There is a small station every 10 kilometers on the railway line from city a to city B. one slow train runs from city a to city B at the speed of 45 kilometers per hour at 9:00 a.m., and the other fast train runs from city a to city B at the speed of 60 kilometers per hour at 9:30 a.m How far should the train stop at the small station from city a at the latest to let the express pass? At least how long does the local train take? (the length of fast and slow trains is ignored)

There is a small station every 10 kilometers on the railway line from city a to city B. one slow train runs from city a to city B at the speed of 45 kilometers per hour at 9:00 a.m., and the other fast train runs from city a to city B at the speed of 60 kilometers per hour at 9:30 a.m How far should the train stop at the small station from city a at the latest to let the express pass? At least how long does the local train take? (the length of fast and slow trains is ignored)

(1) At this time, the local bus leaves: (45 × 12-8) / (60-45) × 60 + 8, = (452-8) / (15 × 60 + 8, = 292 ／ 15 × 60 + 8, = 58 + 8, = 66 (km); it can't stop at the station of 70km, because the distance between the two buses is less than 8km, so the local bus stops at the station 60km away from city A. A: this local bus should stop at the station 60km away from city a at the latest (2) when the slow train stops at the station, the fast train runs: 6045-12 = 56 hours = 50 (minutes); when the fast train reaches 60 + 8 = 68 km, the slow train can leave the station. When the fast train reaches 68 km, it needs 68 △ 60 = 6860 hours = 68 (minutes); then the slow train stops at the station: 68-50 = 18 (minutes). A: the slow train needs to wait at least 18 minutes

The car has been delayed for 6 minutes in the midway, and then the speed has been increased from 40 km / h to 50 km / h
How many kilometers do you need to walk like this

6 minutes delay: 6 / 60 * 40 = 4 km
Time required to make up for the delay: 4 / (50-40) = 0.4 hours
During walking: 0.4 * 50 = 20 km