For what it’s worth, if you’re trying to break things down to a molecular level, you fail, because unless the tires are stored at absolute 0, there is molecular motion of some sort.
To make it even more fun:
There is always one part of a train that is moving backwards. What is it?
I so want to know the answer to this one!
The only thing my feeble mind can muster is the photons emitted by any lights pointing backwards on the last car. 
The bottom-most part of the wheel flange, which sits below the contact point on the rail. If the point touching the rail is stationary, any point below that is going backwards.
So in fact there are several points as most trains have more than one wheel 
It’s the same question, really. Except that the wheels of a train extend a little below the part that’s actually rolling on the rail. So that part which extends below the rail is moving backwards.
Well now we’re talking about moviephysics, and that’s a whole other thing. If it’s them pesky Duke brothers and the General Lee is in the middle of an implausibly large jump, the answer is “whatever will look good and conform to moviephysics cliche”.
Is there any way we can throw Heisenberg into the pot too?
I’m not certain.
Throw him into the trunk with the cat.
Yes, but once we put him in, we have no way of knowing how fast his car is moving.
Could you provide a cite for that? Or what is the 1-2% figure supposed to be (one to two percent of what)?
If you’re saying that a car which moves forward three inches while a point on the tire is in contact with the ground, will have that point on the ground slide w.r.t. the pavement by 1 to 2% of that three inches, I have to say I can’t believe this without a very convincing explanation or a very authoratative source.
Granted, but suppose we’re talking about an ordinary, real-life car performing a plausible jump - if the wheels happen to be braked below road speed for a moment while it is airborne, then no part of the car is guaranteed to be stationary with respect to the road.
Thank you for this. Before your post my head was hurting. I thought an absence from the Dope had made me blond(er).
This confused me too. I’m not a tire expert by any means, but I poked around and found a couple papers that talk about tire slip (both pdf files):
Dynamic Friction Models for Road/Tire Longitudinal Interaction (Carlos Canudas-de-Wit, Panagiotis Tsiotras, Efstathios Velenis, Michel Basset and Gerard Gissinger)
Modeling of Pneumatic Tires by a Finite Element Model for the Development a Tire Friction Remote Sensor (H. Holscher, M. Tewes, N. Botkin, M. Lohndorf,
K.-H. Hoffmann, and E. Quandt)
From the first:
The [longitudinal] slip rate results from the reduction of the effective circumference of the tire as a consequence of the tread deformation due to the elasticity of the tire rubber. This, in turn, implies that the ground velocity v will not be equal to rω.
From the second:
When a driving torque is applied to a pneumatic tire, a tractive force is developed at the tire-ground patch. Consequently, the tread before and in the contact area is compressed and a shear deformation of the sidewall of the tire is developed. As the tread is compressed before entering the contact region, the distance that the tire travels when subject to a driving torque will be less than that in the free rolling case. This phenomenon is the so-called longitudinal slip…
To be honest, this is still unclear to me, but what these references seem to be saying is that the “slip” of the tire is defined as the difference between the actual ground velocity (v), and the expected tire circumference velocity (rω), and that this “slip” isn’t necessarily from the tire contact patch slipping with respect to the road, but from deformation of the tire.
It’s from an old ABS training manual, I have somewhere. I am on the road this week, so I can’t give you any cite at the moment other than my memory of that text.
But before you dismiss this out of hand, let’s think about it.
Now assume we are driving down the highway at 75 mph in a very large not very areodynamic SUV. It just so happens that we have a tailwind of exactly 75 mph from directly behind us. We have no wind resistance. Tire slip against the pavement will be either very small or non-existant. What happens when the wind dies and we now have to punch a hole through the air? More throttle will have to be applied, and more power will be directed to the drive wheels. The body of the car wants to slow down due to drag, and the wheels want it to speed up to maintain 75 mph. This is going to result in some tire slip. Not much mind you, but some.
I read this as “There is a part of a car that is going oomph at all times…” I was ready to agree with the OP’s friend.
In that case it would be a car driven by one of these guys and the noise would be from the passenger compartment
But from zut’s cite, it appears that “slip” is not slip in the sense of skidding. Rather, due to compression of the tyre due to torque just before the contact patch, the effective rolling diameter reduces under power. Hence the distance the tyre travels per revolution is less (presumably per your figures 1%-2% less) than would a tyre not under power and this is referred to as “slip”.
This makes much more sense to me. I just couldn’t imagine that a tyre would endure an equivalent of 100 to 200 miles of skidding down the road under full load for every 10,000 miles travelled. No way would it survive that.
Tires are not rigid and the contact patch actually flattens. So as a car moves forward, portions of tire – the contact patch — actually goes backwards, too. The ‘parts’ must move backwards briefly before continuing round.
No?
Relative to what, Philster?