3 ms·
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 r
by throw0101a 4mo ago
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 https://www.youtube.com/watch?v=i3D7XYQExt0
- senectus1 4mo agoheh, thats a fun little experiment.
- NamlchakKhandro 4mo agoIn what way is it fun?
- slicktux 4mo agoCool anecdote! Couldn’t help but notice you misspelled car twice but only when talking about the blue car..
- deleted 4mo ago[deleted]
- frogulis 4mo agoPerhaps the beginning of a new vowel harmony phenomenon in English
- Swizec 4mo ago> 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.
- tracerbulletx 4mo agoBut what if the cars are spherical cows?
- BigTTYGothGF 4mo agoCows can't roll that fast.
- zdragnar 4mo agoNot with that attitude
- deleted 4mo ago[deleted]
- elromulous 4mo agoOr on shabbos
- lstodd 4mo agoIt takes a stupid cow, but when they can climb mountains as they do it is not inconcievable that one can roll down. (I was suprised to see a cow jumping up on a ~3m rock ledge like it was nothing)
- ndsipa_pomu 3mo ago
- cucumber3732842 4mo ago>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.
- ThrustVectoring 4mo agoOP 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.
- AlexandrB 4mo agoThere's a great Australian traffic safety ad that makes this same point: https://www.youtube.com/watch?v=7x7c0qNGbv0 https://www.youtube.com/watch?v=7x7c0qNGbv0
- 6510 4mo agoNice 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. You need 150% the distance at 65 vs 60.
- linzhangrun 4mo agoIIHS video shows the relationship between kinetic energy and speed in a very intuitive way: https://www.youtube.com/watch?v=RWwGFDynOHo 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.
- terminalbraid 3mo agoVery upset this didn't rely on doppler shifting of the car colors