2. (10 points) The little rabbit and the little turtle start from location $A$ to the forest amusement park at the same time. The little rabbit jumps forward 36 meters per minute, and after every 3 minutes of jumping, it plays on the spot. The first time it plays for 0.5 minutes, the second time for 1 minute, the third time for 1.5 minutes, $\cdots$, the $k$-th time for $0.5 k$ minutes. The little turtle does not rest or play on the way. It is known that the little turtle arrives at the forest amusement park 3 minutes and 20 seconds earlier than the little rabbit. The distance from $A$ to the forest amusement park is 2640 meters. Then, the little turtle crawls $\qquad$ meters per minute.
【Analysis】First, analyze the time it would take for the rabbit without any rest, then find out the number of rests, and calculate the corresponding time difference which is the time for the tortoise. This will lead to the solution.

【Solution】According to the problem:
The rabbit, without resting, would need $2640 \div 36=73 \frac{1}{3}$.
$24 \times 3=72$ proves that the rabbit rested 24 times.
The 24 rest periods form an arithmetic sequence.
$0.5,1, \cdots, 12$.
Thus, the total rest time is $0.5+1+1.5+2+\cdots+12=150$ (minutes)

The total time for the rabbit is: $150+73+\frac{1}{3}=223 \frac{1}{3}$.
The time for the tortoise is $223 \frac{1}{3}-3 \frac{1}{3}=220$ (minutes)
The speed of the tortoise is: $2640 \div 220=12$ (meters/minute)
Therefore, the answer is: 12