## Sunday, November 14, 2010

### Problem of the Week #2

As always, we welcome you to send in solutions, or even any ideas you have about how to solve the problems to the.physics.virtuosi@gmail.com with “problem of the week” in the subject line. We will keep track of the top Virtuosi problem solvers.

Why did the ant cross the rubber band?

A rubber band is held fixed at one end. The other end is pulled at a velocity v. At time t = 0, the rubber band has a length of L, and an ant starts crawling from one end to the other at velocity u.

Does the ant reach the other side? If so, how long does it take to get there? Assume that the rubber band is able to be stretched indefinitely without breaking.

1. Assuming that the ant is a point and always moving at a velocity v with respect to where its feet are touching...

Brute-force solving the differential equation gives me

t = (L/v) * ( Exp[v/u] - 1)

So the ant always makes it to the end of the band!

In the limit that v->0, this returns the expected result that t = L/u.

In the limit that v >> u, the time it takes is blowing up (approximately) exponentially fast in v. It makes sense to me that the ant would always finish, but that as v>>u it would take a mighty long time.

That it would have this specific functional form isn't obvious to me, but maybe there's a more clever way than just solving the diff-eq.

2. How come ants?

3. Thanks ants.

Thants.

4. @Anon. Coward 8:20
what differential equation did you use?

5. The end of the rubber band is moving at a constant speed. The ant is accelerating. The position of the ned of the band varies with t, while the position of the ant varies with t^2. t^2 > t for large t. The ant always wins.

--Charles

6. Re: Anonymous at Nov 21

Wow. I like your argument a lot. Much more than the differential equations I used.

The details are a teensy bit off: the acceleration of the ant is not constant, so his position doesn't vary as t^2. But, because he is undergoing SOME acceleration, his position will go as some power of t greater than 1, which - as you elegantly point out - will always beat t^1 for sufficiently large t.