5. (2000 World Inter-City League) A bus, starting a 100-kilometer journey at 12:20 PM, has a computer that at 1:00 PM, 2:00 PM, 3:00 PM, 4:00 PM, 5:00 PM, and 6:00 PM, says: “If the average speed from now on is the same as the average speed so far, then it will take one more hour to reach the destination.” Is this possible? If so, how far has the bus traveled by 6:00 PM?
5. The computer is correct.

At $k: 00$ PM, where $1 \leqslant k \leqslant 6$, the distance the car has traveled is $\frac{60 k-20}{60 k+40}$ of the total distance, and the average speed of the car up to this point is $\sqrt{\frac{1}{60 k+40}}$. If it continues to travel the remaining $\frac{60}{60 k+40}$ at the same speed, it will take exactly one hour to reach the destination. Note that between $(k-1): 00$ PM and $k: 00$ PM, the distance traveled is
$$
\frac{60 k-20}{60 k+40}-\frac{60 k-80}{60 k-20}=\frac{3600}{(60 k+40)(60 k-20)},
$$

and these numbers are all determined. At $6: 00$ PM, the distance traveled is $\frac{340}{400}=\frac{17}{20}$, which means the car has traveled 85 kilometers.