9. (12 points) Person A and Person B start from points $A$ and $B$ respectively at the same time, heading towards each other. They meet for the first time 100 meters from point $B$. After the meeting, Person A's speed doubles: Person A reaches $B$ and immediately turns back, catching up with Person B when Person B is still 50 meters away from $A$. What is the distance between $A$ and $B$ in meters?
【Solution】Solution: Let point $C$ be the first meeting point of A and B, and point $D$ be the point where A catches up with B after turning around,

When B travels from $B$ to $D$,
A travels from $A$ to $C$, then speeds up and travels from $C$ to $B$ and then turns around to $D$, the distance traveled is:
$B D \times 2 + 50$ meters (i.e., $A D$)
Assuming A starts from point $A$ and travels at twice the original speed,
it would be 50 meters $+ C D$ more than the current distance (i.e., the green line segment in the diagram),
which is: $B D \times 2 + 50$ meters + 50 meters $+ C D$
$= B D \times 2 + 100$ meters $+ C D$ (from the diagram, it can be seen that $C D + 100$ meters $= B D$)
$= B D \times 3$
From this, we know that when A's speed is increased to twice the original speed, it is 3 times the speed of B,
so A's original speed is: $3 \div 2 = 1.5$ (times)
That is, when they first meet, B travels 100 meters $(B \rightarrow C)$, and A travels $100 \times 1.5 = 150$ meters $(A \rightarrow C)$
Therefore, the distance between $A$ and $B$ is:
$$
\begin{array}{l}
100 \times (1.5 + 1) \\
= 100 \times 2.5 \\
= 250 \text{ (meters) }
\end{array}
$$

Answer: The distance between $A$ and $B$ is 250 meters.