Example 4 A mall has an escalator moving uniformly from the bottom to the top. Two people, A and B, are in a hurry to go upstairs. While riding the escalator, they both climb the stairs at a uniform speed. A reaches the top after climbing 55 steps, and B's climbing speed is twice that of A (the number of steps B climbs per unit time is twice that of A), and he reaches the top after climbing 60 steps. Therefore, the number of steps from the bottom to the top of the escalator is $\qquad$
(1998, Beijing Middle School Mathematics Competition for Grade 8 Preliminary)
Solution: According to the problem, draw the graph, as shown in Figure 5.
$O C, O B, O A$ are the motion graphs of the escalator, person A riding the escalator, and person B riding the escalator, respectively.
Draw $A M \perp t$-axis, intersecting $O B$ and $O C$ at $E$ and $F$, respectively. Draw $B N \perp t$-axis, intersecting $O C$ at $G$.
Assume that the escalator from the ground floor to the upper floor has $x$ steps. Then $A M = B N = x$, $F M, E F, A F$ are the number of steps the escalator rises, the number of steps person A climbs, and the number of steps person B climbs after time $t_1$, respectively. $B G$ is the number of steps person A climbs after time $t_2$. Therefore,
\[
\begin{array}{l}
B C = 55, A F = 60, \\
F M = x - 60, G N = x - 55.
\end{array}
\]

Since person B climbs the stairs at twice the speed of person A, $A F = 2 E F$. Thus, $E F = 30$.
\[
\begin{aligned}
& \because \triangle O E F \sim \triangle O B G, \\
\therefore & \frac{E F}{B G} = \frac{O M}{O N}. \\
\because & \triangle O M F \sim \triangle O N G, \\
\therefore & \frac{F M}{G N} = \frac{O M}{O N}.
\end{aligned}
\]

From (1) and (2), we get $\frac{E F}{B G} = \frac{F M}{G N}$,
which is $\frac{30}{55} = \frac{x - 60}{x - 55}$. Solving for $x$ gives $x = 66$.
Therefore, the escalator from the ground floor to the upper floor has 66 steps.
The key here is equation (3), which can be transformed into $\frac{E F}{F M} = \frac{B G}{G N}$, meaning that when the speed is constant, the ratio of the distances traveled by the two people in the same time is equal.

In summary, using a "distance-time" graph to solve travel application problems can transform implicit conditions into explicit conditions, making abstract concepts more intuitive, thus converting abstract thinking into visual thinking.