14. Two students, A and B, who live in the same community, leave the community gate for school at the same time. At the beginning, A's speed is 40 meters per minute, and B's speed is 60 meters per minute. After A has walked half the distance, A realizes that at this speed, A will be late, so A increases the speed to 60 meters per minute, and B also slows down to 40 meters per minute at this moment. As a result, A arrives 2 minutes later than B. The distance from the community to the school is $\qquad$ meters.
【Analysis】When A arrives at school, if B continues to walk, B will walk 80 meters more than A. It is found that the time A walks at 40 meters/min is the same as the time B walks at 60 meters/min, and the time A walks at 60 meters/min is the same as the time B walks at 40 meters/min. Since B walks 80 meters more than A, it indicates that the time A walks at 40 meters/min is $80 \div (60-40) = 4$ minutes more than the time A walks at 60 meters/min.
Let the time A walks at 60 meters/min be $x$ minutes, then the time A walks at 40 meters/min is $x+4$ minutes. We can set up the equation:
$$
60 x = 40(x+4) \Rightarrow x = 8
$$

Thus, the distance is $40 \times 12 + 60 \times 8 = 960$ meters