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

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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

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For 1st train:
(a)
For 2nd train:
(b)
given:
(c) mi
(d) mi/hr
(e) mi/hr
(f) hrs
-------------------------
Subtract from both sides of (c),

Now rewrite (a) and (b)
For 1st train:
(a)
For 2nd train:
(b)
I have 2 equations and 2 unknowns, so it's solvable
(b)
(b)
(b)
Now add (a) and (b)
(a)
(b)


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


hrs answer
check answer:
(a)
(a)
(a)
and
(b)
(b)
(b)
(c) mi
(c)
OK