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