Take the equation P = FV
For a fixed power of an engine, applying a greater force gives a smaller velocity while a smaller force gives a greater velocity. To me this sounds counter-intuitive.
I guess my question here is what does the 'Force' actually represent in this equation and in a car?
The confusion here is that you are taking $P = F\,v$ as an equation of motion. It is not. The equation of motion is still $$F = m a$$
The power equation provides the accelerating force, as in $$F = \frac{P}{v}$$
so together with the equations of motion you have
$$ a = \frac{P}{m v} $$
So acceleration is just power over momentum.
The above acceleration be used to solve for the motion under constant power. Specifically to go between speeds $v_0$ and $v_1$ you need:
$$ \begin{aligned} t &= \int \frac{1}{a}\,{\rm d}v = \frac{m (v_1^2 - v_0^2)}{2 P} & & \mbox{time} \\ x & = \int \frac{v}{a}\,{\rm d}v = \frac{m (v_1^3 -v_0^3)}{3 P} & & \mbox{distance} \end{aligned}$$
To include air resistance ,for example would produce $$ a = \frac{P}{m v} -\beta v^2$$ since air drag is assumed to be of the form $F_{drag} = m \beta v^2$. This equation has a direct solution for the distance to go from $v_0$ to $v_1$ under constant power with air drag.
$$ x = \frac{1}{3 \beta} \ln \left( \frac{P - m \beta v_1^3}{P - m \beta v_0^2} \right) $$
This comes from the direct integration of $ x = \int \frac{v}{a}\,{\rm d}v $.
If the velocity $V$ is constant, this means that the total force acting on the car is zero. The force $F$ in the formula is what the engine has to produce to counteract the external forces, such as friction and the component of the gravitational force if the car is going uphill.
If the external forces increase (steeper road or more wind against the car), in the ideal case in which the power $P$ of the engine stays constant, the velocity has to decrease.
For a fixed power of an engine, applying a greater force gives a smaller velocity...
That's probably not the way to look at this. Instead, think of the velocity as a given, and either the power or the force is your dependent variable.
The faster the vehicle, the less force the engine at constant power is able to provide. Eventually, the force available drops to equal the drag and acceleration stops.
If the force is large enough, the vehicle will accelerate and change the velocity. But you shouldn't read the velocity here as the output of force and power.
This formula is just derived from the definition of work. We take $dW=F\cdot dx$ and then divide both sides by $dt$. Therefore, we can frame the same question in terms of a constant work done in some amount of time. If our force is larger, then the object needs to move a smaller distance for the force to do the same amount of work as if a smaller force acted over a larger displacement.
We can get more complicated with other external forces, but this formula is still true for just the power supplied by the engine. The velocity might depend on other forces, but the power supplied by the engine is just the force of the engine multiplied by the current velocity of the car.
