So to beat the man, the rabbit needs to do more work than that. Kinda thinking a spiral pattern might work but I’m too lazy to do that math.

Let’s try something simpler. If the rabbit can get himself sufficiently far in one direction, while forcing Secret Service man to be 180 degrees opposite, from there he clearly can win the race. If the pond has radius r, and the rabbit is x away from the center, then a straight swim to the edge is distance (r-x), in which time the man can run 4(r-x) and has to cover pi*r distance. So the rabbit has to be within pi*r/4 of the shore, or almost 1/4 of the way to the shore or closer, to win. The remaining question is can the rabbit get to that point, and force Secret Agent Man to the diametrically opposite side?

Well, rabbit swims 1/4 as fast as Man can run. So if rabbit is anywhere closer to the center than r/4, he can begin to swim in a circle and cover the angle faster than the man can run, and therefore swim in that circle until he forces man to be 180 degrees opposite. Then he can bolt for the shore, and to Rabbit Freedom.

So rabbit can swim to some point in between r/4 from the center and (1 – pi/4)r from the center. Regardless where the man is, Rabbi can then begin swimming his loop until the man is 180 opposite, then dash.

Ha!

“The rabbit would be able to swim from the center to the south edge in 2 seconds, while the agent would only be able to run from the north end to the south end in 3.14 seconds.”

Actually the agent would be able to get from the north end to the south end in 1/2(pi) seconds; with a 2-meter radius, the pond’s circumference is 4(pi), so half that is 2(pi). At 4 meters/second, 2(pi) meters would take about 1.57 seconds, and he would get there 0.43 seconds before the rabbit does.

This distance will vary on the speed of the agent , getting smaller as the agents speed increases. By casual observation we can see that that rabbit can swim out to the fraction 1/max to the shore based on a max speed for the agent. This would leave the rabbit a distance of (max-1)/max to swim at rate (x) while the agent was moving at rate (max)(x).

So the time take should be equal to find the max rate of the agent.

(((max-1)/max)R)/x = (R(pi))/((max)(x))

If multiply both sides by x we get
((max-1)/max)R = (R(pi))/(max)

We can divide both side by R and get
(max-1)/max = (pi)/(max)

We can multiple both sides by (max) and get
max-1 = (pi)

add one to both side and your done
max=(pi)+1

For the maximum agent speed, set (3/4)(R/x) = (pi/max)(R/x). Solve for max = (4/3)(pi)

If the pond is 4 meters in diameter, it would be roughly 12.56 meters in circumference. If the rabbit can swim 1 meter per second, then the agent could run 4 meters per second. The rabbit would be able to swim from the center to the south edge in 2 seconds, while the agent would only be able to run from the north end to the south end in 3.14 seconds.

But on second thought all the rabbit has to do is be able to get out to a radius away from the center that the rabbit can cover in the same time it takes the agent to run around the pond. This will take some time for the rabbit since it need to ensure the agent is always directly behind it. Using MadMolecule’s idea. Once it reaches that radius (distance) from the center of the pond it can make a mad dash at the shore line.

MATH::
Pond has Radius = R
Agents Speed = 4x
Rabbit Speed = x

If the rabbit moves outward to a distance of (1/4)R from the center always keeping the agent behind the rabbit can swim around in circles and the agent will always remain at the opposite shore assuming the agent never stop pursuit.

This is because the rabbit here will be covering a distance of 2(1/4)(pi)R or (1/2)(pi)R at speed x taking ((1/2)(pi)R)/x or (1/2x)(pi)R and the agent will traveling 2(pi)R at 4x which takes (2(pi)R)/4x = (1/2x)(pi)R

Now the rabbit is R – (1/4)R away from the edge or (3/4)R and can cover that in ((3/4)R)/x or (3R)/(4x). The agent is half way around the pond and will need to cover (1/2)(2(pi)R) or (pi)R at a rate of 4x. That will take him ((pi)R)/(4x). Since (pi) is about 3.14 the agent will take longer to reach the other side of the pond then the rabbit in this scenario.

The agent will have to run slower than((pi)+1) times faster than the rabbit. Hopefully I will write back the process later, its time to go home.

Kisses are great!

I think, though, that if while the rabbit swam it kept turning so that it was always swimming directly away from the agent, the rabbit would pretty quickly find itself in a spot that is only .78 (or pi/4) of the distance from the point that was at that moment directly opposite the agent. From there, the rabbit makes a break for it and escapes.

I could be visualizing this improperly, though. Somebody should set up a computer model to figure this out.

Figure the radius of the lake (the direct path for the rabbit to reach the edge) is 10 meters. The agent must travel 10*? — or half the lake’s circumference — in order to reach the point the rabbit reaches land: 31.41 meters.

Considering, the agent runs 4 times as fast as the rabbit can swim — 4 > 3.141 — he’ll still be able to catch it.

The rabbit can only escape if the agent runs less than 3.141 times as fast as the rabbit.

Of course, I could have all these calculations wrong.

So if the rabbit needs to run a distance of r and the agent needs to run a distance pi*r, the agent needs to a run a distance that’s greater by a factor of pi, since pi is 3.14, and the agent can run four times faster then the rabbit then the rabbit will always be caught.

By the same logic any agent that runs slower then a ratio of pi:1 will not be able to catch the rabbit if it takes off in the opposite direction.