A horse has a rubber band attached to it which can expand infinitely and is tied to a pole on the other end. At first the length of the rubber band is $l$. on the pole-side of the rubber band there is a snail. If both start walking at the same time: The horse at speed $u$ and the snail at speed $v$ with $u>v$ when will the snail catch the horse?
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What does "on the pole side of the rubber band" mean? Does that mean next to the pole? – recursive recursion Feb 07 '14 at 23:44
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@recursiverecursion yes, it means he is at distance l from the horse, on the other end of the rubber band. – Asinomás Feb 07 '14 at 23:46
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see also: 1, 2. – Caleb Stanford Jul 12 '15 at 14:05
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The length of the rubber band at time $t$ is $l(t)=l+vt$. The fraction of the total band covered by the snake in one second at time $t$ is $\frac u{l+vt}$. We are going to integrate this from $t=0$ to $t=\infty$. If the integral is less than $1$, the snail will never reach the horse. If it is $1$ or larger, it will reach the horse. Since we have $$ \int\frac u{l+vt}\,dt=\frac uv \ln(l+vt) $$ and $\lim_{t\to \infty}\log(1+vt)=\infty$, the snail will catch the horse. This happens at time $t=T$ for which $$ \int_{t=0}^{t=T}\frac u{l+vt}\,dt=1\\ \frac uv(\ln(l+vT)-\ln (l))=1\\ \ln\left(\frac{l+vT}l\right)=\frac vu\\ T=\frac{l(e^{\frac vu}-1)}v $$
Ragnar
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1@user4140 Apparently, I messed up more than I thought. Mathematica's $1$ and $l$ aren't as distinguishable as they could be. – Ragnar Feb 07 '14 at 23:59