2. As shown in Figure 1, two moving points, Jia and Yi, start moving along the sides of square $ABCD$ from vertices $A$ and $C$ simultaneously. Jia moves in a clockwise direction, while Yi moves in a counterclockwise direction. If Yi's speed is 5 times that of Jia, then they will meet for the 2010th time on side ( ).
(A) $AB$
(B) $BC$
( ($\mathrm{C}$) $CD$
(D) $DA$
2. A.

Let the side length of the square be $a$. Then, when they meet for the first time, A and B have traveled a total of $2a$. From the first meeting to the second meeting, A and B travel a total of $4a$, and for each subsequent meeting, A and B travel a total of $4a$. Therefore, when they meet for the 2010th time, A and B have cumulatively traveled
$2009.5 \times 4a$.
Since B's speed is 5 times that of A, A has cumulatively traveled
$$
2009.5 \times 4a \times \frac{1}{6}=\left(334 \times 4+3 \frac{2}{3}\right) a .
$$

Therefore, they meet on side $AB$.