<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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