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			Solutions of Savchenko Problems in Physics <br>
			<i><b>knowledge must be free</b></i>
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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 $x(0) = 0$.
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			For problem $1.5.15^*$
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			<h3>Solution</h3>
			<p>
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			<img src="https://savchenkosolutions.com/1/1.5.15/draw.png"
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			<figcaption>
			Velocity distribution on threads
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		@@ -79,7 +79,7 @@
			<p>NO: Before viewing the solution to this problem, I advise you to familiarize yourself with the solution <a href="../1.5.14">1.5.14</a></p>

			<p>At time $t$, the height to which the point dropped</p>
		−	$$ x = v_A t\;(1) $$
		+	$$ x = v_A t\quad(1) $$
			<p>Let's consider the change in the length of the thread over a small period of time $dt$
			</p>
			$$ dl = \sqrt{L^2 + (x+dx)^2}-\sqrt{L^2 + x^2} $$

			$$ dl = \sqrt{L^2 + x^2}\cdot \left(\sqrt{1 + \frac{2xdx}{L^2 + x^2}}-1\right) $$
			<p>We will use the formula for small quantities $(1+x)^\alpha \approx 1+\alpha x$, where $x\rightarrow 0$:</p>
			$$ dl = \frac{xdx}{\sqrt{L^2 + x^2}} $$
			<p>Given that $v_B = \frac{dl}{dt}$ and $v_A = \frac{dx}{dt}$</p>
			$$ v_B = \frac{x}{\sqrt{L^2 + x^2}} \frac{dx}{dt} $$

			$$ v_B = v_A\frac{x}{\sqrt{L^2 + x^2}} $$
			<p>Substitute $(1):$ </p>
			$$ \fbox{$v_B = \frac{v_A^2t}{\sqrt{L^2 + v_A^2t^2}}$} $$
			<p>NO: A more detailed and beautiful problem with a similar idea 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 2)</p>
			</p>

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


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	<p class="author">		<p class="author">
	Solutions of Savchenko Problems in Physics <br>		Solutions of Savchenko Problems in Physics <br>
	<i><b>knowledge must be free</b></i>		<i><b>knowledge must be free</b></i>
	</p>		</p>
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	<h3 id="back-link"><a href="../../#1.5">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../../#1.5">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<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 $x(0) = 0$.		$1.5.15^*.$ 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 $x(0) = 0$.
	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="https://savchenkosolutions.com/1/1.5.15/statement.png"		<img src="https://savchenkosolutions.com/1/1.5.15/statement.png"
	loading="lazy" width="250" />		loading="lazy" width="250" />
	<figcaption>		<figcaption>
	For problem $1.5.15^*$		For problem $1.5.15^*$
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	<center>		<center>
	<figure>		<figure>
	<img src="https://savchenkosolutions.com/1/1.5.15/draw.png"		<img src="https://savchenkosolutions.com/1/1.5.15/draw.png"
	loading="lazy" width="300" />		loading="lazy" width="300" />
	<figcaption>		<figcaption>
	Velocity distribution on threads		Velocity distribution on threads
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>

	@@ -79,7 +79,7 @@
	<p>NO: Before viewing the solution to this problem, I advise you to familiarize yourself with the solution <a href="../1.5.14">1.5.14</a></p>		<p>NO: Before viewing the solution to this problem, I advise you to familiarize yourself with the solution <a href="../1.5.14">1.5.14</a></p>

	<p>At time $t$, the height to which the point dropped</p>		<p>At time $t$, the height to which the point dropped</p>
	$$ x = v_A t\;(1) $$		$$ x = v_A t\quad(1) $$
	<p>Let's consider the change in the length of the thread over a small period of time $dt$		<p>Let's consider the change in the length of the thread over a small period of time $dt$
	</p>		</p>
	$$ dl = \sqrt{L^2 + (x+dx)^2}-\sqrt{L^2 + x^2} $$		$$ dl = \sqrt{L^2 + (x+dx)^2}-\sqrt{L^2 + x^2} $$

	$$ dl = \sqrt{L^2 + x^2}\cdot \left(\sqrt{1 + \frac{2xdx}{L^2 + x^2}}-1\right) $$		$$ dl = \sqrt{L^2 + x^2}\cdot \left(\sqrt{1 + \frac{2xdx}{L^2 + x^2}}-1\right) $$
	<p>We will use the formula for small quantities $(1+x)^\alpha \approx 1+\alpha x$, where $x\rightarrow 0$:</p>		<p>We will use the formula for small quantities $(1+x)^\alpha \approx 1+\alpha x$, where $x\rightarrow 0$:</p>
	$$ dl = \frac{xdx}{\sqrt{L^2 + x^2}} $$		$$ dl = \frac{xdx}{\sqrt{L^2 + x^2}} $$
	<p>Given that $v_B = \frac{dl}{dt}$ and $v_A = \frac{dx}{dt}$</p>		<p>Given that $v_B = \frac{dl}{dt}$ and $v_A = \frac{dx}{dt}$</p>
	$$ v_B = \frac{x}{\sqrt{L^2 + x^2}} \frac{dx}{dt} $$		$$ v_B = \frac{x}{\sqrt{L^2 + x^2}} \frac{dx}{dt} $$

	$$ v_B = v_A\frac{x}{\sqrt{L^2 + x^2}} $$		$$ v_B = v_A\frac{x}{\sqrt{L^2 + x^2}} $$
	<p>Substitute $(1):$ </p>		<p>Substitute $(1):$ </p>
	$$ \fbox{$v_B = \frac{v_A^2t}{\sqrt{L^2 + v_A^2t^2}}$} $$		$$ \fbox{$v_B = \frac{v_A^2t}{\sqrt{L^2 + v_A^2t^2}}$} $$
	<p>NO: A more detailed and beautiful problem with a similar idea 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 2)</p>		<p>NO: A more detailed and beautiful problem with a similar idea 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 2)</p>
	</p>		</p>

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


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