Question 167614: Two planes start from the same airport and fly in opposite directions. The second plane starts 45 minutes after the first plane, but travels 50mph faster than the first plane. Find the speed of each plane if, 2 hours and 15 minutes after the first plane takes off, the planes are 2325 miles apart.
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! rate * time = distance
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plane 1 travels x miles per hour for 2.25 hours.
rate for plane 1 = x miles per hour.
since rate * time = distance,
distance for plane 1 = 2.25*x = d1.
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plane 2 travels (x+50) miles per hour for 1.5 hours.
rate for plane 2 = (x+50) miles per hour.
since rate * time = distance,
distance for plane 2 = 1.5*(x+50) = d2.
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since they are traveling in opposite directions, distance between them is d1 + d2.
total distance between them is given as 2325 miles.
therefore:
d1 + d2 = 2325.
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since d1 = 2.25*x
and
since d2 = 1.5*(x+50)
then
d1 + d2 = 2325 = 2.25*x + 1.5*(x+50)
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equation to solve is:
2.25*x + 1.5*(x+50) = 2325
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simplifying, this equation becomes:
2.25*x + 1.5*x + 1.5*(50) = 2325
simplifying further and combining like terms:
3.75*x + 75 = 2325
subtracting 75 from both sides of equation:
3.75*x = 2250
dividing both sides of equation by 3.75:
x = 600
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since d1 = 2.25*x, then
d1 = 2.25*(600) = 1350 miles.
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since d2 = 1.5(x+50), then
d2 = 1.5 * (650) = 975 miles.
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1350 miles + 975 miles = 2325 miles.
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plane 1 travels at 600 miles per hour.
plane 2 travels at 650 miles per hour.
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