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Question 250640: A pilot wants to make an 840-mile round trip from Cleveland to Peoria and back in 5 hours flying time. There will be a headwind of 30 mph going to Peoria and it is estimated that there will be a 40 mph tailwind returning to Cleveland. At what constant engine speed should the plane be flown?
Answer by checkley77(12844)   (Show Source):
You can put this solution on YOUR website! D=RT
840/2=420 MILES EACH TRIP.
420=(R-30)T THE TRIP TO PEORIA.
T=420/(R-30) REPLACE T BY (420/(R-30) IN THE NEXT EQUATION.
420=(R+40)(5-T) THE RETURN TRIP.
420=(R+40)(5-(420/(R-30)))
420=(R+40)(5R-150-420)/(R-30)
420=(R+40)(5R-570)/(R-30)
420(R-30)=(R+40)(5R-570)
420R-12,600=5R^2-370R-22,800
5R^2-370R-420R-22,800+12,600=0
5R^2-790R-10,200=0
5(R^2-158R-2,040)=0
5(R-170)(R+12)=0
R-170=0
R=170 SPEED OF THE PLSNE .
PROOF:
420=(170-30)T
420=140T
T=420/140
T=3 HOURS FOR THE FIRST TRIP.
420=(170+40)(5-3)
420=210*2
420=420
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