Achilleus and the turtle

[SimVig]

There's a story (I don't know how to call it actually) about Achilleus racing a turtle. It goes something like this..

 

Achilleus wants to race a turtle. He runs 10 times faster than the turtle, so the turtle gets a head start of 10 meters (forgive the measuring system, all you US folks, but I can't do it any other way). By the time Achilleus runs the 10 meters, the turtle has crawled forward 1 meter. By the time Achilleus finishes that, the turtle has crawled 1dm. By the time he's done with that, the turtle has moved another 1cm. And so on. Thus, Achilleus will never catch up with the turtle.

 

Now, this logic is obviously flawed, as can be seen from real life. The problem is, I can't spot the flaw :P . Another thing: if such a situation would arise, at what point would Achilleus actually catch up with the turtle? (I know the answer to that one).

 

Forgive my poor choise of forum, but I just couldn't place it anywhere else. To make it a little debatable, what do you think about this logic?

[Darnoc]

I don't remember, who solved this paradox and how this person did it (we just had it in physics, but I'm not good at physics). But I think the logical flaw of the paradox is that the question is just asked the wrong way. What do we have?

 

we've got the velocity "v" of botht the turtle and Achilles. v1 (turtle speed) is one tenth of v2 (Achilles speed).

 

we can calculate the way "s" of the turtle and Achilles, by assuming a certain amount of time. With this we can acctually calculate the point where both ways are the same in the same time. Problem solved.

 

If anyone knows the physical formula, just write it down. I don't remember by heart, have to look it up.

[Shakkara]

Assuming Achilles would have V=5m/s and the turtle have V=0.5m/s the formula would be 10+0.5X = 5X, which is at X = 2.22, so after 2.22 seconds Achilles would catch up, at 11.11 meters.

[Marxist ßastard]

The flaw? As Achilles runs the ten meters' head start, the turtle travels a single meter. If the race were to end at that point, the turtle would win. However, once the turtle travels another meter, Achilles travels ten. Assuming that the turtles' speed is a tenth of a meter per second and that the velocity for both parties would be constant, the race would play out as follows:

 

00:00 -- Start

 

Achilles: 0m

 

Turtle: 10m

 

00:01

 

Achilles: 1m

 

Turtle: 10.1m

 

00:11.2 -- Achilles has surpassed turtle, turtle cannot possibly regain lead

 

Achilles: 11.2m

 

Turtle: 11.11m

 

00:12

 

Achilles: 12m

 

Turtle: 11.2

 

The most logical conclusion is that whoever thought that up mistook the turtle's distance fromt he starting point for the turtle's distance from Achilles. Based on the previous assumptions, one could make the following formulas to calculate the distance from the starting point of both racers:

 

a = t

 

u = 10 + ( t / 10 )

 

u = 10 + ( a / 10 )

 

...Where a is Achilles's distance, u is the turtle's distance, and t is the elapsed time. To figure out the exact moment that Achilles's distance from the starting point equals the turtle's, find the value of either variable in the second or third equation for u = a = t.

 

EDIT: Damn, Shak beat me to the punch.

[SimVig]

Yes, that's right, he would catch up at 11.1111 meters. As I see it the flaw is that the logic doesn't work in reverse. When the turtle has done 1 meter, Achilles has done 10 and everything else is as Marxist ßastard said.