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| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> |
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| 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> |
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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> |
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| <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 $x(0) = 0$. |
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| <figcaption> | | <figcaption> |
| For problem $1.5.15^*$ | | For problem $1.5.15^*$ |
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| <h3>Solution</h3> | | <h3>Solution</h3> |
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| <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> |
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| $$ dl = \sqrt{L^2 + x^2}\cdot \left(\sqrt{1 + \frac{2xdx}{L^2 + x^2}}-1\right) $$ | | $$ dl = \sqrt{L^2 + x^2}\cdot \left(\sqrt{1 + \frac{2xdx}{L^2 + x^2}}-1\right) $$ |
| <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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