I would argue that an ant cannot be controlled in a way such as to walk from 1 end of a rubber string to the other, by this logic the riddle is invalid.

Also depending on perspective and a much simpler version the answer the string could be tied together in which case if the ant walks back it has reached the end.

Now heres a riddle
Sorry if it has an ad

I'd go with the previously-posted answer that the ant reached the end at the very beginning of the trip. If the prof doesn't accept that answer then he probably shouldn't be teaching.

The tease here is the length stretched. It throws you off that it's such a huge number.

Here's how to make the thought exercise into a practical test. Use a ruler, a piece of paper, and a pencil.

Use the centimeter portion of the ruler. Mark two points on the paper at the halfway point, one centimeter part. At one line , put a a circle.

Move the two lines one centimeter apart in opposite directions (for example, if the midpoint was 4 and 5 centimeters, you'd move it to 3 and 6 centimeters). Move the dot forward one centimeter toward the other opposite line every time you move the lines.

...42....

What class is this? Wordplay skullduggery aside, sounds like this could be modeled with calculus.

EDIT: yup, calculus solution exists. No wordplay here.

At t = 100,000 s, the ant will have traveled 100,000 cm (1km) and the rope will have stretched to 100,000 km. Some streching will have occurred behind the ant (virtually moving it forward), but most will have occurred in front of the ant (greatly increasing the distance it has to travel.

EDIT1: After 100,000 seconds the ant will have traveled 1 km under his own engine. But any uniform stretching of the rope AT ALL will make it so he does NOT reach the end of the rope at t=100,000 s. The distance behind him will have increased dramatically, but much, much more rope will still be in front of him.

The answer will be >>100,000 s.

EDIT2: Calculus method:
any distance along the rope = x
start (0 km) = x0
end of the rope = x1
(x1 changes based on time steps)
x1 travels away from x0 at speed = v
(v = 1km/s)
t = seconds
total length of the rope is x1(t) = x1(t=0) + v*t

the intermediate lengths of rope will be traveling at different fractions of v based on position:
speed of rope at point x = vx

vx = total speed * fraction = v * (x/x1(t))

Verify: vx0 = 0 km/s, vx1 = 1 km/s

EDIT3:
basic ant speed (1cm/s) = a
total ant speed under stretching is therefore: a + vx

EDIT4: OK, therefore:
ant speed = a + v * (x/x1(t=0) + v*t) , ant speed is a derivative of its position x

EDIT5: Now I'm just cheating (SPOILERRRRSS):
http://en.wikipedia.org/wiki/Ant_on_a_rubber_rope

...24...

http://www.youtube.com/watch?v=jR1_epbmI2g