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.
I understand now, that the static friction opposes the tendency of a surface to slip over another.
Exactly! Let's remember this point in the next sentences.
When a wheel rotates, static friction pushes the wheel forward to keep the contact surface stationary with respect to the ground.
Not necessarily forward. It can also be backwards. It pulls in whatever direction necessary to keep the contact point stationary. And only if necessary (e.g. there is no static friction when a wheel rolls at constant speed over a horizontal surface. There are no forces at all present, so nothing for the static friction to oppose. The rolling motion just continues effortless until stopped.)
When a car is going at constant speed down a hill, static friction must push the car down the hill where you also have gravity helping too.
No. It pulls uphill.
Think for a moment of a star rolling down. It's legs touches the ground one at a time. While a leg touches the ground, it must not slide. We can think of it as a stationary object in that moment while it is touching. Downwards forces (gravity) must be opposed by static friction upwards, just like for a stationary object resting on the hill.
Then the next leg takes over, and the same thing is the case. Static friction must hold back upwards to avoid sliding of the leg.
Now add more legs to the star. Many, many more. Soon you almost have a round circle, where the legs are the "points" of the circle that touch the ground for only a split-second. Nevertheless, the same is the case; while touching, static friction holds on to the touching point upwards to avoid it from sliding because of gravity.
[...] for the car to go up the hill, where the static friction is again pushing the car forward, countering the slipping of the tires that would result due to the rotation of the tires.
Yes. Again static friction pulls uphill to prevent sliding because gravity pulls down. The direction of static friction does not depend on the rolling direction; it doesn't care if you roll up or down the hill.
If the wheel accelerates at the same time, the static friction might point differently. This is again regardless of the rolling direction but only the acceleration direction comes into play.
How can you get a constant velocity when going down hill, when the static friction and gravity are both creating a net force down the hill (if you ignore rolling friction)?
You are exactly on point here. It can't have constant speed, if all forces pull the same way. Such thinking will lead to the understanding that static friction in fact must point the other way.
A question asks me to find the steepest hill a car can descend at constant speed given the static friction coefficient, but I think what is needed is the rolling friction coefficient. I don't think its possible to answer the question without rolling friction.
As mentioned in a comment, most questions would assume ideal world-models. No deformation of surfaces e.g. So rolling friction will be assumed 0 most often, unless you drive on a clearly non-ideal surface, like a sandy beach or a flexible trampoline.
The idea to solve this question is to remember the formula for maximum static friction:
$$f_s\leq \mu_s n$$
If you have a certain friction coefficient $\mu_s$, you can do your Newtons 2nd law calculations on the car and put this formula in with an $=$ instead of $\leq$, because you are looking for the maximum limit. Then solve it for the slope angle (the angle will be a part of the force components).
Best Answer
It's hard to make the wheels spin at high speeds because you're in a higher gear, so the torque at the wheels is less. So I assume you are only asking about wheel spin in first gear i.e. it's quite easy to spin the wheels when pulling away in first gear but much harder if e.g. you're travelling at 10 mph in first gear.
The reason is that if you're stationary and drop the clutch the angular momentum of the engine contributes to the torque. That is, the torque at the wheels is the torque from the engine plus the torque from angular momentum stored in the flywheel, crankshaft etc. This happens because the engine is spinning faster that it would if the clutch were engaged, so engaging the clutch slows the engine speed. The extra torque is given by:
$$ \tau = I\frac{d\omega}{dt} $$
where $I$ is the moment of inertia of the spinning bits of the engine and $\omega$ is the engine speed, so $d\omega/dt$ is the rate of change of engine speed. If you drop the clutch the engine speed changes rapidly so $d\omega/dt$ is large and the extra torque is large. If you ease the clutch out $d\omega/dt$ is small so the extra torque is small and the wheels won't spin.
When you're driving at (e.g.) a steady 10 mph the engine speed matches the wheel speed, so if you now suddenly stamp on the accelerator it's only the torque from the engine that's available to spin the wheels. You don't get the contribution from $d\omega/dt$.
To see this try driving at 5 mph, then disengage the clutch, rev the engine and drop the clutch. As the clutch bites the wheels will spin just as they do when the car is stationary.
It's worth noting that a powerful car can spin the wheels in first gear even without playing with the clutch. In fact an old sports car I had many years ago would spin the wheels in second gear in the dry and in third gear if the road was wet!