1. W took a trip from Davenport to Omaha, a distance of 300 mi. She traveled part of the way by bus, which arrived at the train station justin in time for W to complete the journey by train. The bus averaged 40mph and the train 60mph. The entire trip took 5 1/2 h. How long did W spend on the train?
2. 2 cyclist, 90mi apart, start riding toward eachother at the same time. Once cycles twice as fast as the other. If they meet 2 h later, at what average speed is each cyclist traveling?
3. A pilot flew a jet from Montreal to LA, a distance of 2500 mi. on the return trip the average speed was 20% faster than the outbound speed. the round trip took 9 h 10min. What was the speed from Montreal to LA?
- Astral WalkerLv 71 decade agoFavorite Answer
1. W traveled B miles by bus @ 40mph and T miles by train @ 60 mph.
B+T = 300 --> the total miles traveled
B/40 + T/60 = 5.5 --> the total time traveled.
120B/40 + 120T/60 = 120*5.5
3B+2T = 660
2B + 2T = 600
B = 60
T = 240
So let's check.
60 miles at 40mph took 60/40 hours = 1.5 hours
240 miles at 60 mph took 240/60 hours = 4 hours
So the total distance is 300 miles and the total time is 5.5 hours.
2. cyclist 1 travels x mph and cyclist 2 travels 2x mph and their average speed is x+2x=3x mph.
If they meet in 2 hours then their average combined speed is 90/2 = 45 mph
3x = 45 --> x=15
so cyclist 1 travels at 15 mph and cyclist 2 traves at 30 mph.
3. 9h 10m = 9 1/6h = 55/6 hours.
the initial speed is x and the return speed is 1.2x so
2500/x + 2500/(1.2x) = 55/6
2500*1.2 + 2500 = 55/6*1.2x
5500 = 11x
x = 550
So the trip to LA was flown at 550 mph.
- Anonymous1 decade ago
just set up two equations and then substitute and solve