newtonian mechanics - The Achilles Paradox

I know that it is a very old question but still I don't find any satisfactory solution for Achilles Paradox. Please explain me the fundamentals of Achilles paradox in terms of stage wise distance covered. Note that it is easily solvable in terms of time, but if you start analysing this event in terms of time, then there is not at all any paradox. So please explain in terms of stage wise distances only.

Answer

This is quite straightforward as long as you break it up into steps.

Let the tortoise travel with speed $v_t$ and Archilles with speed $v$. We split up the steps Archilles takes as \begin{equation} d_A = d_0 + d _1 + ... \end{equation} where $d_i $ is the distance travelled in a single step. The initial distance $d_0 $ is then just the initial separation between Archilles and the tortoise. We denote $\Delta t _i $ as the time of step $i $. The distance $d_1 $ is just how far the tortoise managed to step while Archilles was catching up to the tortoise's initial position. It is given by \begin{equation} d_1=v_t \Delta t_1 =v_t \frac{d_0}{v} \end{equation} Similarly, the distance $d_2 $ is given by \begin{equation} d_2=v_t \Delta t_2 = v_t \frac{d_1 }{v} = \left(\frac{v_t}{v} \right)^2 d_0 \end{equation} Its easy to see that this trend will continue: \begin{align} d_A &= d_0 + \frac{v _t }{v} d_0 +\left(\frac{v_t}{v} \right)^2 d_0 + ... \\ &= d _0 \left( 1 + \frac{ v _t }{ v} + \left( \frac{ v _t }{ v } \right) ^2 + ...\right) \end{align} In the brackets we have a geometric series. If $v_t\le v$ then, \begin{align} d _ A & = d_0\frac{1}{ 1 - v _t / v } \end{align} This is finite (and hence Archilles can catch up with the tortoise) as long as $v_t

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I think it's also interesting to solve this problem the simple way, avoiding splitting it up into steps. For Archilles and the tortoise representatively we have, \begin{equation} d_A=\frac{v}{t}\, , \quad d_t= \frac{v_t}{t}+d_0 \end{equation} Setting $d_A = d_t $ and solving these equations gives, \begin{equation} d_A=d_0 \frac{1}{1-v_t/v} \end{equation} as required.