14. (15 points) Person A and Person B start from locations $A$ and $B$ respectively at the same time, walking towards each other, and meet at location $C$. After meeting, A continues walking towards $B$, while B rests for 14 minutes and then continues walking towards $A$. After A and B reach $B$ and $A$ respectively, they immediately turn back and meet again at $C$. It is known that A walks 60 meters per minute, and B walks 80 meters per minute. How far apart are $A$ and $B$?
【Solution】Solution: Let the time for the first meeting between A and B be $x$ minutes, then A has traveled $60 x$ meters and B has traveled $80 x$ meters; for the second meeting, A has traveled $80 x \times 2$ meters and B has traveled $60 x \times 2$ meters;

According to the fact that A travels 14 minutes more than B, we can get the equation:
$$
\begin{aligned}
(80 x \times 2) \div 60-(60 x \times 2) \div 80 & =14 \\
\frac{8}{3} \times \frac{3}{2} x & =14 \\
\frac{7}{6} x & =14 \\
x & =12
\end{aligned}
$$
The length of $A B$ is: $12 \times(60+80)=1680$ (meters).
Answer: The distance between $A$ and $B$ is 1680 meters.