1.1.14 | revision #11748 | Savchenko Solutions

Statement

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 $$x = a$$ .

Solution

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 $$(1)$$ :

${x}' = 2a - vt_1$

Answer

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: $t_2 = (2a - vt_{1})/v, \, {x}' = 2a - vt_{1}$