SOLUTION: A train leaves the station at 6:00pm traveling west at 80mi/hr on a parallel track, a second train leaves the station 3 hrs later traveling west at 100mi/hr. At what time will the
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: A train leaves the station at 6:00pm traveling west at 80mi/hr on a parallel track, a second train leaves the station 3 hrs later traveling west at 100mi/hr. At what time will the
Log On
Question 100093: A train leaves the station at 6:00pm traveling west at 80mi/hr on a parallel track, a second train leaves the station 3 hrs later traveling west at 100mi/hr. At what time will the second train catch up with the first? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let t=amount of time it takes second train to catch up with the first after the second train leaves at 9:00pm
time the second train catches up with the first=9:00pm+t
distance(d)=rate(r) times time(t) or d=rt;t=d/r and r=d/t
When the second train leaves, the first train has already travelled 80*3 or 240 mi
Distance first train travels =240+80t
Distance second train travels=100t
Now we know that the second train will have caught up with the first when the distance the second train travels equals the distance the first train travels. So our equation to solve is:
100t=240+80t subtract 80t from both sides
100t-80t=240 collect like terms
20t=240 divide both sides by 20
t=12 hrs------amount of time it takes the second train to catch up
time the second train catches up =9:00pm + 12 hrs=9:00am the next day
CK
80*12+240=100*12
1200=1200
Hope this helps---ptaylor
