Example 1: In a 10000-meter race on a 400-meter circular track at a school sports meet, two athletes, A and B, start running at the same time. B runs faster than A. At the 15th minute, A speeds up. At the 18th minute, A catches up with B and begins to overtake B. At the 23rd minute, A catches up with B again, and at 23 minutes and 50 seconds, A reaches the finish line. Therefore, the time B takes to complete the entire course is $\qquad$ minutes.
(Ninth Hope Cup National Mathematics Invitational Competition)
Solution: According to the problem, draw the figure, as shown in Figure 2.
The broken line $O A B$ represents the movement of A, and the line segment $O C$ represents the movement of B. Since A catches up with B again at the 23rd minute, $E F=400$. And $B G=C K=10000$.

Since $\triangle D E F \sim \triangle D H B$, according to the ratio of corresponding heights of similar triangles equals the similarity ratio, we get
$\frac{E F}{B H}=\frac{M V}{M G}$, which is $\frac{400}{B H}=\frac{5}{5 \frac{5}{6}}$.
Solving this, we get $B H=\frac{1400}{3}$.
Thus, $G H=B G-B H=10000-\frac{1400}{3}$
$$
=\frac{28600}{3} \text {. }
$$

From $\triangle O G H \sim \triangle O K C$, we get
$\frac{G H}{C K}=\frac{O G}{O K}$, which is $\frac{28600}{10000}=\frac{23 \frac{5}{6}}{O K}$.
Therefore, $O K=25$.
Hence, the time B takes to complete the entire course is 25 minutes.