Re: the ant on a rubber rope
Consider the problem in coordinates which get stretched together with the rubber band:
dy(t)/dt = c/exp(H t) = c exp(-H t)
This is easy to solve, and the total distance of the ant approaches c/H. For c>H, it reaches the other side, for c<=H, it does not.
Re: the ant on a rubber rope
Quote:
Originally Posted by
mfb
Consider the problem in coordinates which get stretched together with the rubber band:
dy(t)/dt = c/exp(H t) = c exp(-H t)
This is easy to solve,
.
thank you for your answer; I don't see precisely the equation to be solved, would you be kind enough writing it
Quote:
Originally Posted by
mfb
and the total distance of the ant approaches c/H. For c>H, it reaches the other side, for c<=H, it does not.
for any value of c and H, does the ant win the race if D<c/H ?
Re: the ant on a rubber rope
dy(t)/dt = c exp(-H t)
This can be solved by integration of both sides from t=0 to t=T (or infinity).
I missed the additional D in your formula, but this is just a scaling of c:
dy(t)/dt = c/D exp(-H t)
The ant reaches the other side for c/(DH)>1. In other words, the initial velocity of the ant has to be larger than the initial extension velocity of the rubber band.
Re: the ant on a rubber rope