6. As shown in Figure 1, two skaters, Jia and Yi, are at points $A$ and $B$ on a circular skating rink, with $OB=20 \, \text{m}$, $OA=15 \, \text{m}$, and $OA \perp OB$. Yi starts from point $B$ and skates along the circumference of the circle $\odot O$ in a clockwise direction at a speed of $5 \, \text{m/s}$. At the same time, Jia starts from point $A$ and skates along a straight line at a speed of $5 \, \text{m/s}$. This straight line is chosen so that Jia and Yi meet as soon as possible under the given speeds. The earliest time (s) for their meeting is in ( ).
(A) $(0,5)$
(B) $(5,6)$
(C) $(6,7)$
(D) $(7,10)$
6.C.

As shown in Figure 6, rays $AO$ and $BO$ intersect
$\odot O$ at points $A'$ and $B'$, respectively.
Points $C$ and $D$ are taken sequentially on $\overparen{B A'}$.
$AC$ and $AD$ intersect $\odot O$ at points $C'$ and $D'$, respectively.
Perpendiculars are drawn from point $O$ to chords $CC'$ and $DD'$, with feet at $C''$ and $D''$, respectively.
Let $OC''$ intersect $DD'$ at point $E$. Then
$OC'' > OE > OD''$.
Thus, $AC + AC' > 0, AD - AD' > 0$.
Therefore, $AC - AC' < AD - AD' = \sqrt{(15 + 10 \sqrt{2})^2 + (10 \sqrt{2})^2} \\
= 5 \sqrt{25 + 12 \sqrt{2}}$.
Hence, $t > \sqrt{25 + 12 \sqrt{2}} > 6.3$.
Therefore, person B has traveled at least $31.5 \, \text{m}$.
Since $\frac{2 \pi \times 20}{4} = 10 \pi < 31.5$, person B has passed point $A'$ by time $t$.
Also, $AA' = 35 \, \text{m}$, so person A takes less than $7 \, \text{s}$ to slide along a straight line to any point on $\overparen{B A' B'}$ (except point $A'$). Thus, $t < 7$.
Therefore, $t \in (6, 7)$.