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A friend of mine posed a riddle to me:

A man swims upstream in a river, which is flowing at an unknown rate. He is wearing swimming goggles. At a certain point he loses his goggles. 10 minutes later he realizes he lost them, so he immediately turns around and swims back downstream to get them. When he finds his goggles, floating in the water, he finds himself at a point 500 meters downstream from the point where he lost his goggles (with respect to the ground, of course, not the water).

Question: At what speed (km/h) is the river flowing?






-------------- SPOILER -------------
My solution is one and a half kilometres per hour, but another friend does not agree with me, he says swimming upstream is more efficient (with respect to the water) than swimming downstream.

Is he correct? If so, why?

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3  
This looks almost like homework, but if you look at the structure of the question, the swimmer has nothing to do with the answer. The question says that the goggles moved 500m in 10min. Its a simple calculation to solve for the speed. –  Hal Swyers Sep 30 '12 at 14:01
    
@HalSwyers: you haven't counted the time the swimmer takes to get back downstream to the goggles, so the river speed is less than 500/600 m/sec. –  John Rennie Sep 30 '12 at 14:41
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I get 1.5 km/hour as well. I think Galileo would disagree with your friend. –  John Rennie Sep 30 '12 at 14:59
    
@JohnRennie your correct, that's what I got once I removed the infinite solutions. –  Hal Swyers Sep 30 '12 at 17:59
    
Infinite solutions? What forces come to play in swimming up/downstream? @HalSwyers "Homework"? I have seen people use this term to criticize questions before, mostly because they pose their question in a lazy way. Please explain the nature of your criticism. If I am wrong, please excuse me. –  pancake Sep 30 '12 at 21:37

3 Answers 3

Since this is not marked as homework... An equation for this problem can be created as follows:

Velocity of the swimmer is: $V_{sw}$

Velocity of the stream is: $V_{st}$

The ratio of velocity of swimmer and stream is: $k = \dfrac{V_{sw}}{V_{st}}$

The time the swimmer goes upstream is: $t_{u}$

The time the swimmer goes downstream is: $t_{d}$

The distance traveled by the goggles is: $d_{g}$

The distance the swimmer travels upstream is: $(k-1)V_{st}t_u$

The distance the swimmer travels downstream is: $(k+1)V_{st}t_d$

The equation then is:$$(k+1)V_{st}t_d - (k-1)V_{st}t_u = d_g$$

If one sets $k = 1$ and $t_u = t_d = \dfrac{1}{6}hr$ and $d_g = 0.5km$ then one gets the solution of $V_{st} = 1.5 \dfrac{km}{hr}$

For the sake of argument, lets set $k=2$ and $d_g = 0.5km$ and $t_u = \dfrac{1}{6}hr$. Let's first rewrite the equation: $$(k+1)t_d - (k-1)t_u = \dfrac{d_g}{V_{st}}$$

then as: $$(k+1)t_d - (k-1)\dfrac{1}{6}hr = \dfrac{0.5km}{V_{st}}$$

and: $$6t_d - \dfrac{1}{3} = \dfrac{1}{V_{st}}$$

This can be plugged into wolfram alpha to find the set of solutions.

Update for Berhard's sake (where now all you need to do is input the values in for $t_d$ and $V_{st}$ since I have already their units):

$$6t_d\dfrac{hr}{km} - \dfrac{1}{3}\dfrac{hr}{km} = \dfrac{1}{V_{st}}\dfrac{hr}{km} $$

Note: For future ref, the general form of the equation is:$$x + y + k(x-y) = \dfrac{d}{z}$$

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Ok, so I take it my friend telling me when swimming upstream a person will go faster with respect to the water than when swimming downstream is full of it? –  pancake Oct 1 '12 at 0:27
    
Thank you, by the way, for the extremely thorough explanation! –  pancake Oct 1 '12 at 0:30
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Yes, unless there is some other engineering aspect, but that is not part of the question. In the real world there are multiple effects that can complicate these sorts of things. There may be some sort of circumstances one can read into the question, but those have nothing to do with solving the question as stated. –  Hal Swyers Oct 1 '12 at 0:33
    
@HalSwyers At least your last equation is dimensionally not correct. –  Bernhard Oct 1 '12 at 5:59
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@Bernhard it is actually, I dropped the dimensions, all factors are of the form hr/km, so I didn't see the need to carry the dimensions since I thought it was self-evident –  Hal Swyers Oct 1 '12 at 9:25

he says swimming upstream is more efficient (with respect to the water) than swimming downstream.

Go to a swimming pool and try swimming in various directions, the water in any pool on planet Earth is moving at 67,000 miles per hour around the sun. If it is easier "upstream" you should soon find out.

If you move the water in this pool to the middle of the Amazon river, the body of water in which you are swimming is still travelling at about the same 67,000 MPH around the sun.

If there is some property of the water's motion that makes swimming in one direction more efficient, your friend should explain why it doesn't apply to movement of that water around the Sun

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You are also travelling at the same speed as the water around the sun so it becomes more to do with the motion of the water itself 'on top' of the 67,000 miles an hour If you had to swim against a current you are clearly going to cover less distance going against the current than with it A river with a heavy current could easily carry someone downstream without them even needing to swim so then you add their swimming motion to that and they will travel faster than the river's actual current Swim against the current and you will have a force against the body of the swimmer so if they use the same swimming motion then they will not get as far Simple

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