$1.5.15^*.$ Plot an approximate graph of the speed of point $B$ as a function of time, if the speed $v_A$ of point $A$ is constant. Find the formula for this relationship if $x(0) = 0$.
NO: Before viewing the solution to this problem, I advise you to familiarize yourself with the solution 1.5.14
At time $t$, the height to which the point dropped
$$ x = v_A t\;(1) $$Let's consider the change in the length of the thread over a small period of time $dt$
$$ dl = \sqrt{L^2 + (x+dx)^2}-\sqrt{L^2 + x^2} $$ $$ dl = \sqrt{L^2 + x^2}\cdot \left(\sqrt{1 + \frac{2xdx}{L^2 + x^2}}-1\right) $$We will use the formula for small quantities $(1+x)^\alpha \approx 1+\alpha x$, where $x\rightarrow 0$:
$$ dl = \frac{xdx}{\sqrt{L^2 + x^2}} $$Given that $v_B = \frac{dl}{dt}$ and $v_A = \frac{dx}{dt}$
$$ v_B = \frac{x}{\sqrt{L^2 + x^2}} \frac{dx}{dt} $$ $$ v_B = v_A\frac{x}{\sqrt{L^2 + x^2}} $$Substitute $(1):$
$$ \fbox{$v_B = \frac{v_A^2t}{\sqrt{L^2 + v_A^2t^2}}$} $$NO: A more detailed and beautiful problem with a similar idea can be found in "Very Long Physics Problems" by A.I. Slobodyanyuk (Problem 2)
$$v_B = \frac{v_A^2t}{\sqrt{L^2 + v_A^2t^2}}$$