Question 100783: A train leaves the station at 6:00 P.M. traveling west at 80 mi/h. On a parallel track, a second train leaves the station 3 hours later traveling west a 100 mi/h. At what time will the second train catch up with the first?
A passenger train's speed is 60 mi/h, and a freight train's speed is 40 mi/h. The passenger train travels the same distance in 1.5 h less time than the freight train. How long does each train take to make the trip?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A train leaves the station at 6:00 P.M. traveling west at 80 mi/h. On a parallel track, a second train leaves the station 3 hours later traveling west a 100 mi/h. At what time will the second train catch up with the first?
:
Let t = traveling time of the 1st train, when the 2nd train catches up
Then
(t-3) = traveling time of the the 2nd train.
:
When this occurs, both trains will have traveled the same distance;
Write a distance equation: Dist = speed * time
:
100(t-3) = 80t
:
100t - 300 = 8t
:
100t- 80t = +300
:
20t = 300
:
t = 300/20
:
t = 15 hrs
:
Time = 6 PM + 15 hrs = 9 AM the next day
:
Check solution by find the distance traveled by each train:
80 * 15 = q1200 mi
100 * (15-3) = 1200 mi
:
:
A passenger train's speed is 60 mi/h, and a freight train's speed is 40 mi/h. The passenger train travels the same distance in 1.5 h less time than the freight train. How long does each train take to make the trip?
:
Use another distance equation:
Let t = pass train time
then
(t+1.5) = freight train time
:
60t = 40(t+1.5)
60t = 40t + 60
60t - 40t = 60
t = 60/20
t = 3 hrs for the pass train, and 4.5 hrs for the freight
:
Check solution see if the distance is the same:
60 * 3 = 180 mi
40 * 4.5 = 180 mi
|
|
|
