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+ <meta name="description" content="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?">
+ <meta name="author" content="Aliaksandr Melnichenka">
+ <meta name="date" content="2023-10" scheme="YYYY-MM">
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+ <title>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?</title>
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+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <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>
+
+</p>
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+ <img src="https://savchenkosolutions.com/1/1.5.14/draw.png"
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+<p>
+<p>Consider the change in the coordinates of the turtles over a short period of time $dt$</p>
+
+<p>Over time $dt$ the distance between neighboring turtles changed from $a$ to $a'$</p>
+
+<p>Express $a'$ using the Pythagorean theorem</p>
+$$ a' = \sqrt{(a-dx)^2 - d^2x} $$
+<p>Considering the smallness of the value $dx$</p>
+$$ a' = \sqrt{a^2 - 2a\, dx} $$
+
+$$ a' = a\sqrt{1 - \frac{2dx}{a}} $$
+<p>We will use the formula for small quantities $(1+x)^\alpha \approx 1+\alpha x$, where $x\rightarrow 0$:</p>
+$$ a' = a - dx $$
+<p>Thus, the increment of the coordinate $a$ is</p>
+$$ da = a' - a = dx $$
+<p>Hence, the rate of change of distance between the turtles is</p>
+$$ u = \frac{da}{dt} = -\frac{dx}{dt}=-v $$
+<p>From this it follows that after $a=0$, after a period of time</p>
+$$ t = a/v $$
+<p>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.</p>
+
+<p>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 <a href="https://belphol.github.io/books/LongProblemsPart1.pdf" target="_blank">"Very Long Physics Problems"</a> by A.I. Slobodyanyuk (Problem 1)</p>
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ <p>At the center of the square after time $t = a/v$.</p>
+ </p>
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