We let the distance function $x(t)=100\cdot\frac{40+t}{100+t}$ where $t$ the number of minutes since the bus started calculating, 40 minutes in. Suppose that during the first 40 minutes, the bus does whatever it wants to get from 0 km to 40 km, then follows this position function for the rest of the (indefinitely long) trip. Note that $x(t)=100-\frac{6000}{100+t}$ which is increasing, but less than 100. Therefore the bus is moving forward the whole time. The average speed is $\frac{x(t)}{t+40}=\frac{100}{100+t}$ because the bus has traveled $x(t)$ km in $t+40$ minutes. Now note that from the rewritten statement of $x(t)$ above, the total distance left to travel is $\frac{6000}{100+t}$ because the bus has already traveled 100km minus this distance. So the estimated time to travel this far at the given average speed is $\frac{\left(\frac{6000}{100+t}\right)}{\left(\frac{100}{100+t}\right)}=60$ minutes. Therefore, the computer will always say 1 hour remaining. In those 5 hours, the bus traveled $x(300)-x(0)=100\cdot\frac{340}{400}-100\cdot\frac{40}{100}=85-40=45$ kilometers.