Just for fun of it, let's see. Police Crown Victoria tops out at about 130 mph. Let's assume that at that speed power dissipated to drag equals power delivered by engine. Let's also assume that drag is purely quadratic in speed and that engine always operates at peak power during the 0-110mph acceleration. Yes, I know, that last one is going to make it a bit off, but without going into actual gear ratios of the cruiser, which I can't be bothered to look up, that's the best we can do.
So all in all, very rough estimate, but there is some physics in it.
Engine power: P = 250HP = 186kW
Vehicle mass: m=1700 kg.
Top speed: 130mph = 58m/s
Target speed: 110mph = 49m/s
Power dissipated to drag is given by kv³, and we know it to be equal to engine power at top speed.
Power coefficient: k = 0.953 kg/m.
At any given speed, the force is given by available power over speed.
F = (P - kv³)/v
Hence, the acceleration:
a = P/(mv) - kv²/m
The analytic solution to this equation is so ugly that I want nothing to do with it. On the plus side, numerical solution is easy.
I'm getting 545m traveled in 15s by the time it hits 110mph. That's almost exactly 1/3rd of a mile. Though, 15s is just a touch too fast for 0-110mph. It should take closer to 20s. So the actual number will be a bit higher. Maybe 1/2 of a mile or so.
Regardless, surely it's irrelevant. If you were still accelerating away from them once they hit 110, you were clearly going faster than them. All the math in the world won't help you.
I think he's hoping that they'd knock off 10mph or so if he can "prove" that the cop car couldn't get to 110. Difference in fines at 100 and 110 tends to be significant. (I don't know exact rates in Cali.) Problem is, he's likely to dig himself in deeper if he tries to defend. If it's just a fine, he should go with no-contest, pay the fine, and be glad that's all it cost him. I have a feeling that if the cops actually show up at the hearing, this guy won't be keeping his license.
