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.
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<h3> Statement </h3>
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$1.1.14.$ 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 $x = a$.
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<h3>Solution</h3>
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The law of motion of the first particle:
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$x_1(t) = vt \\; (1)$
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The law of motion of the second particle:
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$x_2(t) = \frac{v}{2}(t-t_1) + a$
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Crossing condition at the moment of time $t_2$:
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$x_1(t_2)=x_2(t_2) \Leftrightarrow vt_2 = \frac{v}{2}(t_2-t_1) + a$
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From where
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$t_2 = \frac{2a}{v} - t_1 \\; (2)$
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Substituting $(2)$ into $(1)$:
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${x}' = 2a - vt_1$
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<h4>Answer</h4>
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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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