# Another ant riddle

Let me come back with a totally different ant riddle I hope this one has not been posted already. As far as I know, it should be attributed originally to Martin Gardner.

This time, let us put a (point-like) ant on the edge (mathematically idealized) one meter long elastic band. The ant walks at a constant speed of 1 cm/s towards the other end of the elastic band. Every second, the elastic band is streched one meter longer.
Will the ant reach the other edge ?

Does the midpoint remain fixed in position relatively to the ground?

Really good question.

I'd say the elastic band would reach its modulus eventually and burst, killing the ant in the process and giving the observer a nasty snap. So the ant wouldn't reach the end... alive anyway.

That or:

I think no. A centimeter is 1/100th of a meter. Assuming the band is fixed at one end, and stretched from the other, each section stretches relative to its position from the fixed end. This would allow the ant to only make it to the 1/100th mark at any given time, which is not even half way there.

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Let me come back with a totally different ant riddle I hope this one has not been posted already. As far as I know, it should be attributed originally to Martin Gardner.

This time, let us put a (point-like) ant on the edge (mathematically idealized) one meter long elastic band. The ant walks at a constant speed of 1 cm/s towards the other end of the elastic band. Every second, the elastic band is streched one meter longer.
Will the ant reach the other edge ?
This is poorly worded. The ant's progress is stated to be at a constant speed, but the stretching is not. So for instance, it may be that the band is stretched by 1 meter nearly instantaneously at the end of each second. Also, we may interpret the words '1 cm/s towards the other end of the band' to mean that each second brings the ant 1 cm closer to the end point regardless of the stretching, that is, 1 meter and 1 cm per second, and so will finish the course in 100 seconds.

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Defennder
Homework Helper
I think what he means is that the band is stretched uniformly to its new length and the "fractional position" (the new position of the ant over the new length of the band) remains the same before and after stretching provided the ant hasn't moved during the stretching.

So suppose the ant is at 0.01m of the band(after having moved 1cm). The band is then stretched by 1 metre uniformly ie. every single part of the band expands. The new length of the band is 2m. The position of the ant after it has been stretched once is $$\frac{1}{100} \times (100+100)$$

This may be generalised as follows. Let $$L_{k}$$ be the length of the band at any one time. Let $$x_{k}$$ be the position of the ant at any one time.

$$L_{k+1} = L_{k} + 100$$
$$x_{k+1} = \frac{x_{k}}{L_{k}}(L_{k+1}) + 1$$. The +1 term for x_k is because the ant moves 1 cm each second.

Ok so I was thinking of writing it out as a matrix and then see whether it's diagonalisable. But the expression for x_{k+1} isn't linear; it has $$L_{k}$$ as its denominator. Ugh I'm stuck.

There appears to be a problem with my formulation of the length and position of the band as given above; given that the band is stretched continuously and uniformly while the ant is walking, the recurrence relation should be continuous rather than discrete as I have given. But I don't see how to do it that way...

I refute my answer. The ant will reach the end. I can't put it on formal grounds. Something along the lines of its always making progress. I sort of visualized it, Ill try to formalize this tomorrow.

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Does the midpoint remain fixed in position relatively to the ground?
Not necessarilly. You may suppose it does if it helps you. The problem of the ant walking on an elastic band is very similar to conceptual problems one encounters when first dealing with general relativity-like concepts. The elastic band may be thought of as a space by itself independent of another embedding.
This is poorly worded.
Apologies if this is the case, I certainly cannot render justice to Martin Gardner. Buy his books ! On the other hand, you may want to try to word riddles in french, and I'll tell you how good it is. :tongue2:
The ant's progress is stated to be at a constant speed, but the stretching is not.
If this is the reason you complain about my wording, let me make it clear that this is exactly what I meant.
So for instance, it may be that the band is stretched by 1 meter nearly instantaneously at the end of each second.
Yes, this is a math but not physics if you will.
Also, we may interpret the words '1 cm/s towards the other end of the band' to mean that each second brings the ant 1 cm closer to the end point regardless of the stretching, that is, 1 meter and 1 cm per second, and so will finish the course in 100 seconds.
No, I do not see any possible confusion : the ant walks at a constant speed of 1cm/s locally with respect to the elastic band.
I think what he means is that the band is stretched uniformly to its new length and the "fractional position" (the new position of the ant over the new length of the band) remains the same before and after stretching provided the ant hasn't moved during the stretching.
Very good observation ! jimmysnyder said:
So for instance, it may be that the band is stretched by 1 meter nearly instantaneously at the end of each second.
This is physical, I put the word 'nearly' in there for a reason. I have googled the answer so I have no more to say about this puzzle. However, you have some nerve distinguishing between what is and what is not physical in my post.

