But if the cars produce downforce this is no longer true because you brake harder (more friction available) at higher speeds!
This is how F1 cars pull 4G when breaking. Some custom cars (like one of Ken Block’s last monsters or the Valkyre) use active aero braking to even greater effect.
2. I know you know this, but for the sake of others, it's when _braking_ (applying the brakes), not _breaking_ (becoming broken).
I'm not a pedant. But these errors jump out at me and I'm always a bit surprised and dismayed at this dichotomy; in our field, somehow the requisite attention to detail, the precision inherent to communicating scientific concepts, code, algorithms and formulae, is so often just abandoned when it comes to prose.
Honestly that was a typo and I noticed too late to edit. Thanks for catching
(I was suprised to see a cow jumping up on a ~3m rock ledge like it was nothing)
Point is that’s not always true. If they are the same type of car, and the car happens to be the kind with downforce, then their rate of deceleration greatly depends on air speed. A downforce car decelerates faster at higher speeds.
This is why you often see race cars lock their wheels towards the end of the braking zone, never at the beginning. The driver has to release the brakes as the car decelerates because there’s less friction available. You go from pulling 4G at the beginning of the braking zone to pulling the usual 1G once your speed drops enough for downforce to become negligible.
Alos! Many non-race cars actualy produce lift. Meaning the faster car decelerates at a slower rate than the slower car (0.8G vs 1G), making the effect from OP even more pronounced.
That’s not the only reason, and I’m not even sure it’s the majority reason.
Braking in a straight line offers more braking traction than braking while turning. What happens towards the end of a braking zone? The turn in. (Which also shifts weight to the outside tire and away from the inside tire.)
https://www.youtube.com/watch?v=RWwGFDynOHo
For these basic virtual car experiments, BeamNG.drive is a pretty good physics simulator. You can open its built-in tools and run braking tests directly.
5 km/h = 0.13 meter
30 km/h = 4.5 meter
60 km/h = 14 to 18 meter
65 km/h = 21 to 24 meter
You need 150% the distance at 65 vs 60.
In Dutch its remweg (something like brakeway) and my mind was occupied not finding the English word for it.
It cannot be both. It mathematically cannot be both. They can brake at the same rate (acceleration) or intensity (conversion of kinetic energy into heat) but because they are traveling different speeds those two values cannot be the same for both cars.
The math you did was for intensity, not force/acceleration, which because of the ^2 in the KE equation exaggerates the difference. Whereas if you did the math based on force you'd get a mild, linear, difference.
> and braked at the same rate,
You're being a bit sly with word choice here. You're doing the math for conversion of KE into heat whereas in common parlance "rate" means force/acceleration.
Braking "at the same rate" [of energy conversion] is way less actual braking force for the faster car.
This is basically the same kinetic energy into heat math wherein you can descend a grade at a low speed, apply a force and be fine and descend the same grade at a higher speed and apply the same force and cook the brakes. Or you can apply less force, and get the same amount of energy conversion into heat (i.e. your wording trick in the proposed scenario)
You've taken what's basically the math behind trucks descending a grade (rate of energy conversion is actually limited by ability of brakes to shed heat, not friction) and re-framed it as cars stopping to create a trick question.
You are right that the faster car is converting kinetic energy into heat faster per unit time. It also has less time to do so. The work formulation of the problem makes it obvious that these have to cancel out exactly.
Couldn’t help but notice you misspelled car twice but only when talking about the blue car..