newtonian mechanics - Simple friction formula for a car

I am making a 2D driving video game, and I would like to know the "simple" formula for calculating the friction force between the car and the road. I have read lots of friction diagrams involving balls rolling down inclined planes, but I'm having trouble applying it to this problem.

Consider a car driving on a horizontal road, with some coefficient of friction, $\mu$. Its velocity is $v$, mass is $m$, and a constant driving force is accelerating the car at $a\ \mathrm{m/sec}^2$ in the forwards direction. What is the friction force, $F$, that acts on the car in the backwards direction?

I guess that the friction force is proportional to the car's speed, but I can't explain this guess. The reason I think that is because of a handful of case studies. Say that the driver's foot is on the pedal such that the car will get up to $16\ \mathrm{m/sec}$:

• If the car's velocity is $0\ \mathrm{m/sec}$, there is no force accelerating it backwards (it is stationary).


• If the car's velocity is $8\ \mathrm{m/sec}$, the friction force accelerating it backwards must be less (in magnitude) than the driving force, and hence the car keeps accelerating.


• If the car's velocity is $16\ \mathrm{m/sec}$, the friction force accelerating it backwards is equal (in magnitude) and opposite to the driving force, and hence the car maintains a constant velocity.


• When the driver releases the pedal, the same friction force accelerates the car backwards, but now there is no driving force, so the car drifts slowly to a halt.

It would therefore make sense that $F \propto -\mu v$, such that in the first case, $F = 0$; in the second case $F = -\frac{1}{2}ma$; in the third and fourth cases, $F = -ma$, where $a$ is whatever acceleration is required to maintain a constant velocity of $16\ \mathrm{m/sec}$. But I can't figure out the exact relationship between $F$ and $v$, or explain why they are proportional.