Leaking Tire: Better be Slow or Fast?
van.physics.illinois.edu
“I do not know the exact answer, but here is what I think. Firstly, I will be assuming that your tire is a pressurized container of ideal gas. The number of escaping molecules will be proportional to number of particles that collide the hole in the tire, so the rate of leakage is proportional to the pressure inside. Second, tire is somewhat flexible and mechanical stresses or pressure may change the shape of the whole a bit, but it depends on everything including the exact profile of the road and the shape of the cut, so I will ignore it. If the road is perfectly smooth, then the pressure will be monotonically decreasing as time passes regardless of your speed. So it could be advantageous to reach the nearest service station as quickly as possible.
If the road is bumpy, each bump will slightly squeeze your tires, so the pressure will be increasing a bit, increasing the escape rate for some time. Hitting a bump will not will not cause a smooth change: there will be some pressure waves that will eventually relax and potentially a change in temperature slightly, but let's ignore those again, and assume that the pressure everywhere within the tire rapidly gets equilibrated according to a new volume of the tire, which will depend on the height profile of the road, say h(x). x here is a measure of the distance from your start point to the destination. Under these assumptions, the volume of the tire will depend on the roads shape, say by a function V(h(x)).
Let N be the number of particles at the start and dN/dt denote the rate of change of N. The reason why increased pressure result in increased escape rate is because it is proportional to number density of molecules inside, this can be written mathematically: dN/dt α -N(t) / V(h(x)) or n = -A N/V, where A is a constant of proportionality that we do not know. Rearrange this: dN/N = -A dt/V(h(x)). We can solve this by taking an integral from your initial to final point: ln(N(f)/N(i)) = -A int [(dx dt/dx) / (V(x) ] = - A int [ dx / ( s(x) * V(x)), where s(x) is your speed at every point along the route.
If you travel at a constant speed, we end up with: ln(N(f) / N(i) ) = -A/s int [dx/V(x)] Now please note that the integral is over your route, so it is independent of your speed. A is a constant, number of initial gas particles is also predetermined by how much you filled your tires at the first place. We can therefore conclude that number of molecules still within the tire, N(f) α exp(-constant/s). This means under our assumptions, the faster you go (s large), the smaller the exponent will be so the more pressure remains in your tire as you reach the service station.“
