14. (10 points) Person A departs from location $A$ at a constant speed to location $B$. When A departs, person B departs from location $B$ at a constant speed to location $A$. They meet at location $C$ on the way. After meeting, A continues for another 150 meters and then turns around to chase B. A catches up with B 540 meters from $C$. When A catches B, A immediately turns around to head to $B$. As a result, when A arrives at $B$, B also arrives at $A$. The distance between $A$ and $B$ is $\qquad$ meters.
【Solution】Solution: According to the problem, we know:
When A catches up with B from the meeting point, A's distance is $150+150+540=840$ (meters); the ratio of the distances covered by A and B is $840: 540=14: 9$; the first meeting point is at $C$, so $A C: B C=14: 9$; the total length is divided into 23 parts. Assuming the second catch-up point is at position $D$, then $A D: B D=9: 14$;
Comparing the two situations, we know that $C D$ accounts for 5 parts, and the total length of $C D$ is 540 meters; thus, $540 \div 5 \times 23=2484$ meters;
Therefore, the answer is: 2484.