SOLUTION: A freight train leaves Kansas and travels due west. A passenger train departs at exactly the same time on a parallel track and travels due east at a speed that is twice the speed

Question 29256: A freight train leaves Kansas and travels due west. A passenger train departs at exactly the same time on a parallel track and travels due east at a speed that is twice the speed of the freight train. After 7 hours the trains are 945 miles apart. How fast is each train traveling?
Found 2 solutions by Paul, venugopalramana:
Answer by Paul(988) (Show Source): You can put this solution on YOUR website!
Passegner's train speed = 2x
Fireight train's speed =x

Equation:
7(x+2x)=945
21x=945
x=45

2(45)=90

Hence, the speed of the freight train is 45mph and the speed of the passenger train is 90mph.
Paul.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
A freight train leaves Kansas and travels due west. A passenger train departs at exactly the same time on a parallel track and travels due east at a speed that is twice the speed of the freight train. After 7 hours the trains are 945 miles apart. How fast is each train traveling?
TRAINS ARE TRAVELLING IN OPPOSITE DIRECTIONS..SO THEIR RELATVE SPEED =SUM OF THEIR INDIVIDUAL SPEEDS
SPEED OF ONE TRAIN IS X MPH....SAY..SECOND TRAIN SPEED =2X..MPH....SO
RELATIVE SPEED =X+2X=3X...MPH
TIME =7 HRS
INTIAL DISTANCE BETWEEN THEM =0
FINAL DOISTANCE BETWEEN THEM =945 MILES
SO DISTANCE CHANGEW =945-0=945 MILES..HENCE
945 =7*3X=21X
X=945/21=45 MPH
SO THE 2 TRAINS TRAVELLED AT 45 AND 90 MPH.
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