2. (10 points) It is known that the distance between $A$ and $B$ is 300 meters. Two people, Jia and Yi, start from $A$ and $B$ respectively at the same time, walking towards each other, and meet at a point 140 meters from $A$; if Yi walks 1 meter more per second, then the meeting point is 180 meters from $B$. What was Yi's original speed in meters per second?
A. $2 \frac{3}{5}$
B. $2 \frac{4}{5}$
C. 3
D. $3 \frac{1}{5}$
【Analysis】This problem is a typical example of solving a travel problem using direct and inverse proportions. First, determine the direct or inverse proportion based on the constant. In the two meetings, the time for both individuals is the same, so the ratio of distances traveled is equal to the ratio of speeds. In the two processes, A's speed remains unchanged. By comparing B's speed through a common denominator, the problem can be solved.

【Solution】Solution: In the first meeting, the ratio of the distances traveled by A and B is $140: (300-140) = 7: 8$. Since the time is the same, the ratio of distances is the same as the ratio of speeds.

In the second meeting, the ratio of the distances traveled is $(300-180): 180 = 2: 3$, and the speed ratio is also 2: 3.
In the two meeting problems, A's speed remains constant. By finding a common denominator, the first speed ratio is $7: 8 = 14: 16$. The second speed ratio is $2: 3 = 14: 21$.

The speed increases from 16 parts to 21 parts, with an increase of 1 meter per second, i.e., $1 \div (21-16) = \frac{1}{5}$. B's original speed is $16 \times \frac{1}{5} = 3.2$ meters/second.

Therefore, the answer is: $D$.