Example 3 On a street $AB$, person A walks from $A$ to $B$, and person B rides a bike from $B$ to $A$. Person B's speed is 3 times that of person A. At this moment, a public bus departs from the starting station $A$ and heads towards $B$, and a bus is dispatched every $x$ minutes. After some time, person A notices that a bus catches up to him every 10 minutes, while person B feels that he encounters a bus every 5 minutes. Therefore, the interval time $x=$ $\qquad$ for the dispatch of buses at the starting station.
(9th Hope Cup National Mathematics Invitational for Junior High School)
Solution: According to the problem, draw the graph as shown in Figure 4.
$A C, A_{1} C_{1}, A_{2} C_{2}, A_{3} C_{3}$ represent the motion graphs of buses departing every $x$ minutes, $A D$ and $B . V$ are the motion graphs of person A and person B, respectively. $E_{1}, E_{2}$ are the points where a bus catches up with person A every 10 minutes. $F_{1}, F_{2}, F_{3}, F_{4}$ are the points where person B encounters a bus every 5 minutes.

Draw $E_{1} G \perp t$-axis through $E_{1}$, draw $F_{2} H / / t$-axis intersecting $O C$ at $H$, and draw $F_{1} M \perp H F_{2}$, then $A G=$ 10. $M F_{2}=5, H F_{2}=A A_{1}=x$.

Let $\angle E_{1} A G=\alpha, \angle F_{1} H M=\angle E_{1} A_{1} G=$ $\beta, \angle F_{1} F_{2} M=\gamma$. Also,
$\tan \alpha=\frac{E_{1} G}{A G}=\frac{E_{1} G}{10}$.
In $\triangle E_{1} A_{1} G$,
$\tan \beta=\frac{E_{1} G}{A_{1} G}=\frac{E_{1} G}{A G-A A_{1}}=\frac{E_{1} G}{10-x}$.
In $\triangle F_{1} H M$,
$\tan \beta=\frac{F_{1} M}{H M}=\frac{F_{1} M}{H F_{2}-M F_{2}}=\frac{F_{1} M}{x-5}$.
$\tan \gamma=\frac{F_{1} M}{M F_{2}}=\frac{F_{1} M}{5}$.
(1) $\div$ (2) gives $\frac{\tan \alpha}{\tan \beta}=\frac{10-x}{10}$.
(3) $\div$ (4) gives $\frac{\tan \beta}{\tan \gamma}=\frac{5}{x-5}$.
Since person B's speed is 3 times that of person A, then $\tan \gamma=3 \tan \alpha$.
(3) $\times$ (6) gives $\frac{1}{3}=\frac{10-x}{2(x-5)}$. Solving for $x$ yields $x=8$.
Therefore, the interval between departures at the starting station is 8 minutes.
The key here is equations (3) and (6). Their practical significance is the same as in Example 2.