B3. We observe two bodies $A$ and $B$, which are $450 \mathrm{~m}$ apart from each other. At the same time, they start moving towards each other (in a straight line). Body $A$ moves $5 \mathrm{~m}$ in the first minute, and then in each subsequent minute, it moves $15 \mathrm{~m}$ more than in the previous minute. Body $B$ moves $100 \mathrm{~m}$ in the first minute, and then in each subsequent minute, it moves $10 \mathrm{~m}$ less than in the previous minute. After how much time and at what distance from the initial position of body $A$ do the bodies meet?

## 21st Mathematics Knowledge Competition for Students of Secondary Technical and Vocational Schools  National Competition, May 15, 2021  Solutions to Problems for 1st Year

![](https://cdn.mathpix.com/cropped/2024_06_07_5bf9d67aee417dc7ba6ag-18.jpg?height=137&width=232&top_left_y=477&top_left_x=929)
B3. We find that the total distance traveled by both bodies is $450 \mathrm{~m}$. The distance traveled by body $A$ until the meeting is the sum of an arithmetic sequence $\mathrm{z}$ with the first term $a_{1}=5 \mathrm{~m}$ and the common difference $d_{1}=15 \mathrm{~m}$. The distance traveled by body $B$ until the meeting is the sum of an arithmetic sequence $\mathrm{z}$ with the first term $b_{1}=100 \mathrm{~m}$ and the common difference $d_{2}=-10 \mathrm{~m}$. Let $n$ be the time in minutes that elapses from the start of the journey until the meeting. We substitute the data into the formula for the sum of the first $n$ terms of an arithmetic sequence and write the distances traveled by both bodies.

Body $A$ travels the distance

$$
S_{A}=\frac{n}{2}(2 \cdot 5+(n-1) \cdot 15)
$$

body $B$ travels the distance

$$
S_{B}=\frac{n}{2}(2 \cdot 100+(n-1) \cdot(-10))
$$

Considering that the total distance traveled by both bodies $S_{A}+S_{B}$ measures $450 \mathrm{~m}$. We get the equation

$$
\frac{n}{2}(2 \cdot 5+(n-1) \cdot 15)+\frac{n}{2}(2 \cdot 100+(n-1) \cdot(-10))=450
$$

We rearrange the equation and get a quadratic equation

$$
n^{2}+41 n-180=0
$$

We solve the equation and get $n_{1}=4, n_{2}=-45$. The negative solution has no meaning, so the bodies have spent 4 minutes traveling until the meeting. We also calculate the distance traveled by body $A$ in this time.

$$
\begin{gathered}
S_{A}=\frac{n}{2}(2 \cdot 5+(n-1) \cdot 15) \\
S_{A}=\frac{4}{2}(2 \cdot 5+(4-1) \cdot 15)=2(10+45)=110
\end{gathered}
$$

The bodies meet after 4 minutes. Body $A$ travels a distance of $110 \mathrm{~m}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_5bf9d67aee417dc7ba6ag-25.jpg?height=68&width=1742&top_left_y=1714&top_left_x=171)

Writing the distance of body $A a_{1}=5, d_{1}=15$ and the distance $S_{A}=\frac{n}{2}(2 \cdot 5+(n-1) \cdot 15) \ldots \ldots \ldots \ldots \ldots . . . . .1$ point

Writing the distance of body $B b_{1}=100$ and $d_{2}=-10$ and the distance $S_{B}=\frac{n}{2}(2 \cdot 100+(n-1) \cdot(-10)) \ldots \ldots . .1$ point

![](https://cdn.mathpix.com/cropped/2024_06_07_5bf9d67aee417dc7ba6ag-25.jpg?height=52&width=1727&top_left_y=1873&top_left_x=176)

![](https://cdn.mathpix.com/cropped/2024_06_07_5bf9d67aee417dc7ba6ag-25.jpg?height=57&width=1727&top_left_y=1919&top_left_x=176)

![](https://cdn.mathpix.com/cropped/2024_06_07_5bf9d67aee417dc7ba6ag-25.jpg?height=55&width=1728&top_left_y=1970&top_left_x=174)

Determination that they travel the distance in 4 minutes...................................................... 1 point

Calculation of the distance traveled by body $A$ from the start of the journey until the meeting $S_{A}=\frac{4}{2}(2 \cdot 5+(4-1) \cdot 15)=$ $2(10+45)=110$.

1 point