# Homework Help: Mechanics - Swimmer going upstream at an angle

1. Sep 6, 2008

### CaptainSFS

1. The problem statement, all variables and given/known data

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to the water.

1) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?

which I answered by first find the time (125 seconds) and then using a kinematics and finding that it was 1000m. It is correct.

...but the second question has been plaguing me all day.

2) If she wants to be swept a smaller distance downstream, she heads a bit upstream. If she heads 36° upstream, how far downstream is she swept before reaching the opposite bank?

2. Relevant equations

kinematics (X = X(initial) + V(initial) * t + (.5) * a * t(squared)) ...although acceleration is 0...
V (naught x) = V * Cos(theta)
V (naught y) = V * Sin(theta)

3. The attempt at a solution

well... I have tried many things. um... everything i can think of like using kinematics, to doing it just with geometry. I wont go into detail with all the process's because frankly I've lost track of them all. I have pages of notebook paper scribbled with math and what not. Here are some answers that i came up with that don't work. These are all rounded.
677ft, 1013ft, 239ft, 311ft, 403ft, 595ft, 540ft, 688ft, 701ft, 840ft, 845ft, 850ft, & 925ft.
Yes I understand that's quite an array of answers. I thought this was gonna be an easy problem but when i input the answer into my online homework, they're all wrong.

Could someone please explain how to do this one exactly? I wouldn't think it to be too complicated of a problem. Thanks for any help. :)

2. Sep 7, 2008

### Chi Meson

FIrst problem is we have to establish what "36º upstream" means. Is it 36º away from the line that points directly across the river? I assume so.

So the time it takes to cross the river depends only on the component of the velocity that points directly across the river. A general rule for components: the cosine of the angle away from any axis gives you the component along that axis. So the v cos 36º will lead you to the time it takes to cross the river.

Now the net speed down river will be the difference between the speed of the river and the appropriate component of the swimmers velocity.

I get an answer that is different from all of your attempts.

3. Sep 7, 2008

### CaptainSFS

ah, well. if you are right about it being 36 degrees away from the line perpendicular to the river then that will probably be the problem. I read 36 degrees upstream as 36 degrees off the side of the river. I'll work it out again in a bit and let you know my results.

Thanks for you help so far! :)

4. Sep 7, 2008

### CaptainSFS

Hey I wanna say thanks for your help. I made it 36 degrees from the original path of the swimmer, and not the river, and I got the answer on the first try. Haha. oh well. Thanks for the help! :D

-SFS
peace.

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