- 23 Oct '05 19:13im studying physics now and we get all kind of funny problems to solve there. here is one:

an oldtimer is driving from A to B. there is no wind but its raining nonstop. the oldtimer has 2 problems: there's a big hole in the horizontal roof where its raining through, and the vertical windshield (90° to the roof) is missing. so through both of these openings its raining.

how does the amount of water that gets into the car depend on the constant speed the car goes with?

ps: sorry for bad translation - 23 Oct '05 21:58

Are you looking for help for your homework? If so then we shouldn't do more than give some hints.*Originally posted by crazyblue***im studying physics now and we get all kind of funny problems to solve there. here is one:**

an oldtimer is driving from A to B. there is no wind but its raining nonstop. the oldtimer has 2 problems: there's a big hole in the horizontal roof where its raining through, and the vertical windshield (90° to the roof) is missing. so through both of these open ...[text shortened]... into the car depend on the constant speed the car goes with?

ps: sorry for bad translation - 24 Oct '05 06:32

Well it seems to me that the rate of rain entering the car from the roof is constant no matter how fast the car is going. That rate would depend on the density of water per unit area (assume an infinitesimally thick horizontal layer) and the rate at which the rain is falling.*Originally posted by crazyblue***no, i solved it already. i just posted it here for your entertainment**

and for the hole sizes, you can do that in general as A1 and A2 or whatever you want to call it. but it does not really matter for the solution.

The amount coming in through the broken windshield would depend on how far the car can move in the time it takes the rain to fall from the roof's height to the hood's height; in other words, from the top of the missing windshield to the bottom of it (assuming that the water will run off the hood once it hits the hood and not do so into the car itself but instead onto the street). This quantity of water should increase linearly with the speed of the car. - 24 Oct '05 20:54 / 1 edit

Yes, but surely the oldtimer wants to know how fast he should drive in order to get the least wet? Assume he can also measure the slant angle of the rainfall from vertical.*Originally posted by AThousandYoung***Well it seems to me that the rate of rain entering the car from the roof is constant no matter how fast the car is going. That rate would depend on the density of water per unit area (assume an infinitesimally thick horizontal layer) and the rate at which the rain is falling.**

The amount coming in through the broken windshield would depend on how f ...[text shortened]... ad onto the street). This quantity of water should increase linearly with the speed of the car. - 25 Oct '05 03:37

The fact that there's no wind implies that the rain is coming straight down. Therefore the faster the car moves the more water gets in.*Originally posted by iamatiger***Yes, but surely the oldtimer wants to know how fast he should drive in order to get the least wet? Assume he can also measure the slant angle of the rainfall from vertical.** - 25 Oct '05 11:18

The amount of rain that gets in the top is proportional to time spent in the rain. Therefore as we are travelling a distance X (say) it is proportional to X/v (which is time spent in rain).*Originally posted by AThousandYoung***Can you clarify where I am making my mistake?**

A) The amount of rain that gets in from the top is constant with respect to velocity.

B) The amount of rain that gets in from the front increases as forward velocity increases.

Therefore:

a) proportional to X/v

b) proportional to v

Give these both constants and:

R = a*X/v + b*v

To find the minimum simply find the derivitive and solve for dR/dv=0. - 25 Oct '05 13:32

are you sure about b) being proportional to v?*Originally posted by XanthosNZ***The amount of rain that gets in the top is proportional to time spent in the rain. Therefore as we are travelling a distance X (say) it is proportional to X/v (which is time spent in rain).**

Therefore:

a) proportional to X/v

b) proportional to v

Give these both constants and:

R = a*X/v + b*v

To find the minimum simply find the derivitive and solve for dR/dv=0. - 25 Oct '05 20:00 / 1 edit

A) is correct. The amount from top depends on the time. Therefore the faster you go, the less rain gets in from the top.*Originally posted by AThousandYoung***Can you clarify where I am making my mistake?**

A) The amount of rain that gets in from the top is constant with respect to velocity.

B) The amount of rain that gets in from the front increases as forward velocity increases.

B) is wrong. Think about it some more, the answer might be unexpected.

Some 2D-sketches are very helpful here (at least they helped me solve it hehe).

edit: Oh, maybe your A was wrong too. Not sure how you meant it... - 26 Oct '05 01:38

I didn't take into account a finite distance travelled. If I take this into account, then the slower the car goes, the more rain comes in the top before the car gets there. However the amount that comes in the front will be the same no matter what speed for a given distance. I took time travelled as a constant instead of distance travelled.*Originally posted by crazyblue***A) is correct. The amount from top depends on the time. Therefore the faster you go, the less rain gets in from the top.**

B) is wrong. Think about it some more, the answer might be unexpected.

Some 2D-sketches are very helpful here (at least they helped me solve it hehe).

edit: Oh, maybe your A was wrong too. Not sure how you meant it...

Therefore, the*faster*the car goes, the less rain gets in - because the car gets out of the rain faster!

Xanthos, I think your b) is incorrect. It is proportional to v, but it's also inversely proportional to the time spent at v. Basically all the rain directly in front of the windshield along the path will get into the car no matter how fast the car goes given a constant distance.

So the answer is - go as fast as possible! Right?