This is physical, I put the word 'nearly' in there for a reason. I have googled the answer so I have no more to say about this puzzle. However, you have some nerve distinguishing between what is and what is not physical in my post.
Any time you have been posting about the riddles that I cared to share with people on PF, for the mere reason that I love riddles and enjoy them, you have been either throwing away a solution without explaining it (which is not very fair and interesting, anybody can google the solutions to 99.99% of the riddles posted here anyway, what is interesting is how you got your own), or you have been posting negative comments. Thank you. I get e^100 steps before he hits the other side. Thanks to mathematica, some fitting, and soem simple numerical methods.

edit: i think im off by a factor somewhere, but at least i got the idea.

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what is interesting is how you got your own
I definitely have difficulties with english. At this point, I really meant that I am not interested in the solution but in how to get it. I never meant to imply anybody would google the problem, and throw away the solution here later, which would be completely pointless.

I publicly apologies to Jimmysnyder for my own difficulties of communication, and want to state that I know he is an excellent puzzle-solver on this forum. I really meant I am frustrated if somebody gives me a solution without explaining how he got it, especially if I think this person really has an alternative solution to the one I know.

Defennder
Homework Helper
I get e^100 steps before he hits the other side. Thanks to mathematica, some fitting, and soem simple numerical methods.

edit: i think im off by a factor somewhere, but at least i got the idea.
How did you arrive at that answer?

This is a relatively straightforward analogue to a simple GR sort of problem.

To solve this problem, it's convenient to put down a coordinate system on the band that stretches with it. That way, we can always uniquely specify a point, without having to know at what time we're looking. So, let x run from 0 (the point where the ant starts) to 1 (the point the ant is trying to get to). At the initial time, x=0 and x=1 are separated by a distance of 1 m. To keep this independent of units, for the moment, call the initial length of the band $$l_0$$. Then, the distance at the initial time from the starting end to a point x is $$l(x)=l_0x$$.

The length of the band increases by 1 meter every second. Call this rate of stretching s. Then, the length of the band at time t is $$l(t)=l_0+st$$. So, the distance from the starting end to point x at time t is $$l(x,t)=(l_0+st)x$$. At this time, the distance from point x to point x+dx will be $$dl=(l_0+st)dx$$.

Now, if the ant moves from x to x+dx in a time dt sufficiently short that the stretching is negligible, we get the relation between physical and coordinate speeds:

$$\frac{dl}{dt} = (l_0 + st)\frac{dx}{dt}$$

Now, we're given that the ant's physical speed is constant - call it v. So, the ant's coordinate speed at time t must be:

$$\frac{dx}{dt} = \frac{v}{l_0 + st}$$

This can be integrated rather simply and, given that the ant is defined be at x=0 at t=0, we find that $$x(t)=\frac{v}{s} \ln (1 + \frac{st}{l_0})$$

Since ln(x) has no upper limit, the ant must eventually reach x=1. We can solve for the time at which this will happen:

$$t_1 = \frac{l_0}{s} \left (e^{\frac{s}{v}} - 1 \right )$$

Plugging in numbers, this gives $$t_1 = e^{100} - 1\ \mathrm{s}$$.

Defennder
Homework Helper
Now, we're given that the ant's physical speed is constant - call it v. So, the ant's coordinate speed at time t must be:

$$\frac{dx}{dt} = \frac{v}{l_0 + st}$$
I don't understand this part. Your formulation assumes that x=0 is always the start and x=1 is always the end. But where did this expression come from?

I don't understand this part. Your formulation assumes that x=0 is always the start and x=1 is always the end. But where did this expression come from?
This just comes from recognizing that $$\frac{dl}{dt}$$ is the physical velocity relative to the elastic band, which is given to be constant, and which I've called v for convenience. I then use the expression above to solve for the coordinate velocity.

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Defennder
Homework Helper
Took me some time, but I think I got it, thanks.

This is basically Zeno's paradox with an ant and a elastic band, before the and can reach the end he will have to reach the middle, but before he reaches the middle he has to reach the 1/4 point, but he will never reach any of those points because the elastic is constantly stretching at a speed that is greater than the ants.

Defennder
Homework Helper
Actually the conclusion was that the ant will reach the end of the elastic band. The ant is travelling at a constant speed, and bear in mind that the every part of the elastic band stretches at the same rate, including the portion which the ant has already traversed. If the stretching was limited only to the endpoint of the band, then of course the ant will never reach the end.

in one the and and moves 1cm then band length doubles instantly. ant is now 2cm along band of 2m. or 0.01 of the distance

the ant moves 1cm (total 3 cm) and same again but band increases by 1/3 so ant is now 2cm+1cm+2/3cm+4/3cm or 5cm along a 3m band. or 0.16 of the way.

so the ant moves then the band streches adding a proportionate distance to all the ants movementa in previous seconds.

so, he will reach the end.

in one the and and moves 1cm then band length doubles instantly. ant is now 2cm along band of 2m. or 0.01 of the distance

the ant moves 1cm (total 3 cm) and same again but band increases by 1/3 so ant is now 2cm+1cm+2/3cm+4/3cm or 5cm along a 3m band. or 0.16 of the way.

so the ant moves then the band streches adding a proportionate distance to all the ants movementa in previous seconds.

so, he will reach the end and appear to accellerate