I stumbled on a reference which summarizes what I think is the most correct answer to this question. You have several options, but I think we should only use terms that we have a physical reason to write. Here's the reference:
http://usna.edu/Users/physics/schneide/Buick.htm
They use a lot of unnecessary details like time between stopping that we're not interested in. I agree with their graph but not their equation. So here is my equation to explain their graph. I also limited it to flat roads (no hills).
$$F = \frac{P_0}{v} + \mu mg + c \rho A \frac{v^2}{2} $$
From the reference, known values about their car are:
- The weight, $m=1800 kg$
- The front area $A = 3 m^2$
I report these because there is no direct measurement available. I would then use their data to evaluate the three coefficients that determine the relative friction from each thing.
They put a constant term in the transmission factor (the 1/v term). I don't like this, because I want clean mathematics, so I'm bunching that long tail of transmission with the friction coefficient.
$$ \mu (1800 kg) (9.8 m/s^2) = 200 N + 250 N = 450 N \rightarrow \mu = 0.026 $$
$$ \frac{P_0}{15 m/s} = 500 N - 250 N = 250 N \rightarrow P_0 = 3750 W$$
$$ c (3 m^2) (1.3 kg/m^3) \frac{(31 m/s)^2}{2} = 500 N \rightarrow c = 0.27 $$
These are all consistent with what the link claimed, aside from the cases where I willfully used a different kind of definition. Another good thing that these all have physical interpretations, which those units are suggesting. I will avoid getting into the exact interpretation because I feel like there's space for quibbling.
I plotted this on Wolfram alpha. This is my altered version of that link.
This fits expectations fairly well. Take note, however, of the 1/v term. That represents fuel consumption due to constant loads (thus, units of power, of course). That might not be relevant if you're looking for a force, but it can be kind of interpreted as a force. It's a force the engine is exerting against itself (to some fraction of that number) due to idling. It is also the constant electronic loads on the battery... and the charging of the battery itself. It's not the friction of the wheels on the road of air on the car. If you're only interested in those then you might do well to just take the last two terms. If you do that, however, there is no concept of maximum gas mileage, nor should there be. What you're going to do with these terms depends on the application. I just believe this to be the best available option so I posted it.
The slip ratio depends on the speed for the car you would calculate based circumferential speed of the wheel in the frame of the car (angular velocity of wheel times radius), and the actual linear speed of the car.
The reason that you get slip at even the smallest forces results not from the fact that the tire is slipping against the ground, but that the tire is elastic. Let's see how this could happen.
To measure the slip, lets put little green splotches of die on the circumference of the tire spaced 1cm apart. From this we can tell how much the wheel has rotated. Now imagine a car that is accelerating. What happens to the tires? Well the road is providing a force on the tire. What does that do to the bottom of the tire? Well just imagine a stationary tire that can't rotate and you apply a force tangent to the tire. This will cause the tire to deform and the part of the tire you are apply the force to will get scrunched up in the direction of the force. Now if you force the tire to rotate against the force, the scrunched up part will go to where the "ground" (the thing applying the force) is.
This means that our little green splotches, instead of being 1 cm apart, they will be .8 cm apart. Suppose there are a total of 11 splotches. Then by the time tire turns enough for each splotch gets to its original position, the wheel has rotated a full revolution. On the other hand, the car has only moved 8cm, because each of the splotches is .8 cm apart and there are 11 of them (so 10 intervals). Now when we compare this 8cm that the car has actually moved while the wheel rotated once to the full circumference of the wheel, which is 10 cm, we conclude that the wheel has slipped.
Since there will always be some scrunching given a non-zero tangent force, you will always get slip for a nonzero tangent force.
Of course this scrunching goes to zero in the limit of an infinitely rigid wheel, which is the sort of wheel used in physics homework problems.
Now for high enough slip ratios, the wheel will actually slide across the pavement, but until you get to the this point, static friction is still in play, so the car is accelerating from static friction.
Best Answer
There is an approximate law that states that the frictional force, $F$, is given by:
$$ F = \mu W $$
where $W$ is the normal force and $\mu$ is the coefficient of friction; $\mu$ is normal taken to lie in the range zero to one. For a car the weight $W$ is given by $mg$, and the acceleration would be $F/m$, so dividing through by $m$ we get th acceleration to be:
$$ a = \mu g $$
If $\mu$ has the maximum value of one then the acceleration cannot exceed $g$ because trying to accelerate faster would just make the tyres spin.
But ...
The equation we started with is an approximation, and friction is actually a far more complicated phenomenon that that simple law suggests. For example car tyres deform and can key into irregularities on the road to increase the friction. Racing car tyres can have effective values of $\mu$ far greater than one, so they can accelerate at more than 1g. In fact a drag racer can accelerate at around 4g.