Originally posted by: CraKaJaX
Originally posted by: Jephph
#2, speed doesn't have anything to do with it. I would say it would be the size of the windshield and how fast the rain is coming down...
Hence the disclaimer of identical cars and the rain coming down at the same rate, same pattern, etc.
Rate at which raindrops hit the windshield is speed dependent.
Imagine your hand passing through the rain. Liek imagine bitchslapping a raindrop. Do it at a million miles an hour. In 1 second you can slap let's say 200,000 raindrops. If you do this at 0.0001 mph, you might hit 1 drop in 1 second.
Flux is dependent on relative speed, and so if your relative speed between the windshield and drops is different, you will have different # of drops hitting.
Now, a lot of people are reading "rain falls at the same rate." Ok, this isn't clear. If it's saying the rain falls at the same rate relative to passengers in both cars, then the answer is obvious. It's like saying person A observed 500 drops, and person B observed 500 drops. Duh it's the same, but I don't htink that's what the question is getting at. It's more like saying the rain is falling at the same rate in both environments, and so will varying the speed affect the # of drops/second hitting?
I say yes. But please someone point out the flaw in my thinking before I start viciously defending my view like some people do with teh airplane/treadmill problem.
Originally posted by: DrPizza
Problem 2: both cars sweep out an equal volume, therefore each will be hit by the same number of raindrops. However, this ignores aerodynamic effects.
But volume is not the question here. We're dealing with drops hitting in a time period, which involves flux. You're no longer dealing with constants but with rates and time dependent things. Just based on the fact that speed is a time dependent variable, it's enough to say that changing this time dependent variable will alter the rate at which drops hit the car windshields.
Edit. I misread the speed. I thought it was 60 and not 65 for Car B. From initial analysis it may seem that the drops are teh same, but if integration is required to calculate this, we might be dealing with something different.