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		@@ -33,7 +33,7 @@
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		−	<h2>Solutions of Savchenko P~~hysics Textbook~~</h2>
		+	<h2>Solutions of Savchenko Problems in Physics</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>
			$1.1.14.$ Using coordinate-time graphs, find the point in time and place of collision of particles moving along one straight line. The speed of the first particle is $v$, the speed of the second particle is $v/2$. The first particle at time $t = 0$ had coordinate $x = 0$, the second particle at time $t_1$ had coordinate $x = a$.
			</p>

			<h3>Solution</h3>
			<p>
			<center>
			<figure>
			<img src="graph.svg"
			loading="lazy" alt="1.1.14" width="350" />
			</figure>
			</center>
			<p>
			The law of motion of the first particle:
			</p>
			<p style="text-align: center;">
			$x_1(t) = vt \; (1)$
			</p>
			<p>
			The law of motion of the second particle:
			</p>
			<p style="text-align: center;">
			$x_2(t) = \frac{v}{2}(t-t_1) + a$
			</p>
			<p>
			Crossing condition at the moment of time $t_2$:
			</p>
			<p style="text-align: center;">
			$x_1(t_2)=x_2(t_2) \Leftrightarrow vt_2 = \frac{v}{2}(t_2-t_1) + a$
			</p>
			<p>
			From where
			</p>
			<p style="text-align: center;">
			$t_2 = \frac{2a}{v} - t_1 \; (2)$
			</p>
			<p>
			Substituting $(2)$ into $(1)$:
			</p>
			<p style="text-align: center;">
			${x}' = 2a - vt_1$
			</p>
			</p>

			<h4>Answer</h4>
			<p>
			The ordinate and abscissa of the point of intersection of the graphs $x_{1} = vt$ and $x_{2} = a + v(t - t_{1})/2$ give the time and coordinate of the point of impact of the particles: $t_2 = (2a - vt_{1})/v, \, {x}' = 2a - vt_{1}$
			</p>


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	@@ -33,7 +33,7 @@
	</head>		</head>
	<body style="">		<body style="">
	<header style="text-align:center;">		<header style="text-align:center;">
	<h2>Solutions of Savchenko P~~hysics Textbook~~</h2>		<h2>Solutions of Savchenko Problems in Physics</h2>
	<p class="author">		<p class="author">
	Aliaksandr Melnichenka <br/>		Aliaksandr Melnichenka <br/>
	October 2023		October 2023
	</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>
	$1.1.14.$ Using coordinate-time graphs, find the point in time and place of collision of particles moving along one straight line. The speed of the first particle is $v$, the speed of the second particle is $v/2$. The first particle at time $t = 0$ had coordinate $x = 0$, the second particle at time $t_1$ had coordinate $x = a$.		$1.1.14.$ Using coordinate-time graphs, find the point in time and place of collision of particles moving along one straight line. The speed of the first particle is $v$, the speed of the second particle is $v/2$. The first particle at time $t = 0$ had coordinate $x = 0$, the second particle at time $t_1$ had coordinate $x = a$.
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	<center>		<center>
	<figure>		<figure>
	<img src="graph.svg"		<img src="graph.svg"
	loading="lazy" alt="1.1.14" width="350" />		loading="lazy" alt="1.1.14" width="350" />
	</figure>		</figure>
	</center>		</center>
	<p>		<p>
	The law of motion of the first particle:		The law of motion of the first particle:
	</p>		</p>
	<p style="text-align: center;">		<p style="text-align: center;">
	$x_1(t) = vt \; (1)$		$x_1(t) = vt \; (1)$
	</p>		</p>
	<p>		<p>
	The law of motion of the second particle:		The law of motion of the second particle:
	</p>		</p>
	<p style="text-align: center;">		<p style="text-align: center;">
	$x_2(t) = \frac{v}{2}(t-t_1) + a$		$x_2(t) = \frac{v}{2}(t-t_1) + a$
	</p>		</p>
	<p>		<p>
	Crossing condition at the moment of time $t_2$:		Crossing condition at the moment of time $t_2$:
	</p>		</p>
	<p style="text-align: center;">		<p style="text-align: center;">
	$x_1(t_2)=x_2(t_2) \Leftrightarrow vt_2 = \frac{v}{2}(t_2-t_1) + a$		$x_1(t_2)=x_2(t_2) \Leftrightarrow vt_2 = \frac{v}{2}(t_2-t_1) + a$
	</p>		</p>
	<p>		<p>
	From where		From where
	</p>		</p>
	<p style="text-align: center;">		<p style="text-align: center;">
	$t_2 = \frac{2a}{v} - t_1 \; (2)$		$t_2 = \frac{2a}{v} - t_1 \; (2)$
	</p>		</p>
	<p>		<p>
	Substituting $(2)$ into $(1)$:		Substituting $(2)$ into $(1)$:
	</p>		</p>
	<p style="text-align: center;">		<p style="text-align: center;">
	${x}' = 2a - vt_1$		${x}' = 2a - vt_1$
	</p>		</p>
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	The ordinate and abscissa of the point of intersection of the graphs $x_{1} = vt$ and $x_{2} = a + v(t - t_{1})/2$ give the time and coordinate of the point of impact of the particles: $t_2 = (2a - vt_{1})/v, \, {x}' = 2a - vt_{1}$		The ordinate and abscissa of the point of intersection of the graphs $x_{1} = vt$ and $x_{2} = a + v(t - t_{1})/2$ give the time and coordinate of the point of impact of the particles: $t_2 = (2a - vt_{1})/v, \, {x}' = 2a - vt_{1}$
	</p>		</p>


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