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The endless chasing scene between Tom and Jerry comes to a big, perfectly round lake. Jerry manages to escape Tom by a hair by plunging himself into the water. Tom can't swim - he can only run at a maximum speed of $v$ along the shoreline to try to catch Jerry. How fast must Jerry swim in order to land safely?

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  • $\begingroup$ What does it mean to land safely? Where is safe, and how far away from Tom must Jerry be? $\endgroup$
    – Qfwfq
    Commented Jun 24, 2016 at 16:31
  • $\begingroup$ Assume both Tom and Jerry are "points" and safe means Tom is not at the exact same position where Jerry is landing. Of course a mathematical description would be more fun, and probably preferred by mathematicians (like myself, in fact) $\endgroup$
    – user24855
    Commented Jun 24, 2016 at 16:34
  • $\begingroup$ @f" looks like one, but I am not sure if the accepted answer there is correct :/ $\endgroup$
    – user24855
    Commented Jun 24, 2016 at 16:36
  • $\begingroup$ @Pigpag A duplicate really applies to the question - if the answer on the duplicate is wrong, why not supply your own correct answer? (And explain why the existing answer is wrong in a comment on it) $\endgroup$
    – Milo Brandt
    Commented Jun 24, 2016 at 16:42
  • $\begingroup$ @Milo I understand that. I wasn't aware of the other similar problem at the time of posting. $\endgroup$
    – user24855
    Commented Jun 24, 2016 at 16:59

1 Answer 1

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Started solving based on the (probably erroneous) assumption that Jerry would swim in a straight continuous line.

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Jerry needs to swim the chord length 2*Rsin(a/2) where a is the angle that contains the chord
Tom needs to run the arc length aR
Toms time will be (a*R)/v this will also be Jerry's target time
So Jerry's speed needs to be 2*R*sin(a/2)v/(aR)

This chart shows the relationship between angle and necessary speed so Jerry's best bet is to swim across the lake when he has to swim faster then v*2/pi enter image description here

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    $\begingroup$ When Jerry swims to the center, he should at least try to swim in the opposite direction of where Tom is at that time. Therefore keep swimming in the same direction does not seem to be a good strategy :) $\endgroup$
    – user24855
    Commented Jun 24, 2016 at 17:03