2. (10 points) Cars A and B start from points $A$ and $B$ respectively at the same time, heading towards each other. At the start, the speed ratio of cars A and B is 5:4. Shortly after departure, car A gets a flat tire, stops to change the tire, and then continues, increasing its speed by $20 \%$. As a result, 3 hours after departure, the two cars meet at the midpoint of $A B$. After the meeting, car B continues forward, while car A turns around. When car A returns to point $A$, car B reaches the location where car A had the flat tire. How many minutes did car A spend changing the tire?
【Analysis】First, analyze the process of the second half of the sprint from the midpoint to $A$. By finding the speed ratio of the two, we can determine the distance ratio and locate the tire blowout position. Then, based on the original speed ratio, calculate the normal driving time and subtract the time before the blowout. Finally, based on the speed ratio before and after the blowout for person A, find the time ratio and take the difference.

【Solution】According to the problem:
The speed ratio of cars A and B later: $5(1+20 \%) : 4 = 3 : 2$, A travels 3 parts while B travels 2 parts. This indicates that the blowout position of car A is at $\frac{1}{3}$ of $A C$.

If car A does not blow out, the time and speed are inversely proportional: Let the time for car A to travel be $x$, then: 4: $5 = x: 3, x = \frac{12}{5}$
The normal time for car A to travel from the blowout position to the midpoint of $A C$ is: $\frac{12}{5} \times \left(1 - \frac{1}{3}\right) = \frac{24}{15} = \frac{8}{5}$ (hours); the speed ratio of car A before and after the blowout is: $5 : 5(1+20 \%) = 5 : 6$;
With a fixed distance, time and speed are inversely proportional: Let the time from the blowout to the midpoint be $y$, then: $6 : 5 = \frac{8}{5} : y, y = \frac{4}{3}$; the repair time is: $3 - \frac{12}{5} \times \frac{1}{3} - \frac{4}{3} = \frac{13}{15}$ (hours)
$$
\begin{array}{l}
\frac{13}{15} \times 60 = 52 \text { (minutes) } \\
\text { Therefore, the answer is: } 52 \text { minutes }
\end{array}
$$