15. (15 points) Two cars, A and B, travel from City A to City B at a speed of 55 kilometers/hour. At 10:00 AM, the distance Car A has traveled is 5 times the distance Car B has traveled; at noon 12:00 PM, the distance Car A has traveled is 3 times the distance Car B has traveled. How many hours later did Car B start compared to Car A?
【Analysis】It is known that during the 2 hours from 10:00 to 12:00, both car A and car B are driving. Therefore, the time that car B starts later than car A is the time that car B drives less than car A before 10:00. Thus, the problem can be solved by finding the time each car has driven before 10:00.
(1) The speed of both cars is 55 kilometers/hour. Here, we can assume that the distance car B has driven by 10:00 is $x$ kilometers, then the distance car A has driven is $5x$ kilometers. Therefore, the time car A has driven before 10:00 is $\frac{5x}{55}$ hours, and the time car B has driven is $\frac{x}{55}$ hours. Thus, the total time car A has driven is $\frac{5x}{55}+2$ hours, and the total time car B has driven is $\frac{x}{55}+2$ hours.
(2) According to the principle that distance is directly proportional to time when speed is constant, since the distance car A has driven is 3 times the distance car B has driven, the time car A has driven is 3 times the time car B has driven. We can set up an equation to solve the problem.

【Solution】Solution: Let the distance car B has driven by 10:00 be $x$ kilometers, then the distance car A has driven is $5x$ kilometers. Therefore, the time car A has driven before 10:00 is $\frac{5x}{55}$ hours, and the time car B has driven is $\frac{x}{55}$ hours. According to the problem, we can set up the equation:
$$
\begin{aligned}
\frac{5x}{55}+2 & =3 \times\left(\frac{x}{55}+2\right), \\
2x & =220, \\
x & =110,
\end{aligned}
$$

Therefore, the time car A has driven before 10:00 is: $\frac{5x}{55}=\frac{5 \times 110}{55}=10$ (hours),
the time car B has driven is $\frac{x}{55}=\frac{110}{55}=2$ (hours),
$10-2=8$ (hours),
Answer: Car B started 8 hours later than car A.