Four turtles are located at the vertices of a square with side a. They start moving simultaneously at a constant modulo velocity v. Each turtle moves clockwise in the direction of its neighbor. Where will the turtles meet and after what time?

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    <h3 id="back-link"><a href="/#1.5">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $1.5.14^*.$ Four turtles are located at the vertices of a square with side $a$. They start moving simultaneously at a constant modulo velocity $v$. Each turtle moves clockwise in the direction of its neighbor. Where will the turtles meet and after what time?
    </p>

    <h3>Solution</h3>
    <p>

Consider the change in the coordinates of the turtles over a short period of time $dt$

Over time $dt$ the distance between neighboring turtles changed from $a$ to $a'$

Express $a'$ using the Pythagorean theorem

$$a' = \sqrt{(a-dx)^2 - d^2x}$$

Considering the smallness of the value $dx$

$$a' = \sqrt{a^2 - 2a\\, dx}$$

$$a' = a\sqrt{1 - \frac{2dx}{a}}$$

We will use the formula for small quantities $(1+x)^\alpha \approx 1+\alpha x$, where $x\rightarrow 0$:

$$a' = a - dx$$

Thus, the increment of the coordinate $a$ is

$$da = a' - a = dx$$

Hence, the rate of change of distance between the turtles is

$$u = \frac{da}{dt} = -\frac{dx}{dt}=-v$$

From this it follows that after $a=0$, after a period of time

$$t = a/v$$

From the symmetry of the problem, it follows that all turtles will go the same way and end up in the center of the square.

NO: It would be interesting to know what would happen if it were not a square? If there were not $4$ turtles, but $n$ ones? A more detailed version of the problem can be found in "Very Long Physics Problems" by A.I. Slobodyanyuk (Problem 1)

    <h4>Answer</h4>
    <p>
        <p>At the center of the square after time $t = a/v$.</p>
    </p>


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