SOLUTION: At what air speed must a plane be flown to complete a trip between two airports 420 miles apart in 5 hours if the flight going has a head wind (blowing against the nose of the plan
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Question 885699: At what air speed must a plane be flown to complete a trip between two airports 420 miles apart in 5 hours if the flight going has a head wind (blowing against the nose of the plane) of 40 mph and the flight returning has a tail wind of 30 mph? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! At what air speed must a plane be flown to complete a trip between two airports 420 miles apart in 5 hours if the flight going has a head wind (blowing against the nose of the plane) of 40 mph and the flight returning has a tail wind of 30 mph?
:
let s = air speed required for this scenario
then
(s-40) = ground speed against the wind
and
(s+30) = ground speed with the wind
:
Write a time equation; time = dist/speed + = 5
multiply by (s-40)(s+30)
(s-40)(s+30)* + (s-40)(s+30)* = 5(s-40)(s+30)
:
Cancel the denominators, FOIL the right side, resulting in:
420(s+30) + 420(s-40) 5(s^2+30s-40s-1200)
:
420s + 12600 + 420s - 16800 5(s^2 - 10s - 1200)
:
840s - 4200 = 5s^2 - 50s - 6000
0 = 5s^2 - 50s - 840s - 6000 + 4200
0 = 5s^2 - 890s - 1800
simplify divide by 5
s^2 - 178s - 360 = 0
You can us the quadratic formula here but this will factor to
(s-180) (s+2) = 0
the positive solution
s = 180 mph is the air speed of the aircraft
:
:
Check this by finding the flight time for each way
420/140 = 3 hr (40 mph head wind)
420/210 = 2 hr (30 mph tail wind)
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total time 5 hrs
