9. (12 points) Three people, A, B, and C, depart from location $A$ to location $B$. A departs at 8:00, B at 8:20, and C at 8:30. They all travel at the same speed. 10 minutes after C departs, the distance from A to $B$ is exactly half the distance from B to $B$. At this moment, C is 2015 meters away from $B$. Therefore, the distance between $A$ and $B$ is $\qquad$ meters.
9. (12 points) Three people, A, B, and C, depart from location $A$ to location $B$. A departs at 8:00, B at 8:20, and C at 8:30. They all travel at the same speed. 10 minutes after C departs, the distance from A to $B$ is exactly half the distance from B to $B$. At this time, C is 2015 meters away from $B$, and the distance between $A$ and $B$ is 2418 meters.
【Solution】Solution: According to the problem, we know:
Let the distance B and C travel in 10 minutes be 1 unit;
The distance A and B travel in 20 minutes is 2 units;
The distance from A to $B$ is the same as the distance from A to B, which is also 2 units.

Therefore, the distance from C to $B$ is 5 units, corresponding to 2015 meters, so 1 unit is $2015 \div 5 = 403$ meters.
The distance between B and C is also the distance C travels in 10 minutes, which is 1 unit. Therefore, the total distance is 6 units, which is: $403 \times 6 = 2418$ meters.

Thus, the answer is: 2418.