On a circular track with a circumference of 40 meters, two points are moving: $A$ and $B$, at speeds of 3 meters and 1 meter per second, respectively. $B$ always moves in the same direction on the track, while $A$ changes its direction whenever it gets within 10 meters of $B$ (measured along the track). Write down the movement of $A$ if the two points start simultaneously from the endpoints of a diameter of the track, in the same direction.
Any point $P$ on the circular track is characterized by the length of the arc $O P$, which $B$ travels from its starting point $O$ to $P$ during the first lap. Thus, for the starting point $K$ of $A$, $O K=20 \mathrm{~m}$. It is sufficient to examine the turning points of $A$, as between two consecutive turning points, $A$ moves uniformly.

![](https://cdn.mathpix.com/cropped/2024_05_02_8e06b62db3a91c31d53eg-1.jpg?height=515&width=528&top_left_y=298&top_left_x=728)

Figure 1

Until the first turning point, $A$ chases $B$. At the start, $A$ was $20 \mathrm{~m}$ behind $B$, and this decreases to the allowed $10 \mathrm{~m}$ by the first turning point. $A$ reduces this gap by $3-1=2 \mathrm{~m}$ per second, so it takes $10: 2=5 \mathrm{~s}$ to reach the turning point. During this time, $A$ covers $15 \mathrm{~m}$, so it reaches the point $A_{1}$ on the track, which is $20+15=35 \mathrm{~m}$ from $O$. Meanwhile, $B$ has traveled $5 \mathrm{~m}$, so it is at the point $B_{1}$, which is $5 \mathrm{~m}$ from $O$.

In the second segment of the movement, $A$ moves towards $B$. The initial distance of $30 \mathrm{~m}$ between them decreases by $3+1=4 \mathrm{~m}$ per second, and it takes $5 \mathrm{~s}$ to reduce this distance to the allowed $10 \mathrm{~m}$. During this time, $A$ returns to $K$, while $B$ moves $5 \mathrm{~m}$ past $B_{1}$ to a new position $B_{2}$, where $O B_{2}=10 \mathrm{~m}$.

In the third segment, $A$ chases $B$ again, as in the first segment. However, this segment lasts twice as long, $10 \mathrm{~s}$, because $A$ needs to reduce its gap from $30 \mathrm{~m}$ to $10 \mathrm{~m}$, a difference of $20 \mathrm{~m}$. During these $10 \mathrm{~s}$, $A$ passes through $O$ and reaches the point $A_{3}$, where $O A_{3}=K A_{3}-K O=10 \mathrm{~m}$. $B$ moves $10 \mathrm{~m}$ ahead to the point $B_{3}$, where $O B_{3}=20 \mathrm{~m}$.

This gives us two consecutive turning points of $A$ with the same characteristics, i.e., when $A$ chases $B$ and then turns to face $B$. The durations of the subsequent segments, where $A$ alternates between chasing and facing $B$, are also 5 and $10 \mathrm{~s}$, respectively. Thus, $A$ first moves in the same direction as $B$ for $5 \mathrm{~s}$, then alternates between moving backward for $5 \mathrm{~s}$ and forward for $10 \mathrm{~s}$.

Another question is whether, considering the movement relative to the track, there is a position of $A$ and $B$ that has already occurred. From this point onward, the segments of the movement will repeat relative to the track.

$B$'s movement on the track repeats every $40 \mathrm{~s}$, which is the time it takes to complete one lap; $A$'s turning points of the same type (e.g., from chasing to facing $B$) occur every $15 \mathrm{~s}$. The least common multiple of these two time intervals is $120 \mathrm{~s}$, which is the smallest time interval after which both the position of $B$ relative to the track and the position of $A$ relative to $B$ are the same. After $120 \mathrm{~s}$, $B$ is indeed at $O$, having completed 3 full laps. However, during the $115 \mathrm{~s}$ of $A$'s movement after the initial $5 \mathrm{~s}$ segment, the $5 \mathrm{~s}$ backward and $10 \mathrm{~s}$ forward movements repeat 7 times, and the remaining $10 \mathrm{~s}$ consists of $5 \mathrm{~s}$ backward and $5 \mathrm{~s}$ forward. Thus, $A$ moves backward for $7 \cdot 5 + 5 = 40 \mathrm{~s}$ and forward for $5 + 7 \cdot 10 + 5 = 80 \mathrm{~s}$. Therefore, it moves forward as if it had only moved forward for $40 \mathrm{~s}$. The distance covered in this time is $120 \mathrm{~m}$, which is three times the length of the track, so after $120 \mathrm{~s}$, $A$ is also at its starting point $K$. (However, at this moment, $K$ is not a turning point for $A$.)

This result can also be read from the graph of the movements. The condition that $A$ can only approach $B$ to within $10 \mathrm{~m}$ can be replaced by the following condition: $B$ is a $20 \mathrm{~m}$ long object (measured along the arc). Thus, the graph shows the start and end of $B$ ($B'$ and $B''$) and the segment between them is filled by $B$.

![](https://cdn.mathpix.com/cropped/2024_05_02_8e06b62db3a91c31d53eg-1.jpg?height=520&width=1174&top_left_y=2106&top_left_x=454)

Figure 2

The passages through the point $O$ are indicated by the graphs of $A$, $B$, $B'$, and $B''$ breaking and continuing from the point corresponding to $O$ on the line representing the $40 \mathrm{~m}$ track length. (The segments of the movement near $O$ are repeated above and below for better clarity.) The line segments representing the forward and backward movements of $A$ rise and fall at the same angle relative to the time axis and always remain between the graphs of $B'$ and $B''$. The graph also shows that after $120 \mathrm{~s}$, both $B$ and $A$ are back at their starting points, but they do not coincide at any earlier time.

Similarly, it can be shown that the movement also repeats every $120 \mathrm{~s}$ if $A$ starts moving in the opposite direction to $B$.