SOLUTION: one train leaves a station heading due west. two hours later a second train leaves the same station heading due east. the second train is traveling 15 mi/h faster than the first. s

Algebra ->  Systems-of-equations -> SOLUTION: one train leaves a station heading due west. two hours later a second train leaves the same station heading due east. the second train is traveling 15 mi/h faster than the first. s      Log On


   

Question 5005: one train leaves a station heading due west. two hours later a second train leaves the same station heading due east. the second train is traveling 15 mi/h faster than the first. six hours after the second train leaves, the two trains are 580 miles apart. find the rate at which each train is traveling.
Answer by xcentaur(357) About Me  (Show Source):
You can put this solution on YOUR website!
1st train going west has speed x mph
second train going east has speed (x+15)mph


distance covered by first train in 2 hours=s*t
speed=x mph,time=2 hours
distance=(2x)miles


after 2 hours,the second train leaves


it is given that six hours after the second train leaves,the two trains were 580 miles apart.
that means that 8 hours after the first train leaves,the two trains were 580 miles apart.


distance travelled by first train due west in 8 hours=s*t
=x*8
=(8x) miles


distance travelled by secnd train going east in 6 hours=s*t
=(x+15)*6


Then total distance between the trains is distance travelled by first train added to distance travelled by second train
ie.
8x+6(x+15)=580
8x+6x+90=580
14x=580-90
14x=490
x=35


Hence the first train was travelling at 35 mph
while the second train was traveliing at (35+15)=50 mph


Hope this helps,
good luck.