A blue care is travelling along at 70 units, and a red car (exact same make and model) is catching up to it going 100. When they're both right beside each other a bend in the road reveals an obstacle blocking both lanes, so both cars brake at the same intensity and deceleration.
The blue care stops right before the obstacle. Since the red car was going at a faster speed, and braked at the same rate, it doesn't managae to stop: but what speed is it going when it hits the obstacle?
The blue car, using ½mv², shed (~70²=) 4900 units of energy (we'll hand wave away the constants). So the red car, which had (100²=) 10000 units of kinetic energy to start, also shed 4900 units, which means it had 5100 units of energy when it collided, and so was going (√5100~) 71.
> The blue car, using ½mv², shed (~70²=) 4900 units of energy (we'll hand wave away the constants). So the red car, which had (100²=) 10000 units of kinetic energy to start, also shed 4900 units, which means it had 5100 units of energy when it collided, and so was going (√5100~) 71
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.
1. +1 insightful, thanks for sharing your physics knowledge
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.
Considering that where I saw it they were doing that unsupervised since at least ~200BC (this being the Illyricum and then Dalmatia), I don't think they have any problems climbing down now.
While it is true that some cars can brake harder due to downforce etc, the point from GP was that both cars brake/ decelerate at the same rate. Regardless of how exactly that deceleration is achieved.
> the point from GP was that both cars brake/ decelerate at the same rate
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.
> This is why you often see race cars lock their wheels towards the end of the braking zone, never at the beginning.
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.)
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.
OP wasn't explicit about taking the work = force * distance approach to dissipating energy. Two cars with the same mass and braking force (and thus deceleration) will put the same amount of work into the vehicle per unit distance, so will dissipate the same amount of energy in the braking maneuver.
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.
Nice bit of camera trickery. He says "both drivers react and a moment later they break", but the cars are still side by side. It (apparently) takes drivers 1.5 seconds to respond, the 5 km/h speed difference cuts the distance by 2 meter. Which apparently is a big deal. Rough estimate breaking distance:
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
The +5 km/h adds 6 to 7 meters or 8 to 9 if you account for response time.
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.
A blue care is travelling along at 70 units, and a red car (exact same make and model) is catching up to it going 100. When they're both right beside each other a bend in the road reveals an obstacle blocking both lanes, so both cars brake at the same intensity and deceleration.
The blue care stops right before the obstacle. Since the red car was going at a faster speed, and braked at the same rate, it doesn't managae to stop: but what speed is it going when it hits the obstacle?
The blue car, using ½mv², shed (~70²=) 4900 units of energy (we'll hand wave away the constants). So the red car, which had (100²=) 10000 units of kinetic energy to start, also shed 4900 units, which means it had 5100 units of energy when it collided, and so was going (√5100~) 71.
* Numberphile: https://www.youtube.com/watch?v=i3D7XYQExt0