1. I am a grade 12 student. This is a problem that was assigned by my physics teacher. However, neither of the other 2 physics teachers in the school were able to solve it.

A sunbather, drifting downstream on a raft, dives off the raft just as it passes under a bridge and swims against the current for 15 min. She turns and swims downstream, making the same total effort and overtaking the raft when it is 1.0km downstream from the bridge. What is the speed of the current in the river?

We were given that the answer is 2km/h but we have to show how to get that answer. Can anyone solve this?  2.

3. Look at it this way. Making the same effort upstream as downstream places her in the same relative position to the moving water. The only fact that matters for the 2 km. is that 30 minutes past and her and the raft traveled 1 km.

What is 1 km divided by 1/2 hour?

Look for a formula that adds the unknown moving water speed with the plus and minus of her unknown swimming speed. Her speed doesn't matter, because it becomes zero when equal time for equal and opposite speed is added together.  4. i think i get it. if she goes upstream at velocity x for 15 min and downstream with an equal effort of velocity x for 15 minutes then the velocity x cancels and all that is left is 30 mins elapsed. the velocity of the raft is equal to the velocity of the current so Vc= d/t= 1km/.5h= 2km/h

Thanks  5. The intuitive solution given by Wild cobra is excellent.
But not all problems can be solved this way.
Which is why Mathematics is so important to Physics.

Here is a Mathematical solution ......

Basic governing equation is S = R T
S = distance, R = rate, T = time

Two equations can describe the given problem,
1/ one for swimming upstream; u = [ x - c ] 1/4
x = swimmers rate, c = current rate, 1/4 hr = 15 min
2/ one for swimming downstream; d = [ x + c ] 1/4

We also know that d - c = 1 km because the swimmer
swam 1km more downstream than upstream

Subtracting Eq 1 from Eq 2 we get .....
d - c = 1/4 [ (x + c) - (x - c) ]
1 = 1/4 [2 c]
c = 2

So the current c is 2 km/hr  6. Originally Posted by paulfr2
Two equations can describe the given problem,
1/ one for swimming upstream; u = [ x - c ] 1/4
x = swimmers rate, c = current rate, 1/4 hr = 15 min
2/ one for swimming downstream; d = [ x + c ] 1/4
You seem to be assuming that the swimmer took the same time to swim upstream as downstream; that is not correct.

Suppose the swimmer speed in still water is and that the speed of the current is The distance covered by the swimmer going upstream is and the raft will have drifted downstream a distance of As the swimmer heads back, the time taken to cover the distance back to the bridge is and so during this time the raft would have drifted a further downstream. Now suppose is the time taken for the swimmer to catch up with the raft from that point onwards. The swimmer would swim a distance of while the raft would drift a further downstream. Thus

• From this, it should follow by patient algebraic manipulation that   7. If the swimmer swims 2km/h (equal to current), he/she gains no distance when swimming against current (.5km current cancels .5km swimmer). .5km = Distance in 15 mins.

At this moment, 15 mins, the raft is .5km away from swimmer, while swimmer is still near/under the bridge.

Another 15 minutes passes...

Raft is at 1km, and elapsed time is 30 mins (2km/.5h)

Swimmer is at 1km in 15 minutes, because 2km/.25hCurrent (.5km) + 2km/.25hSwimmer (.5km) = 1km. 2km/.5 (1km)

They're at the same distance in 30mins elapsed.  8. Interesting  9. Originally Posted by Sophie Originally Posted by paulfr2
Two equations can describe the given problem,
1/ one for swimming upstream; u = [ x - c ] 1/4
x = swimmers rate, c = current rate, 1/4 hr = 15 min
2/ one for swimming downstream; d = [ x + c ] 1/4
You seem to be assuming that the swimmer took the same time to swim upstream as downstream; that is not correct.

Suppose the swimmer speed in still water is and that the speed of the current is The distance covered by the swimmer going upstream is and the raft will have drifted downstream a distance of As the swimmer heads back, the time taken to cover the distance back to the bridge is and so during this time the raft would have drifted a further downstream. Now suppose is the time taken for the swimmer to catch up with the raft from that point onwards. The swimmer would swim a distance of while the raft would drift a further downstream. Thus

• From this, it should follow by patient algebraic manipulation that Can you explain this (I understand v=d/t but your explanation is like a foreign language to me) to someone like me who hasn't taken Physics yet and who got a C in high school Algebra (taking College Algebra next semester and will be paying verrry close attention as I need to learn up to Calculus for my Bio degree)?  10. [quote="Sophie"] Originally Posted by paulfr2

You seem to be assuming that the swimmer took the same time to swim upstream as downstream; that is not correct.

From this, it should follow by patient algebraic manipulation that ================================================== ======

I would like to see that patient algebraic manipulation.

And while your solution is interesting, I think the words "with the same effort" would imply the swimmer made the same strokes for the same period of time ..... not an unreasonable assumption.

Nevertheless, your more general solution is quite interesting.  11. By with the same effort I understand it to mean that the swimmer was swimming with the same speed that she would in still water, not that she took the same time.

Okay.

•  Sub into •     12. With the same effort does indeed mean that she would be swimming with the same speed as in still water. But with the same total effort, means that the same amount of energy was expended going up river as down, and its easiest to assume, if it doesn't atleast suggest, that it's the same speed for the same time as opposed to half the speed for twice the time or twice the speed for half the time.  Posting Permissions
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