW$$
The excess power accelerates the vehicle:
$$ P_x = P-P_0 = (Ma)v,$$
where $M=1500\,$kg is the total mass of the vehicle (and $P=30\,$kW).