There are 4 beatles (John, Paul, George, and Ringo) on the vertices of a square with side length a, each beatle is moving in a constant speed v and in the direction of the beatle that started at the verice clockwise to the one the beatle started at. Their motion will follow a spiral. The question is how long will it take for them to meet?

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    $\begingroup$ Beatles as in John, Paul, George, and Ringo? $\endgroup$ – noah Mar 14 at 11:22
  • $\begingroup$ This is a great update for the Problem and Im adding it $\endgroup$ – sean python Mar 14 at 11:23
  • $\begingroup$ Lmao that give me a good laugh @noah $\endgroup$ – Buraian Mar 14 at 12:03

This is quite a commonly posed problem. The easiest way to find a solution is to notice that for each individual Beatle at every moment in time, their own velocity vector is perpendicular to the velocity vector of the Beatle they are chasing. Therefore, in an infinitesimal timeslice $\mathrm{d}t$, the distance to the Beatle in front decreases by $v\,\mathrm{d}t$, if they are moving with velocity $v$ (the sideways motion by the Beatle in front is not changing the distance in an infinitesimal time step). So if the initial distance is $a$, and the Beatle separation decreases at a constant rate $v$, then the time for them to meet in the center and play their concert is $t = a/v$.

  • $\begingroup$ But why would the velocity vectors be perpendicular? $\endgroup$ – sean python Mar 14 at 11:36
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    $\begingroup$ From symmetry; the setup is symmetric under rotation by 90°. They obviously start out perpendicular, and then whatever change in angle one of them makes, the one in front will have done the same, because they see the same initial situation and follow the same rules. $\endgroup$ – noah Mar 14 at 11:39

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