Fun little anecdote:

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

> 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.

IIHS video shows the relationship between kinetic energy and speed in a very intuitive way:

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.

There's a great Australian traffic safety ad that makes this same point: https://www.youtube.com/watch?v=7x7c0qNGbv0

>same intensity and deceleration.

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.

Cool anecdote!

Couldn’t help but notice you misspelled car twice but only when talking about the blue car..

heh, thats a fun little experiment.