$1.1.14.$ 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 $x = a$.
The law of motion of the first particle:
$x_1(t) = vt \; (1)$
The law of motion of the second particle:
$x_2(t) = \frac{v}{2}(t-t_1) + a$
Crossing condition at the moment of time $t_2$:
$x_1(t_2)=x_2(t_2) \Leftrightarrow vt_2 = \frac{v}{2}(t_2-t_1) + a$
From where
$t_2 = \frac{2a}{v} - t_1 \; (2)$
Substituting $(2)$ into $(1)$:
${x}' = 2a - vt_1$
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: $t_2 = (2a - vt_{1})/v, \, {x}' = 2a - vt_{1}$