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