9. (10 points) At 8:10 AM, Feifei walked from home to school. 3 minutes later, the dog started running to catch up with her, and caught up with her 200 meters away from home; after catching up, it immediately ran back home, and then immediately turned around to catch Feifei again, catching up with her 400 meters away from home. After catching up, it ran back home again, and then went to catch Feifei, finally catching up with her at school. Feifei arrived at school at 8: $\qquad$ minutes.
【Solution】Solution: According to the problem, we have:
From 200 meters away from home to 400 meters, as shown in the figure, within the same time, Feifei walked 200 meters, and the dog walked 600 meters, $V_{\text {dog }}: V_{\text {person }}=S_{\text {dog }}: S_{\text {person }}=3: 1$;
The distance from Feifei's home to the school is $400 \times 2=800$ (meters).
When the dog catches up with Feifei for the first time, the speed ratio is $3: 1$, and the time ratio is inversely proportional to the speed, which is $1: 3$, with a time difference of 3 minutes.

The time it takes for Feifei to walk 200 meters is: $3 \div(3-1) \times 3=4.5$ minutes.
The entire distance is 800 meters, which is 4 times 200 meters, so the time is 4 times $4.5 \times 4=18$ (minutes), $10+18=28$ (minutes).
Therefore, the answer is: 28