Car A and Car B start from locations A and B simultaneously (Car A starts from A), heading towards each other. The speeds of Car A and Car B are 40 km/h and 50 km/h, respectively, and the distance between A and B is 900 km. When Car A reaches B, it immediately turns around and heads back to A at a speed of 50 km/h. When Car B reaches A, it immediately turns around and heads back to B at a speed of 40 km/h. When Car A reaches A, it immediately turns around and heads back to B, restoring its speed to 40 km/h. When Car B reaches B, it immediately turns around and heads back to A, restoring its speed to 50 km/h. This process continues, with the two cars shuttling back and forth between A and B, and their speeds switching between 40 km/h and 50 km/h. When the two cars meet for the 2016th time, how many kilometers has Car A traveled?
Jia travels from $A$ to $B$ at a speed of 40 km/h, and then returns from $B$ to $A$ at a speed of 50 km/h.
Yi travels from $B$ to $A$ at a speed of 50 km/h, and then returns from $A$ to $B$ at a speed of 40 km/h.
Therefore, Jia and Yi each take the same amount of time to complete one round trip.
During the time Jia and Yi each complete one round trip, they meet 2 times.
$2016 \div 2=1008$, if Jia and Yi each complete 1008 round trips, they meet 2016 times.
From the 2016th meeting to the final return of Jia and Yi to their respective starting points A and B,
Jia's speed is 50 km/h, and Yi's speed is 40 km/h.
Thus, the speed ratio of Jia to Yi is $50: 40=5: 4$.
The distance ratio of Jia to Yi is also $5: 4$.
Among them, Jia travels $900 \times \frac{5}{5+4}=500 \mathrm{~km}$.
Therefore, when the two vehicles meet for the 2016th time, the distance traveled by Jia is equivalent to 1008 round trips minus these 500 kilometers, which is $900 \times (1008 \times 2)-500=1813900 \mathrm{~km}$.