Example 1: Xiao Zhang is riding a bicycle on a road next to a double-track railway. He notices that every 12 minutes, a train catches up with him from behind, and every 4 minutes, a train comes towards him from the opposite direction. If the intervals between each train are constant, the speeds are the same, and both the train and bicycle speeds are uniform, find the interval in minutes at which trains depart from the stations in front of and behind Xiao Zhang.
(1990, Xianyang City Junior High School Mathematics Competition Selection)
Analysis: Let the interval be $x$ minutes at which trains depart from the stations in front of and behind Xiao Zhang. Let Xiao Zhang's cycling speed be $v_{1}$, and the train speed be $v_{2}$. Let $AB$ be the distance between two adjacent trains in the same direction, then $|AB| = x v_{2}$.
(1) First, consider the trains coming towards Xiao Zhang from the opposite direction. Assume Xiao Zhang meets the first train at point $A$, then when he meets the next train, both the train and Xiao Zhang have traveled the distance $|AB|$, so we have $4(v_{1} + v_{2}) = |AB|$. As shown in Figure 1.
(2) Similarly,

consider the trains
catching up with
Xiao Zhang. At point
$B$, he meets the
first train, and the next train is at point $A$ at that moment. To catch up with Xiao Zhang, the next train travels an additional distance of $|AB|$, so we have $12(v_{2} - v_{1}) = |AB|$. As shown in Figure 2.
Solution: Let the trains depart from the station ahead and behind Xiao Zhang every $x$ minutes, Xiao Zhang's cycling speed be $v_{1}$, and the train speed be $v_{2}$, then
$$
\left\{\begin{array}{l}
4\left(v_{1}+v_{2}\right)=x v_{2}, \\
12\left(v_{2}-v_{1}\right)=x v_{2} .
\end{array}\right.
$$

Solving, we get $x=6$ (minutes).
Answer: Omitted.
(2) Time equivalence relationship