9. (3 points) Cars A and B start from locations $A$ and $B$ simultaneously and travel back and forth between $A$ and $B$ at a constant speed. If after the first meeting, Car A continues to drive for 4 hours to reach $B$, while Car B only drives for 1 hour to reach $A$, then when the two cars meet for the 15th time (meetings at $A$ and $B$ are not counted), they have driven $\qquad$ hours.
【Analysis】Let the two cars meet after $t$ hours, the speed of car A is $v_{1}$, and the speed of car B is $v_{2}$. From the problem, we have: $4 v_{1}=t v_{2}$, $(t+4) v_{1}=(t+1) v_{2}$. Solving these equations, we get $t=2$. Therefore, it takes car A 6 hours to complete the journey, and car B 3 hours. Interestingly, when car A completes one trip, car B completes the round trip. Thus, excluding the meetings at points A and B, the two cars meet every 6 hours. They meet 2 hours after starting in opposite directions, and 4 hours after starting in the same direction. The 15th trip is in opposite directions, so $6 \times 14+2=86$ (hours).

【Solution】Let the two cars meet after $t$ hours, with car A's speed being $v_{1}$ and car B's speed being $v_{2}$. Then:
$$
\begin{array}{l}
4 v_{1}=t v_{2}, \\
(t+4) v_{1}=(t+1) v_{2},
\end{array}
$$

Solving these, we get $t=2$.
So it takes car A 6 hours to complete the journey, and car B 3 hours. Excluding the meetings at points A and B, the two cars meet every 6 hours. They meet 2 hours after starting in opposite directions, and 4 hours after starting in the same direction. The 15th trip is in opposite directions, so the two cars meet for the 15th time after:
$$
\begin{array}{l}
6 \times(15-1)+2 \\
=6 \times 14+2 \\
=84+2 \\
=86 \text { (hours) }
\end{array}
$$

Answer: When the two cars meet for the 15th time, they have been traveling for 86 hours.