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BPO 2019, Теория — Задача 3

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Всички материали от BPO 2019

1st Balkan Physics Olympiad – 2019 BPO · 14 юли 2019 г. · 25 т.

Условие

A man stands at a point $A$ in a field, at a distance $l_1 = 600\ m$ from the road $BC$ . In the field his velocity is $v_1 = 1\ m/s$ , while on the road $BC$ it is $v_2 = 2\ m/s$ . He can walk in the field along $AD$ and on the road along $DC$ in order to reach the final destination $C$ . The distance between $B$ and $C$ is $l_2 = 800\ m$ . Your final assignment is to find his route, so that he can reach the destination in the least possible time and to determine the time elapsed. In order to do this, you need to find the coordinate of the point $D$ , i.e., the distance $x = BD$ . We will lead you through several steps in order to fulfill this assignment.

Right-angle diagram: point A above B at distance l1; the road runs from B through D to C with total length l2 and BD = x; a blue arrowed path goes from A straight to D and then along the road to C; labels FIELD and ROAD.

a) Find the expression for the time $t_1$ for the motion on the field (route $AD$ ) as a function of $x$ . [2 т.]

b) Find the expression for the time $t_2$ for the motion on the road (route $DC$ ) as a function of $x$ . [2 т.]

c) Find the expression for the total time $t$ as a function of $x$ . [2 т.]

d) Let your solutions for the total time $t$ and the position $x$ be $t = t_0$ and $x = x_0$ . Put these solutions in your main expression for the total time. Using elementary algebraic manipulations (squaring, etc.) you should get rid of all the square root(s) and fraction(s) in your expression. [3 т.]

e) Now, if the solution for $x$ is a little bit different from the real solution $x_0$ (in other words, if we put the point $D$ a little bit closer to $B$ or to $C$ ), then the solution for $t$ would also be a little bit different from the real solution $t_0$ . So, we need to try to “shake” the solutions to see if they really are the correct solutions. To do this, we suggest you to add small temporal and spatial displacements (perturbations) $\Delta t$ and $\Delta x$ , i.e., instead of $t_0$ and $x_0$ put $t_0 + \Delta t$ and $x_0 + \Delta x$ in your main expression obtained in the previous step. [3 т.]

f) Rewrite the expression from the previous step neglecting all terms containing very small products, such as $\Delta t \cdot \Delta t$ , $\Delta t \cdot \Delta x$ and $\Delta x \cdot \Delta x$ . [4 т.]

g) Using the result from the previous step, express the ratio between small temporal and spatial displacements $\Delta t$ and $\Delta x$ , i.e., express the quantity $\Delta t / \Delta x$ . [3 т.]

h) Your final job in this assignment is to calculate the route and the total time. To do this, equate $\Delta t / \Delta x$ to zero, and with the help of some previous steps you will be in the position to proceed towards obtaining $x_0$ and $t_0$ . [6 т.]

Отговори

Решение

Data:

l_1 = 600\ m, \; v_1 = 1\,\frac{m}{s},

l_2 = 800\ m, \; v_2 = 2\,\frac{m}{s}

Expression for time $t_1$ for the motion on the field as a function of $x$ :

t_1 = \frac{AD}{v_1} = \frac{\sqrt{x^2 + l_1^2}}{v_1} \qquad (1)

Expression for time $t_2$ for the motion on the road as a function of $x$ :

t_2 = \frac{DC}{v_2} = \frac{l_2 - x}{v_2} \qquad (2)

Expression for the total time $t$ as a function of $x$ :

t = t_1 + t_2 = \frac{1}{v_1}\sqrt{x^2 + l_1^2} + \frac{1}{v_2}(l_2 - x) \qquad (3)

Solutions for the total time and position, $t = t_0$ and $x = x_0$ :

t_0 = \frac{1}{v_1}\sqrt{x_0^2 + l_1^2} + \frac{1}{v_2}(l_2 - x_0) \qquad (4)

