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I came across this little riddle:

An amateur runner trains on a 1 250 m long track at a speed of $v_L = 2 m/s$. His dog runs ahead to the target, turns around and then runs back to him, then he runs back to the target, back, etc. His speed is when running forth $v_H = 5 m/s$ and when running back (because of the headwind) $v_Z = 4 m/s$. Calculate the distance traveled by the dog.

It is little bit like Zeno's paradoxes but with the additional complication that here the dog is running back and forth at different speeds.

Could anyone please give some hints how to elegantly tackle the problem or references where a solution can be found (Perhaps this riddle even has a common name under which it can be found in the literature)? Thank you!

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up vote 2 down vote accepted

A variation on the classic elegant approach works. The runner will take $625$ seconds to complete the course. The dog will take $250$ seconds for the first leg. After that, the dog averages $\frac{2}{\frac{1}{5}+\frac{1}{4}}=\frac{40}{9} \frac{m}{sec}$ and you can apply the usual approach. The main points are 1)use the harmonic mean to find average speed, as you are averaging over common distances 2)after the first leg, the dog makes round trips so you can average the speed. As the dog is making round trips for $375$ seconds, it covers $375*\frac{40}{9}=1666\frac{2}{3}$ meters in that time, which when added to the $1250$ it already did makes $2916\frac{2}{3}$ meters total.

Added after the comment to Tim van Beek's answer: the classic version has the dog at a single speed, say $5$ m/sec. Then the simple answer comes from the runner taking $625$ seconds, so the dog travels $625*5=3125$ meters. The complex answer is to sum the infinite series of legs traveled by the dog. The first leg is $1250$ meters, after which the runner has covered $250$ meters. So the dog runs back $\frac{5}{7}$ of $1000$ meters and returns to the end. Then you figure out the next leg of the dog. It is a geometric series which you can sum.

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+1 - why can you use the same average speed after the first leg? The length of the round trips is reducing constantly (or the other way round: why do you have to wait for the first leg to be completed?) In a way the situation seems to be structurally the same after the first leg. – vonjd May 13 '11 at 13:59
Because after the first leg the dog travels the same distance at 4 m/sec as at 5 m/sec. So over each cycle we can calculate the average speed, and given the average speed we don't have to worry about the length of each trip. The first leg isn't symmetric the same way, so we have to handle it separately. If the dog started at the far end of the track we could just use the average speed for the whole time. – Ross Millikan May 13 '11 at 14:04
Ah, now I understand - Thank you! – vonjd May 13 '11 at 14:13

This is a very famous problem, there is a hard way to solve it (formulate an infinite sum and calculate what it converges to) and a very easy way to do it (think like a physicist and apply the relation velocity = way per time).

I've heard the anecdote that a psychologist noticed that mathematicians always did it the hard way and needed several minutes to solve the puzzle, while physicists always did it the easy way and needed only seconds. When he asked John von Neumann to solve the puzzle, von Neumann answered within seconds. The psychologist asked "But you are a mathematician, you're supposed to do it the hard way and evaluate the sum!" von Neumann answered "That's what I did..."

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+1 for the anecdote! When it is so famous under what name does it normally go? And where can I find step-by-step solutions for both ways? – vonjd May 13 '11 at 13:23
gave the easy way a thought: do you think that it works when the dog is running back and forth at different speeds? – vonjd May 13 '11 at 13:31
The version I remembered was like the one posted at , which I think is funnier: "You mean, there's another way?" – lhf May 13 '11 at 14:24
@vonjd: I think Halmos made it famous under the name fly puzzle (if I'm not completely mistaken it also appears in his autobiography, I want to be a mathematician). An online version in Halmos's own words is e.g. here. The punch line there is: When the question was put to von Neumann, he solved it in an instant, and thereby disappointed the questioner: "Oh, you must have heard the trick before!" "What trick?" asked von Neumann; "all I did was sum the infinite series." – t.b. May 22 '11 at 20:37

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