SOLUTION: at noon a train leaves washington d.c headed for charleston south carolina a distance of 500 mi. the train travels at a speed of 60 mph. at 1 p.m. a second train leaves charleston

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: at noon a train leaves washington d.c headed for charleston south carolina a distance of 500 mi. the train travels at a speed of 60 mph. at 1 p.m. a second train leaves charleston      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 182489: at noon a train leaves washington d.c headed for charleston south carolina a distance of 500 mi. the train travels at a speed of 60 mph. at 1 p.m. a second train leaves charleston headed for washington d.c. traveling at 50 mph. how long after the train leaves charleston will the two trains pass each other
Found 2 solutions by solver91311, josmiceli:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!

Let be the distance traveled by the southbound train when they met, and be the distance traveled by the northbound train when they met. Let be the time the northbound train was moving when they met, then, since the southbound train left an hour earlier, it must have been moving for hours.

Since we know that distance equals rate times time, for the southbound train we can say that the distance traveled by the southbound train measured from Washington, DC to the meeting point is:




And the distance traveled by the northbound train is:



But we know that

, so:











So the train from Charleston traveled 4 hours to the meeting point.

Check:







Answer checks.




John


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
For 1st train:
(a) d%5B1%5D+=+r%5B1%5D%2At%5B1%5D
For 2nd train:
(b) d%5B2%5D+=+r%5B2%5D%2At%5B2%5D
given:
(c) d%5B1%5D+%2B+d%5B2%5D+=+500 mi
(d) r%5B1%5D+=+60 mi/hr
(e) r%5B2%5D+=+50 mi/hr
(f) t%5B2%5D+=+t%5B1%5D+-+1 hrs
-------------------------
Subtract d%5B1%5D from both sides of (c),
d%5B2%5D+=+500+-+d%5B1%5D
Now rewrite (a) and (b)
For 1st train:
(a) d%5B1%5D+=+60%2At%5B1%5D
For 2nd train:
(b) 500+-+d%5B1%5D+=+50%2A%28t%5B1%5D+-+1%29
I have 2 equations and 2 unknowns, so it's solvable
(b) 500+-+d%5B1%5D+=+50%2A%28t%5B1%5D+-+1%29
(b) -d%5B1%5D+=+50t%5B1%5D+-+50+-+500
(b) -d%5B1%5D+=+50t%5B1%5D+-+550
Now add (a) and (b)
(a) d%5B1%5D+=+60%2At%5B1%5D
(b) -d%5B1%5D+=+50t%5B1%5D+-+550
0+=+60t%5B1%5D+%2B+50t%5B1%5D+-+550
110t%5B1%5D+=+550
t%5B1%5D+=+5 hrs
This is the time it took the DC to Charlston train to meet the other
The problem wants the time for the Charlston to DC train
t%5B2%5D+=+t%5B1%5D+-+1
t%5B2%5D+=+5+-+1
t%5B2%5D+=+4 hrs answer
check answer:
(a) d%5B1%5D+=+r%5B1%5D%2At%5B1%5D
(a) d%5B1%5D+=+60%2A5
(a) d%5B1%5D+=+300
and
(b) d%5B2%5D+=+r%5B2%5D%2At%5B2%5D
(b) d%5B2%5D+=+50%2A4
(b) d%5B2%5D+=+200
(c) d%5B1%5D+%2B+d%5B2%5D+=+500 mi
(c) 300+%2B+200+=+500
OK