Calculate Meeting Time for Equilateral Triangle Turtles | Classic Problem Solved

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In summary, the conversation discusses a problem involving three turtles moving on the sides of an equilateral triangle with a side length of L and a velocity of V. The turtles move towards the next turtle in a continuous loop. The time it takes for the turtles to meet is calculated to be L/(3V). However, another person suggests using radial coordinates and calculates the time to be 2L/(3V). There is also a discussion about the calculation of a radial step, where it is explained that the turtle step is always at a 30-degree angle to the radial direction, and the longer side of a 30-60-90 triangle is equal to √3/2 times the hypotenuse.
  • #1
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This is supost to be a classic. I solved it but my answer isn´t right and I can´t find out why. Hope you guys can help.

Imagine an equilateral triangle with one turtle on each side of lenth L. All turtles move with velocity V. Turtle 1 always directs its motion to turtle 2, turtle 2 to turtle 3, and turtle 3 to turtle 1, again. Calculate the time the turtles take to meet.


The movement will be something like this.

mice3.gif


Relative to turtle 2, turtle 1 will be in uniform rectilinear motion and its velocity will be:

[tex] \vec V_{12}=\vec v_1 - \vec v_2 [/tex]

[tex]\vec v_1[/tex] and [tex]\vec -v_2 [/tex] make an angle of 60º degrees. So,

[tex] V_{12}= \frac{V}{\sqrt{3}} [/tex]

Relative to turtle 2, turtle 1 will only travel the distance that separates it from the center of the triangle, which is [tex] \frac{L}{3}[/tex]

So, [tex] t= \frac{v_{12}}{d} = \frac{L}{3V} [/tex]

P.S.: I guess some of you might have already solved this problem. If not, I recommend it, though I got it wrong.
 
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  • #2
You have an equilateral triangle with side length L. The velocity of each turtle is v. The triangle defined by the 3 points (turtles) is always at any time t an equilateral triangle. Because the distance d between the turtles (the side of the equilateral triangle) at time t is

d=L-vt,

the turtles meet when

L-vt = 0

Bob S
 
  • #3
Thanks for the answer.

Because the distance d between the turtles (the side of the equilateral triangle) at time t is

d=L-vt,

This isn´t true. How did you arrive to this equation?

The real time the turtles take is [tex]\frac{2L}{3V}[/tex] , 1.5 times what you got.

Does anyone have other sugestions? I advise you, if you´ve got nothing to do, this is a fun problem!
 
  • #4
You are correct. The easiest way to get the solution is to use radial coordinates. The direct distance from the apex to the center is (from plane geometry)

R = L/√3

For a radial step dr, the turtle must always travel a distance 2·v·dt/√3

So to travel a radial distance R=L/√3,

v·t = (L/√3)(2/√3) = 2L/3

t=2L/3v

Bob S
 
  • #5
Hi Bob, thanks for the reply.

R = L/√3

Right. On my orginal post I had a typo. I meant this.


For a radial step dr, the turtle must always travel a distance 2·v·dt/√3

Can you explain this step? What is dr and how did you calculate it?
 
  • #6
For a radial step dr, the turtle must always travel a distance 2·v·dt/√3.
jpas said:
Can you explain this step? What is dr and how did you calculate it?
A turtle step dx = v·dt is always at a 30-degree angle to the radial direction dr from the apex of the equilateral triangle (between the 3 turtles) to the center, because the radial direction always bisects the 60-degree angle of the equilateral triangle (even as the turtles move and the equilateral triangle rotates), and the instantaneous direction of v·dt is always parallel to one of the sides of the equilateral triangle. Using plane geometry, the longer side of a 30-60-90 degree triangle is equal to √3/2 times the hypotenuse. The turtle step v·dt is along the hypotenuse and the radial step projection dr is along the longer side.

So a radial step dr = (√3)·v·dt/2

Bob S
 

1. What is the "Chasing Turtles Problem"?

The "Chasing Turtles Problem" is a classic mathematical problem that involves a turtle chasing a rabbit. The turtle starts at the origin and moves at a constant speed, while the rabbit starts at a given point and moves at a faster variable speed. The goal is to determine the minimum distance the rabbit needs to travel in order to be caught by the turtle.

2. What is the origin of the "Chasing Turtles Problem"?

The "Chasing Turtles Problem" was first posed by mathematician L. J. Beghin in 1943. It has since been studied by many mathematicians and has several variations, including the "Hare and Tortoise Problem" and the "Turtle and Rabbit Problem".

3. What is the significance of the "Chasing Turtles Problem"?

The "Chasing Turtles Problem" is a well-known and widely studied problem in mathematics. It helps to illustrate concepts such as distance, rate, and time, and is often used as a teaching tool in introductory calculus courses. It also has practical applications in areas such as optimization and control theory.

4. What are some strategies for solving the "Chasing Turtles Problem"?

There are several strategies for solving the "Chasing Turtles Problem", including using algebraic equations, setting up a system of equations, and using graphical methods. Some common approaches include finding the time at which the turtle and rabbit meet and calculating the distance traveled by each at that time, or finding the point at which the rabbit's distance from the origin is equal to the turtle's distance traveled.

5. Are there any real-world scenarios that can be modeled using the "Chasing Turtles Problem"?

Yes, there are many real-world scenarios that can be modeled using the "Chasing Turtles Problem". For example, it can be used to calculate the minimum distance a police car needs to travel to catch a speeding car, or the minimum distance a predator needs to travel to catch its prey. It can also be applied to optimization problems in fields such as transportation and logistics.

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