Question 140884: Two planes start from LA Internation airport and fly in opposite directions. The second plane starts 30 minutes after the first plane, but is 80 km per hour faster. Find the airspeed of each plane if 2 hours after the 1st plane departs the plane are 3200 km apart.
I went about setting up the equation:
x=speed of p1
x=speed of p2
2x-(3/2y)=3200,
but don't know what to do next... your help would be appreciated!! thanks.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Two planes start from LA Internation airport and fly in opposite directions. The second plane starts 30 minutes after the first plane, but is 80 km per hour faster. Find the airspeed of each plane if 2 hours after the 1st plane departs the plane are 3200 km apart.
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1st plane DATA:
distance = x km ; rate = r kph ; time = x/r hrs
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2nd plane DATA:
distance = 3200-x km ; rate = r+80 kph ; time = (3200-x)/(r+80)
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EQUATIONS:
1st plane time = x/r = 2 hrs; x = 2r
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2nd plane time = (3200-x)/(r+80) = 3/2 hr
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Substitute x=2r into the 2nd equation to get:
(3200-2r)/(r+80) = 3/2
Cross-multiply to get:
6400-4r = 3r+240
7r = 6160
r = 880 kph (speed of the 1st plane)
r+80 = 960 kph (speed of the 2nd plane)
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Cheers,
Stan H.
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