1. Four elevators of a skyscraper, differing in color (red, blue, green, and yellow), are moving in different directions and at different but constant speeds. Observing the elevators, someone started a stopwatch and, looking at its readings, began to record: 36th second - the red elevator caught up with the blue one (moving in the same direction). 42nd second - the red elevator passed the green one (moving in opposite directions), 48th second - the red elevator passed the yellow one, 51st second - the yellow elevator passed the blue one, 54th second - the yellow elevator caught up with the green elevator. At what second from the start of the count will the green elevator pass the blue one, if during the observation period the elevators did not stop and did not change their direction of movement?

(12 points)
Solution. Let's number the elevators: red - first, blue - second, green - third, yellow - fourth. The elevators move at constant speeds, so the distance traveled $S_{i}, i=1,2,3,4$, in some coordinate system depends on time according to the law $S_{i}=k_{i} t+b_{i}$. According to the problem, the red and blue elevators move in the same direction, and the red elevator is catching up with the blue one, so $k_{1} \cdot k_{2}>0, k_{1}>k_{2}$. For definiteness, let $k_{1}>0$, then $k_{2}>0$. The green and yellow elevators move in the opposite direction to the first two, and the yellow elevator is catching up with the green one, so $k_{3}<0, \quad k_{4}<0, \quad k_{3}<k_{4} \cdot$ Let's plot the graphs of the functions $S_{i}(t)$, according to the problem. We need to determine the abscissa of the point $M$ of intersection of the second graph with the third. Point $M$ is the point of intersection of the medians of triangle $A B C$. From this, we have $\frac{x-36}{51-x}=\frac{2}{1}, x=46$.

Answer: at 46 seconds.

![](https://cdn.mathpix.com/cropped/2024_05_06_f3f8bb21871417ffa2deg-1.jpg?height=793&width=853&top_left_y=1234&top_left_x=1047)