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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.
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barjed: I am having some troubles with this riddle I have to solve by tomorrow for my Uni class.

Find the solution for the following: An ant is placed at one end of a rubber string; this rubber string is one kilometer long. The ant starts walking on the string towards the other end with constant speed of one centimeter per second. At the end of each second the string is stretched so its length is extended by additional kilometer.

Here we assume that the string can be stretched indefinitely and that the stretching is uniform. Units of length and time remain constant.

The question is, does the ant ever reach the end of the string?

Thanks a lot, guys!
What class is this? Wordplay skullduggery aside, sounds like this could be modeled with calculus.

EDIT: yup, calculus solution exists. No wordplay here.
Post edited June 06, 2011 by strixo
Okay, it's really late for me and way past bedtime. But I think the solution is pretty simple, see the picture I attached. The ant reaches the end of the string after 100,000 seconds and after traveling 100,000 kilometers (including the stretching of the string). Look at the plot to see why it can reach the end of the string -- basically the length of the string goes linearly with time, while the position of the ant goes quadratically.
Attachments:
gog.jpg (56 Kb)
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spindown: Okay, it's really late for me and way past bedtime. But I think the solution is pretty simple, see the picture I attached. The ant reaches the end of the string after 100,000 seconds and after traveling 100,000 kilometers (including the stretching of the string). Look at the plot to see why it can reach the end of the string -- basically the length of the string goes linearly with time, while the position of the ant goes quadratically.
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
Post edited June 06, 2011 by strixo
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Rohan15: ...42....
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http://www.youtube.com/watch?v=jR1_epbmI2g