10. (10 points) Locations $A$, $B$, $C$, and $D$ are sequentially distributed along the same road. Three people, Jia, Yi, and Bing, start from $A$, $B$, and $C$ respectively, and travel uniformly towards $D$. When Jia catches up with Yi at $C$, Jia's speed decreases by $40\%$. When Jia catches up with Bing, Jia's speed decreases again by $40\%$. Nine minutes after Jia catches up with Bing, Yi also catches up with Bing, at which point Yi's speed decreases by $25\%$. After Yi catches up with Bing and travels an additional 50 meters, the three arrive at $D$ simultaneously. Given that Yi's initial speed is 60 meters per minute, Jia's initial speed is $\qquad$ meters per minute, and the distance between $A$ and $D$ is $\qquad$ meters.
10. (10 points) Locations A, B, C, and D are sequentially distributed along the same road. Three people, Jia, Yi, and Bing, start from A, B, and C respectively, and travel uniformly towards D. When Jia catches up with Yi at C, Jia's speed decreases by 40%. When Jia catches up with Bing, Jia's speed decreases by another 40%. Nine minutes after Jia catches up with Bing, Yi also catches up with Bing, at which point Yi's speed decreases by 25%; after catching up with Bing, Yi travels another 50 meters, and all three arrive at D simultaneously. It is known that Yi's initial speed is 60 meters per minute, and Jia's initial speed is 125 meters per minute, with the distance between A and D being 1880 meters. [Analysis] Since they arrive simultaneously, Jia's speed equals Bing's speed when Jia catches up with Bing, and Yi's speed equals Bing's speed when Yi catches up with Bing. Yi's initial speed is 60 meters per minute, and after catching up with Bing, Yi's speed becomes \(60 \times (1-25\%) = 45\) meters per minute, so Bing's speed is 45 meters per minute. It can be deduced that Jia's speed after catching up with Bing is 45 meters per minute, and Jia's speed between catching up with Yi and catching up with Bing is \(45 \div (1-40\%) = 75\) meters per minute. Jia's initial speed is \(75 \div (1-40\%) = 125\) meters per minute. Jia catches up with Yi at C, and it takes \(t\) minutes to catch up with Bing. When Yi catches up with Bing, Yi also catches up with Jia. At this point, the distance Jia has traveled is \((75t + 45 \times 9)\) meters, and the distance Yi has traveled is \(60 \times (t + 9)\) meters. The equation is: \(75t + 45 \times 9 = 60 \times (t + 9)\), solving for \(t\) gives \(t = 9\). Since they travel another 50 meters to reach D, the distance from C to D is \(75t + 45 \times 9 + 50 = 1130\) meters. Since Jia takes 9 minutes to catch up with Yi at C, the distance Bing has traveled to C is \(75 \times 9 - 45 \times 9 = 270\) meters, which means Jia has traveled from A to C in \(270 \div 45 = 6\) minutes, so the distance from A to C is \(125 \times 6 = 750\) meters. Therefore, the distance from A to D is \(1130 + 750 = 1880\) meters.
[Solution] After catching up with Bing, the speed becomes:
\[
\begin{array}{l}
60 \times (1-25\%) \\
= 60 \times 0.75 \\
= 45 \text{ (meters/minute)}
\end{array}
\]

Jia's speed between catching up with Yi and catching up with Bing is:
\[
\begin{array}{l}
45 \div (1-40\%) \\
= 45 \div 0.6 \\
= 75 \text{ (meters/minute)}
\end{array}
\]

Jia's initial speed is:
\[
\begin{array}{l}
75 \div (1-40\%) \\
= 75 \div 0.6 \\
= 125 \text{ (meters/minute)} \\
75t + 45 \times 9 = 60 \times (t + 9) \\
75t + 405 = 60t + 540 \\
15t = 135 \\
t = 9 \\
\end{array}
\]

Jia catches up with Yi at C, and it takes \(t\) minutes to catch up with Bing:
\[
\begin{aligned}
75t + 45 \times 9 & = 60 \times (t + 9) \\
75t + 405 & = 60t + 540 \\
15t & = 135 \\
t & = 9
\end{aligned}
\]
The distance from C to D is:
\[
\begin{array}{l}
75t + 45 \times 9 + 50 \\
= 75 \times 9 + 405 + 50 \\
= 1130 \text{ (meters)}
\end{array}
\]

Jia takes 9 minutes to catch up with Yi at C, so the distance Bing has traveled to C is:
\[
\begin{array}{l}
75 \times 9 - 45 \times 9 \\
= (75 - 45) \times 9 \\
= 270 \text{ (meters)}
\end{array}
\]

Jia travels from A to C, and Bing travels: \(270 \div 45 = 6\) minutes,
so the distance from A to C is: \(125 \times 6 = 750\) meters,
therefore, the distance from A to D is \(1130 + 750 = 1880\) meters.
Answer: Jia's initial speed is 125 meters per minute, and the distance between A and D is 1880 meters.
The answers are: 125, 1880.