【Question 15】Junjun starts from $\mathrm{A}$ and travels at a constant speed to $\mathrm{B}$. When Junjun starts, Aping starts from $\mathrm{B}$ and travels at a constant speed to $\mathrm{A}$. They meet at point $\mathrm{C}$ along the way. After meeting, Junjun walks another 100 meters and then turns around to chase Aping, catching up to Aping 360 meters from $\mathrm{C}$. When Junjun catches up to Aping, he immediately turns around and heads to $\mathrm{B}$. As a result, when Junjun arrives at $\mathrm{B}$, Aping also arrives at $\mathrm{A}$. What is the distance between $\mathrm{A}$ and $\mathrm{B}$ in meters?
Analysis:
As shown in the figure, C is the first meeting point of Junjun and Pingping.
Junjun walked 100 meters from point C to point E, then turned around and caught up with Pingping at point D, which is 360 meters from C.
The time it took for Pingping to walk CD is the same as the time it took for Junjun to walk CE + ED.
Therefore, \( v_{\text{Jun}} : v_{\text{Ping}} = CD : (CE + ED) = 360 : (100 + 100 + 360) = 9 : 14 \)
At the first meeting, the time it took for Junjun to walk segment AC is the same as the time it took for Pingping to walk segment BC, so the ratio of their distances is \( AC : BC = 14 : 9 \). Assuming the total distance AB is \( 14 + 9 = 23 \) (units), then \( AC = 14 \) (units).
After the second catch-up, the time it took for Junjun to walk segment DB is the same as the time it took for Pingping to walk segment DA, so \( AD : BD = 9 : 14 \), thus \( AD = 9 \) (units). Therefore, \( DC = 14 - 9 = 5 \) (units).
So, \( AB = 360 \div 5 \times 23 = 1656 \) (meters)