3 ms·
> Let's assume that the brakes are working at their limit, and as they do so, they are shedding energy at a maximum rate that doesn't change. The cars are ident
by krukah 3y ago
> Let's assume that the brakes are working at their limit, and as they do so, they are shedding energy at a maximum rate that doesn't change. The cars are identical, so they are both shedding energy at the same rate per unit distance.
Interesting (and I think very reasonable) first-order assumption. Maybe to second-order, the calculation could assume the braking force is proportional to velocity, which I think is roughly true of friction generally, but is also harder to model.
- bell-cot 3y agoIANAE (...not an engineer), but "braking force is proportional to velocity" is not how brakes, or friction, work.
- purpleflame1257 3y agoNo, but it is how first-order atmospheric drag is modeled, which may be what confused the GP comment.
- Karellen 3y ago> > they are both shedding energy at the same rate per unit distance. > Interesting (and I think very reasonable) first-order assumption. Really? That triggered alarm bells in my head immediately. I mean, it might be true, but I'd need to do the math to figure it out one way or another. > the calculation could assume the braking force is proportional to velocity, which I think is roughly true of friction generally Again, not my intuition at all. I'd have gone with the braking force being constant at non-zero velocity. (And force is variable up to the limit of static friction when at zero velocity.)
- calfuris 3y agoYour intuition about braking force matches the thing that triggered alarm bells in your head: work = force x distance, so if the force is approximately constant the kinetic energy dissipated will be approximately constant per unit distance.