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+ <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>
+ $1.1.14.$ 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: $t_2 = (2a - vt_{1})/v, \, {x}' = 2a - vt_{1}$
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