Question 65329
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A plane flies 720 miles against a steady 30 mi/hr headwind and then returns to the same point with the wind. If the entire trip takes 10 hours, what is the plane’s speed in still air?
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Let p be plane speed
Against the wind DATA:
distance= 720 miles ; Rate= p-30 mph ; time = d/r= 720/(p-30) hrs.
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With the wind DATA:
distance =720 miles ; rate = p+30 mph ; time = d/r = 720/(p+30) hrs.
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EQUATION:
time with + time against = 10 hrs
720/(p+30) + 720/(p-30)= 10
Divide thru by 720 to get:
1/(p+30) + 1/(p-30)= 1/72
Multiply thru by (p+30)(p-30) to get:
p-30+p+30=(1/72)(p^2-900)
Simplify the left side to get:
2p=(1/72)(p^2-900)
Multiply both sides by 72 to get:
144p=p^2-900
Rearrange:
p^2-144p-900=0
Use the quadratic formula to get:
p=[144+-sqrt(144^2-4*-900)]/2
p=[144+-156]/2
Select the positive solution to get:
p=[144+156]/2=300/2=150 mph
Cheers,
Stan H.