Added en 1.1.17-1.1.23

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		+	<!DOCTYPE html>
		+	<html lang="en">
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		+	<head>
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		+	<title>Savchenko Solutions</title>
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		+	messageStyle: 'none'
		+	});
		+	</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="../#1.1">$\leftarrow$Back</a></h3>
		+
		+	<h3> Statement </h3>
		+	<p>
		+	$1.1.23^*.$ The shooter tries to hit a disk of radius $R$, which moves from one wall to another at constant modulo velocity so fast that it cannot be tracked. Draw a graph of the probability of hitting the disk as a function of the distance between the aiming point and the left wall.
		+	</p>
		+	<center>
		+	<figure>
		+	<img src="statement.png"
		+	loading="lazy" width="230" />
		+	<figcaption>
		+	For problem $1.1.23^*$
		+	</figcaption>
		+	</figure>
		+	</center>
		+	<p>
		+	The shots are fired at height $R$ from the floor perpendicular to the direction of motion of the disc. At what point is the least and most probable shot? What is their value? Consider the cases $L > 4R$, $4R > L > 2R$, where $L$ is the distance between the walls.
		+	</p>
		+
		+	<h3>Solution</h3>
		+	<p>
		+
		+	<center>
		+	<figure>
		+	<img src="animation.gif" alt="1.1.23"
		+	loading="lazy" width="300" />
		+	<figcaption>
		+	Wheels moving
		+	</figcaption>
		+	</figure>
		+	</center>
		+
		+	<p>
		+	When the target is moving, a bystander is less likely to see the target from the edge, as it stays there for a shorter period of time.
		+	</p>
		+	<p>
		+	Thus, the maximum probability will be observed on the interval:
		+	</p>
		+	<p style="text-align: center;">
		+	$$\left\{\begin{matrix}
		+	0 \leq x < 2R\\
		+	L-2R < x \leq L
		+	\end{matrix}\right.$$
		+	</p>
		+	<p>
		+	Since the velocity of the target is constant, and finding the target in the range $2R \leq x \leq L-2R$ is equally likely, the probability of finding it in this range is equal to the ratio of the diameter to the length of this range: $\frac{2R}{L-2R}$.
		+	</p>
		+	<p>
		+	Graphs corresponding to these modes:
		+	</p>
		+
		+	<center>
		+	<figure>
		+	<img id="sol" src="sol.png" alt="1.1.23"
		+	loading="lazy" width="350" />
		+	<figcaption>
		+	Probability of a hit as a function of aiming distance
		+	</figcaption>
		+	</figure>
		+	</center>
		+	</p>
		+
		+	<h4>Answer</h4>
		+	<p>
		+	<a href="#sol">See Fig.</a> Zero at the walls. The largest at any place at a distance from the walls,
		+	greater than $2R$, and equal to $2R/(L-2R)$ at $L > 4R$; at any place at a distance from the walls greater than $L - 2R$, and equal to one at $4R > L > 2R$.
		+	</p>
		+
		+
		+	<footer class="row container">
		+	<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>
		+
		+	</html>

	@@ -0,0 +1,123 @@
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			<html lang="en">

			<head>
			<meta charset="utf-8">
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			<meta http-equiv="content-language" content="en">
			<meta name="keywords" content="solutions, savchenko, physics problems, olympiad physics, physics book">
			<meta name="description" content="A website with solutions to physics problems from Savchenko Textbook">
			<meta property="og:title" content="Savchenko Solutions">
			<meta property="og:image" content="img/logo.png">
			<meta property="og:description" content="A website with solutions to physics problems from Savchenko Textbook">
			<meta name="yandex-verification" content="6cfda41f74038368">
			<title>Savchenko Solutions</title>
			<link rel="stylesheet" href="https://savchenko-physics.github.io/css/css-latex/style.css">
			<link rel="icon" href="https://savchenko-physics.github.io/img/logo.png" type="image/png">
			<script src="https://savchenko-physics.github.io/js/jquery-1.10.1.min.js"></script>
			<script async src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.7/MathJax.js?config=TeX-MML-AM_CHTML"></script>
			<script type="text/x-mathjax-config">
			MathJax.Hub.Config({
			extensions: ['tex2jax.js'],
			jax: ['input/TeX', 'output/HTML-CSS'],
			tex2jax: {
			inlineMath: [['$', '$'], ['$', '$']],
			processEscapes: true,
			processClass: 'tex2jax',
			ignoreClass: 'html'
			},
			showProcessingMessages: false,
			messageStyle: 'none'
			});
			</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="../#1.1">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$1.1.23^*.$ The shooter tries to hit a disk of radius $R$, which moves from one wall to another at constant modulo velocity so fast that it cannot be tracked. Draw a graph of the probability of hitting the disk as a function of the distance between the aiming point and the left wall.
			</p>
			<center>
			<figure>
			<img src="statement.png"
			loading="lazy" width="230" />
			<figcaption>
			For problem $1.1.23^*$
			</figcaption>
			</figure>
			</center>
			<p>
			The shots are fired at height $R$ from the floor perpendicular to the direction of motion of the disc. At what point is the least and most probable shot? What is their value? Consider the cases $L > 4R$, $4R > L > 2R$, where $L$ is the distance between the walls.
			</p>

			<h3>Solution</h3>
			<p>

			<center>
			<figure>
			<img src="animation.gif" alt="1.1.23"
			loading="lazy" width="300" />
			<figcaption>
			Wheels moving
			</figcaption>
			</figure>
			</center>

			<p>
			When the target is moving, a bystander is less likely to see the target from the edge, as it stays there for a shorter period of time.
			</p>
			<p>
			Thus, the maximum probability will be observed on the interval:
			</p>
			<p style="text-align: center;">
			$$\left\{\begin{matrix}
			0 \leq x < 2R\\
			L-2R < x \leq L
			\end{matrix}\right.$$
			</p>
			<p>
			Since the velocity of the target is constant, and finding the target in the range $2R \leq x \leq L-2R$ is equally likely, the probability of finding it in this range is equal to the ratio of the diameter to the length of this range: $\frac{2R}{L-2R}$.
			</p>
			<p>
			Graphs corresponding to these modes:
			</p>

			<center>
			<figure>
			<img id="sol" src="sol.png" alt="1.1.23"
			loading="lazy" width="350" />
			<figcaption>
			Probability of a hit as a function of aiming distance
			</figcaption>
			</figure>
			</center>
			</p>

			<h4>Answer</h4>
			<p>
			<a href="#sol">See Fig.</a> Zero at the walls. The largest at any place at a distance from the walls,
			greater than $2R$, and equal to $2R/(L-2R)$ at $L > 4R$; at any place at a distance from the walls greater than $L - 2R$, and equal to one at $4R > L > 2R$.
			</p>


			<footer class="row container">
			<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>

			</html>