$1.5.14^*.$ 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?
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)
At the center of the square after time $t = a/v$.