Riddle: can you swim faster upstream than downstream (with respect to the water)?

Question

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?

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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?

Answer 1

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:$$x + y + k(x-y) = \dfrac{d}{z}$$

Answer 2

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