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Two leopard seals, Snap and Snarl, start 210 meters apart. They swim toward each other at a constant speed of 10 km/h each. Gilly, a gentoo penguin, starts at Snap and swims back and forth between the seals continually until the two seals meet. When going from Snap to Snarl, Gilly swims at 15 km/h, but when going from Snarl to Snap, Gilly swims at 20 km/h. What is the total distance that Gilly swims before the seals meet?

Found similar problems at:

Zeno-like riddle with additional complication: Runner with dog running back and forth at different speeds

but i don't think the average speed method gives correct answer, using average speed doesn't quite make sense to me, since the distance differs each time the penguin going back and forth.

And the train problem Apparent paradox for the bird traveling between two trains puzzle doesn't consider the bird could have different speed going back and forth.

How should the above problem be solved? Thanks.

user526427
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  • Not to mention these problems are all so realistic, since they require infinite deceleration/acceleration and instantaneous turn-around. – Ted Shifrin Jan 02 '20 at 21:45
  • you need a weighted average, which is exactly what the Zeno link you've provided does. What's the issue? – Rushabh Mehta Jan 02 '20 at 21:48
  • @DonThousand In the dog problem the lengths of the legs of the dog's run are $d_1>d_2=d_2>d_3=d_3>d_4$ etc. (from second leg onwards, the "back" and "forth" distances are the same) - so the average speed (as harmonic mean) applies. Here, the distances decrease all the time: $d_1>d_2>d_3>d_4$ etc. –  Jan 02 '20 at 22:01

2 Answers2

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Let's switch to an inertial reference frame where Snap stays still and Snarl swims toward Snap at $20$ kph. In this frame, Gilly swims the outbound leg at $5$ kph and swims the inbound leg at $30$ kph. In this frame, the outbound leg and the inbound leg are exactly the same distance, so we know that Gilly spends $6$ times as long swimming outbound as swimming inbound.

Now let's return to our original reference frame. Gilly's average speed is $\frac {6 \cdot 15+20}{7}= \frac{110}{7}$ kph. Gilly will swim at this average speed for $\frac {210}{20000}$ hours. Thus, Gilly will swim $\frac {3300}{20000} \text{ km }= 165 \text{ m}$.

Robert Shore
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To find the average speed of Gilly, we just need one iteration of Gilly swimming To Snarl and back to Snap: the ratio of each subsequent iteration will be exactly the same.

Indeed, to compute the average for one iteration, let's just pick some more calculation friendly numbers: let's say that the initial distance between Snap and Snarl is $300$ kilometres. Since Gilly will approach Snarl at a combined speed of $25$ km/h, it will take $12$ hours for Gilly to get to Snarl. At that point, both Snap and Snarl will have moved $120$ kilometers each, and are therefore still sperated by $60$ kilomters. Now Gilly will approach Snap with a combined speed of $30$ km/h, so that will take $2$ hours.

Ok, so there you have it: Gilly will spend $6$ times as much time swimming towards Snarl than towards Snap. With that, you can compute Gilly's average speed, and the rest is a piece of cake (in fact, you get a $7$ in the denominator, so that'll divide the original distance of $210$ quite nicely).

Bram28
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