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 761847: 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?
I do not know how to solve this algebraically and have only solved it using guess and check. Ive tried setting it up multiple different ways and guessing with some algebra but keep getting crazy answers that never check out to be right. Can anyone help? I really really appreciate all your work!!!! -Abby
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 = the plane's air speed
then
(s-40 = ground speed outward
and
(s+30) = ground speed back
:
Write a time equation: time = dist/speed
:
outbound time + return time = 5 hrs
+ = 5
Multiply equation by (s-40)(s+30) to clear the denominators, results:
420(s+30) + 420(s-40) = 5(s-40)(s+30)
420s + 12600 + 420s - 16800 = 5(s^2 + 30s - 40s - 1200)
840s - 4200 = 5(s^2 - 10s - 1200)
divide both sides by 5
168s - 840 = s^2 - 10s - 1200
Combine to form a quadratic equation
s^2 - 10s - 168s - 1200 + 840 = 0
s^2 - 178s - 360 = 0
Factors to
(s-180)(s+2) = 0
the positive solution all we want here
s = 180 mph is the air speed
:
:
Check this by finding the times each way
420/140 = 3 hrs, subtracted 40 mph
420/210 = 2 hrs, added 30 mph
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total time 5 hrs