SOLUTION: A pilot flew his​ single-engine airplane 90 miles with the wind from City A to above City B. He then turned around and flew back to City A against the wind. If the wind was a
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-> SOLUTION: A pilot flew his​ single-engine airplane 90 miles with the wind from City A to above City B. He then turned around and flew back to City A against the wind. If the wind was a
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Question 1102561: A pilot flew his single-engine airplane 90 miles with the wind from City A to above City B. He then turned around and flew back to City A against the wind. If the wind was a constant 20 miles per hour, and the total time going and returning was 1.3 hours, find the speed of the plane in still air. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! x=speed of plane in still air, t= time in hours
(x+20)*t=90, t1=90/(x+20)
trip back
(x-20)*t=90, t2=90/(x-20)
but t1+t2=1.3
Therefore 90/(x+20)+90/(x-20)=1.3
Multiply both sides by (x+20)(x-20), or x^2-400
90(x-20)+90(x+20)=1.3x^2-520
90x-1800+90x+1800=1.3x^2-520
1.3x^2-180x-520=0
x=(1/2.6)(180+/- sqrt (32400+2704); sqrt term = 187.36
positive root is (367.36/2.6)=141.29 mph ANSWER
Time over is 90 miles at 161.29 mph or 0.558 hours
Time back is 90 miles at 121.29 mph or 0.742 hours. They add to 1.3 hours.
