How far does the fly fly in the end?

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Two cyclists begin cycling towards each other from a distance of 20km at a speed of 10km/h. They will meet after one hour, during which a fly travels back and forth between them at a speed of 15km/h. The fly, therefore, covers a total distance of 15km before the cyclists meet. The discussion also touches on the von Neumann method for solving similar problems with varying speeds. This classic problem serves as an engaging introduction to concepts in calculus and series.
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Two cyclists start cycling towards each other with an even speed of 10km/h. They start of 20km away from each other. At the same time a fly starts from one of them and flies to the other one, back to the first, and so on until they meet. How far has the fly flown in the end if it flew with a speed of 15km/h?

I'm thinking it has to be 15km, since the two guys will meet half way after 1 hour, and the fly flies 15km in one hour. Can someone confirm this?
 
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i confirm, you got it right :biggrin:
 
Just for fun, did anyone try the von Neumann method to solve the problem - with different speeds for the cyclists? :)
 
Not yet, but as you said, it should be done - just for fun.
 
Tide said:
Just for fun, did anyone try the von Neumann method to solve the problem - with different speeds for the cyclists? :)

That is exactly how my calc 2 teacher introduced us to series, starting with that question then telling the story about Von Neumann.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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