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Two boys come out to meet each other from points A and B at a speed of 5 km/h each.

The distance between A and B is 10 km.

At the same time, a dog runs out from point A to point B at a speed of 10 km/h. The dog runs to the second boy, turns around, and then runs to the first boy, and so on.

How far will the dog run before the boys meet if the dog needs X time to turn around.

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closed as off-topic by Moishe Kohan, Paul Frost, cmk, Daniele Tampieri, mrtaurho Jul 21 at 14:52

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  • $\begingroup$ How long does it take for the boys to meet? The dog runs at constant speed during this time. $\endgroup$ – saulspatz Jul 20 at 14:27
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    $\begingroup$ What have you tried? $\endgroup$ – Arthur Jul 20 at 14:28
  • $\begingroup$ @Arthur I tried only dumb solution step by step on a piece of paper. I want to know if there is more elegant solution. $\endgroup$ – vladon Jul 20 at 14:37
  • $\begingroup$ @saulsplatz it's easy if there is immediately turnaround time. But during dog's turnarounding (time X) the boy moves some distance, and dog stays behind him. $\endgroup$ – vladon Jul 20 at 14:38
  • $\begingroup$ @vladon try to adjust the speed of dog based on the turn around time $X$ $\endgroup$ – rsadhvika Jul 20 at 14:41
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I made a spreadsheet taking $X$ to be $1$ second. The dog only makes three turns before the boys meet, so it runs $3$ seconds less than $1$ hour and covers $10\cdot \frac {3597}{3600} \approx 9.99$ km. Shortening the turn to $\frac 12$ second just made one more turn, so the dog covers just about $10$ km for any reasonable turn around time.

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