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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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| </a> | | </a> |
| <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> |
| </header> | | </header> |
| | | |
| <h3 id="back-link"><a href="../../#1.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#1.1">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $1.1.5^*.$ Three microphones located on the same straight line at points $A$, $B$, $C$ recorded successively at the moments $t_A > t_B > t_C$ the sound of an explosion that occurred at point $O$, which lies on the segment $AC$. Find the length of the segment $AO$ if $AB = BC = L$. At what point in time did the explosion occur? | | $1.1.5^*.$ Three microphones located on the same straight line at points $A$, $B$, $C$ recorded successively at the moments $t_A > t_B > t_C$ the sound of an explosion that occurred at point $O$, which lies on the segment $AC$. Find the length of the segment $AO$ if $AB = BC = L$. At what point in time did the explosion occur? |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| For problem 1.1.5 | | For problem 1.1.5 |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| </p> | | </p> |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| | | |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| Let the explosion occurred at time $t_0$, then the time of signal registration at point $A$ is equal to:<br> | | Let the explosion occurred at time $t_0$, then the time of signal registration at point $A$ is equal to:<br> |
| </p> | | </p> |
| $$t_A = t_0 + \frac{L + x}{c}, \, (1)$$ | | $$t_A = t_0 + \frac{L + x}{c}, \, (1)$$ |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| where $x = BO$.<br><br> | | where $x = BO$.<br><br> |
| Similar to point $B$<br> | | Similar to point $B$<br> |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $$t_B = t_0 + \frac{x}{c}, \,(2)$$ | | $$t_B = t_0 + \frac{x}{c}, \,(2)$$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| For point $C$<br> | | For point $C$<br> |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $$t_C = t_0 + \frac{L - x}{c}, \,(3)$$ | | $$t_C = t_0 + \frac{L - x}{c}, \,(3)$$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| Let's subtract the second equation from the first equation:<br> | | Let's subtract the second equation from the first equation:<br> |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $$t_A – t_B = \frac{L}{c}. \,(4)$$ | | $$t_A – t_B = \frac{L}{c}. \,(4)$$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| And from the first equation we subtract the third equation<br> | | And from the first equation we subtract the third equation<br> |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $$t_A – t_C = \frac{2x}{c}. \,(5)$$ | | $$t_A – t_C = \frac{2x}{c}. \,(5)$$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| From equation $(4)$ let's express $c = \frac{L}{t_A – t_B}$, а из $(5)$ $x = \frac{t_A – t_C}{2} \cdot c$. Then the required distance<br> | | From equation $(4)$ let's express $c = \frac{L}{t_A – t_B}$, а из $(5)$ $x = \frac{t_A – t_C}{2} \cdot c$. Then the required distance<br> |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $$AO = L + x = L + \frac{t_A – t_C}{2}\frac{L}{t_A – t_B}$$ | | $$AO = L + x = L + \frac{t_A – t_C}{2}\frac{L}{t_A – t_B}$$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| After the transformation<br> | | After the transformation<br> |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $$AO = \frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \cdot L. \, (6)$$ | | $$AO = \frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \cdot L. \, (6)$$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| To determine the moment of time at which the explosion occurred, we substitute in the expression<br> | | To determine the moment of time at which the explosion occurred, we substitute in the expression<br> |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $$t_A = t_0 + \frac{L + x}{c}, \; c = \frac{L}{t_A – t_B}$$ | | $$t_A = t_0 + \frac{L + x}{c}, \; c = \frac{L}{t_A – t_B}$$ |
| $$\frac{t_A – t_C}{2} = \frac{x}{c}$$ | | $$\frac{t_A – t_C}{2} = \frac{x}{c}$$ |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| after transformations<br> | | after transformations<br> |
| </p> | | </p> |
| <p style="text-align: center;"> | | <p style="text-align: center;"> |
| $$t_0 = t_B - \frac{1}{2} \cdot (t_A – t_C)$$ | | $$t_0 = t_B - \frac{1}{2} \cdot (t_A – t_C)$$ |
| </p> | | </p> |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$AO = \frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \cdot L, \; t_0 = t_B - \frac{1}{2} \cdot (t_A – t_C)$$ | | $$AO = \frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \cdot L, \; t_0 = t_B - \frac{1}{2} \cdot (t_A – t_C)$$ |
| </p> | | </p> |
