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.

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

    <h3> Statement </h3>
    <p>
        $1.5.15^*.$ 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 $x(0) = 0$.

    <h3>Solution</h3>
    <p>
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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

$$x = v_A t\quad(1)$$

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)

    <h4>Answer: </h4>
  <p>$$v_B = \frac{v_A^2t}{\sqrt{L^2 + v_A^2t^2}}$$</p>



    <p style="text-align: right; font-style: italic; font-size: 14;">   
      Aliaksandr Melnichenka<br>
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