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		+	<body style="">
		+	<header style="text-align:center;">
		+	<h2>Solutions of Savchenko Physics Textbook</h2>
		+	<p class="author">
		+	Aliaksandr Melnichenka <br/>
		+	October 2023
		+	</p>
		+	</header>
		+
		+	<h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3>
		+
		+	<h3> Statement </h3>
		+	<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?
		+	<center>
		+	<figure>
		+	<img src="statement.png"
		+	loading="lazy" width="230" />
		+	<figcaption>
		+	For problem 1.1.5
		+	</figcaption>
		+	</figure>
		+	</center>
		+	</p>
		+	<h3>Solution</h3>
		+	<p>
		+
		+	<p class="TxtSolutions">
		+	Let the explosion occurred at time $t_0$, then the time of signal registration at point $A$ is equal to:<br>
		+	</p>
		+	$$t_A = t_0 + \frac{L + x}{c}, \, (1)$$
		+	<p class="TxtSolutions">
		+	where $x = BO$.<br><br>
		+	Similar to point $B$<br>
		+	</p>
		+	<p style="text-align: center;">
		+	$$t_B = t_0 + \frac{x}{c}, \,(2)$$
		+	</p>
		+	<p class="TxtSolutions">
		+	For point $C$<br>
		+	</p>
		+	<p style="text-align: center;">
		+	$$t_C = t_0 + \frac{L - x}{c}, \,(3)$$
		+	</p>
		+	<p class="TxtSolutions">
		+	Let's subtract the second equation from the first equation:<br>
		+	</p>
		+	<p style="text-align: center;">
		+	$$t_A – t_B = \frac{L}{c}. \,(4)$$
		+	</p>
		+	<p class="TxtSolutions">
		+	And from the first equation we subtract the third equation<br>
		+	</p>
		+	<p style="text-align: center;">
		+	$$t_A – t_C = \frac{2x}{c}. \,(5)$$
		+	</p>
		+	<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>
		+	</p>
		+	<p style="text-align: center;">
		+	$$AO = L + x = L + \frac{t_A – t_C}{2}\frac{L}{t_A – t_B}$$
		+	</p>
		+	<p class="TxtSolutions">
		+	After the transformation<br>
		+	</p>
		+	<p style="text-align: center;">
		+	$$AO = \frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \cdot L. \, (6)$$
		+	</p>
		+	<p class="TxtSolutions">
		+	To determine the moment of time at which the explosion occurred, we substitute in the expression<br>
		+	</p>
		+	<p style="text-align: center;">
		+	$$t_A = t_0 + \frac{L + x}{c}, \; c = \frac{L}{t_A – t_B}$$
		+	$$\frac{t_A – t_C}{2} = \frac{x}{c}$$
		+	</p>
		+	<p class="TxtSolutions">
		+	after transformations<br>
		+	</p>
		+	<p style="text-align: center;">
		+	$$t_0 = t_B - \frac{1}{2} \cdot (t_A – t_C)$$
		+	</p>
		+	</p>
		+
		+	<h4>Answer</h4>
		+	<p>
		+	$$AO = \frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \cdot L, \; t_0 = t_B - \frac{1}{2} \cdot (t_A – t_C)$$
		+	</p>
		+
		+
		+
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		+	<br>
		+	<p>
		+	<small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
		+	</p>
		+	<p>
		+	<small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> astrosander01@gmail.com <br></small>
		+	</p>
		+	</footer>
		+	</body>
		+
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			</script>
			</head>
			<body style="">
			<header style="text-align:center;">
			<h2>Solutions of Savchenko Physics Textbook</h2>
			<p class="author">
			Aliaksandr Melnichenka <br/>
			October 2023
			</p>
			</header>

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

			<h3> Statement </h3>
			<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?
			<center>
			<figure>
			<img src="statement.png"
			loading="lazy" width="230" />
			<figcaption>
			For problem 1.1.5
			</figcaption>
			</figure>
			</center>
			</p>
			<h3>Solution</h3>
			<p>

			<p class="TxtSolutions">
			Let the explosion occurred at time $t_0$, then the time of signal registration at point $A$ is equal to:<br>
			</p>
			$$t_A = t_0 + \frac{L + x}{c}, \, (1)$$
			<p class="TxtSolutions">
			where $x = BO$.<br><br>
			Similar to point $B$<br>
			</p>
			<p style="text-align: center;">
			$$t_B = t_0 + \frac{x}{c}, \,(2)$$
			</p>
			<p class="TxtSolutions">
			For point $C$<br>
			</p>
			<p style="text-align: center;">
			$$t_C = t_0 + \frac{L - x}{c}, \,(3)$$
			</p>
			<p class="TxtSolutions">
			Let's subtract the second equation from the first equation:<br>
			</p>
			<p style="text-align: center;">
			$$t_A – t_B = \frac{L}{c}. \,(4)$$
			</p>
			<p class="TxtSolutions">
			And from the first equation we subtract the third equation<br>
			</p>
			<p style="text-align: center;">
			$$t_A – t_C = \frac{2x}{c}. \,(5)$$
			</p>
			<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>
			</p>
			<p style="text-align: center;">
			$$AO = L + x = L + \frac{t_A – t_C}{2}\frac{L}{t_A – t_B}$$
			</p>
			<p class="TxtSolutions">
			After the transformation<br>
			</p>
			<p style="text-align: center;">
			$$AO = \frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \cdot L. \, (6)$$
			</p>
			<p class="TxtSolutions">
			To determine the moment of time at which the explosion occurred, we substitute in the expression<br>
			</p>
			<p style="text-align: center;">
			$$t_A = t_0 + \frac{L + x}{c}, \; c = \frac{L}{t_A – t_B}$$
			$$\frac{t_A – t_C}{2} = \frac{x}{c}$$
			</p>
			<p class="TxtSolutions">
			after transformations<br>
			</p>
			<p style="text-align: center;">
			$$t_0 = t_B - \frac{1}{2} \cdot (t_A – t_C)$$
			</p>
			</p>

			<h4>Answer</h4>
			<p>
			$$AO = \frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \cdot L, \; t_0 = t_B - \frac{1}{2} \cdot (t_A – t_C)$$
			</p>



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			</p>
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