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That's why I said 'I think' and not 'I correctly think'.
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Yes. The actual speeds are irrelevant to the solution as well. Each time the ant moves, more of the rope expands behind the ant than the previous move and less in front. Eventually when the ant reaches the mid-way point the rope is expanding equally in front and behind. As it moves closer to its destination more and more of the rope is expanding behind the ant. This means that eventually the ant will reach its destination because eventually less rope will expand in front of the ant than the ant can travel which will land the ant at his destination
speed of rope at distance x from centre is x/(t+1) (ant starts at x=-50, t=0)
speed of ant is x/(t+1)+1 = dx/dt
x=c(t+1)+(t+1)log(t+1) so -50=c
ant arrives: 50=-50(t+1)+(t+1)log(t+1)
50(u+1)=u log u
if u=exp(50), u log u = 50.u which is near enough
I say the ant will certainly arrive. Even if we say the ant is moving after the poles, even though they are separating at the same speed the ant is covering at least a part of the gap. As the gap is getting larger at the same rate as the ant is moving, because the ant is not at the end part of the expansion must be on the part already travelled. So if the gap increases uniformly the extra distance still to travel each second will always be less than 1cm.
The ant never makes it to the end of the rope. The waveform set up by the flexing of the rope makes the ant dizzy and so he falls off. On his way to the ground he intercepts the path of an arrow that a tortoise has been struggling to run away from for quite some time.
The ant does not even need to move. Just stay there. Eventually, destination pole will touch the origin pole due to stretching and Earth being spherical (almost). At that point, just switch lanes.
The ant will reach the end -- the part of the whole path he covers in a second is 0.01 / (1 + t),
where t is time in seconds (the covered part doesn't decrease since the line extends uniformly).
So the total part he has covered up to the time t is 0.01 * ln(1 + t),
meaning he crosses whole path in exp(100) - 1 seconds (approx. 10^36 years).