After some elementary algebraic manipulations in (4) we get

v_1^2v_2^2t_0^2 + v_1^2l_2^2 + v_1^2x_0^2 - 2v_1^2v_2t_0l_2 + 2v_1^2v_2t_0x_0 - 2v_1^2x_0l_2 - v_2^2x_0^2 - v_2^2l_1^2 = 0 \qquad (5)

“Shaking“ the solutions with small quantities $\Delta t$ and $\Delta x$ :

\begin{aligned} &v_1^2v_2^2(t_0 + \Delta t)^2 + v_1^2l_2^2 + v_1^2(x_0 + \Delta x)^2 - 2v_1^2v_2(t_0 + \Delta t)l_2 + \\ &+2v_1^2v_2(t_0 + \Delta t)(x_0 + \Delta x) - 2v_1^2(x_0 + \Delta x)l_2 - v_2^2(x_0 + \Delta x)^2 - v_2^2l_1^2 = 0 \qquad (6) \end{aligned}

\begin{aligned} (6) \;\Rightarrow\; &v_1^2v_2^2(t_0^2 + 2t_0\Delta t + \Delta t^2) + v_1^2l_2^2 + v_1^2(x_0^2 + 2x_0\Delta x + \Delta x^2) - \\ &-2v_1^2v_2(t_0 + \Delta t)l_2 + 2v_1^2v_2(t_0 + \Delta t)(x_0 + \Delta x) - 2v_1^2(x_0 + \Delta x)l_2 - \\ &-v_2^2(x_0^2 + 2x_0\Delta x + \Delta x^2) - v_2^2l_1^2 = 0 \qquad (7) \end{aligned}

\begin{aligned} (5) \to (7) \;\Rightarrow\; &v_1^2v_2^2(2t_0\Delta t + \Delta t^2) + v_1^2(2x_0\Delta x + \Delta x^2) - \\ &-2v_1^2v_2\Delta t l_2 + 2v_1^2v_2(\Delta t x_0 + t_0\Delta x + \Delta t\Delta x) - 2v_1^2\Delta x l_2 - \\ &-v_2^2(2x_0\Delta x + \Delta x^2) = 0 \qquad (8) \end{aligned}

Neglecting all the terms containing very small products, $\Delta t \cdot \Delta t$ , $\Delta t \cdot \Delta x$ and $\Delta x \cdot \Delta x$ :

(8) \;\Rightarrow\; v_1^2v_2^2t_0\Delta t + v_1^2x_0\Delta x - v_1^2v_2\Delta t l_2 + v_1^2v_2(\Delta t x_0 + t_0\Delta x) - v_1^2\Delta x l_2 - v_2^2x_0\Delta x = 0 \qquad (9)

Expression for $\Delta t / \Delta x$ :

(9) \;\Rightarrow\; \frac{\Delta t}{\Delta x} = \frac{l_2 + (\frac{v_2^2}{v_1^2} - 1)x_0 - v_2t_0}{v_2^2t_0 - v_2l_2 - v_2x_0} \qquad (10)

Equating $\Delta x / \Delta t$ to zero and evaluation of $x$ and $t$ :

\frac{\Delta t}{\Delta x} = 0 \qquad (11)

(10), (11) \;\Rightarrow\; l_2 + (\frac{v_2^2}{v_1^2} - 1)x_0 - v_2t_0 = 0 \qquad (12)

(4) \to (12) \;\Rightarrow\; x_0 = \ldots = \frac{l_1}{\sqrt{\frac{v_2^2}{v_1^2} - 1}} \qquad (13)

(13) \to (4) \;\Rightarrow\; t_0 = \ldots = \frac{1}{v_2}\left[l_2 + l_1\sqrt{\frac{v_2^2}{v_1^2} - 1}\right] \qquad (14)

(13) \;\Rightarrow\; x_0 = \ldots = 346.8\,m \qquad (15)

(14) \;\Rightarrow\; t_0 = \ldots = 920\,s \qquad (16)

Оригинал в Архива: problems.pdf · официални решения: solutions.pdf