How far does the fly fly in the end?

  • Thread starter Thread starter TSN79
  • Start date Start date
AI Thread Summary
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.
TSN79
Messages
422
Reaction score
0
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?
 
Physics news on Phys.org
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 tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

Similar threads

Back
Top