SOLUTION: The speed of an airplane in still air is 214mph. The plane travels 819mi against the wind and 1505 mi with the wind in a total time of 12 hrs. what is the speed of the wind?

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Question 65311: The speed of an airplane in still air is 214mph. The plane travels 819mi against the wind and 1505 mi with the wind in a total time of 12 hrs. what is the speed of the wind?
Found 2 solutions by checkley71, stanbon:
Answer by checkley71(8403)   (Show Source): You can put this solution on YOUR website!
12=819/(214-X)+1505/(214+X)
12=[819(214+X)+1505(214-X)]/(214-X)(214+X)
12=(175266+819X+322070-1505X)/(45796-X^2)
12(45796-X^2)=-686X+497336
549552-12X^2+686X-497336=0
12X^2-686X-52216=0
USING THE QUADRATIC EQUATION WE GET X=100.4745 THE SPEED OF THE WIND
PROOF
12=819/(214-100.4745)+1505/(214+100.4745)
12=819/113.5255+1505/314.4745
12=7.2142382+4.7857616
12=12
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MERRY CHRISTMAS

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
The speed of an airplane in still air is 214mph. The plane travels 819mi against the wind and 1505 mi with the wind in a total time of 12 hrs. what is the speed of the wind
---------
Let "w" be the speed of the wind.
------------
Against the wind DATA:
distance=819 miles ; Rate=(214-w); time= d/r=819/(214-w) hrs
-----------------------
With the wind DATA:
distance=1505 miles ; rate=(214+w); time = d/r 1505/(214+w) hrs
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EQUATION:
time + time = 12 hrs
819/(214-w) + 1505/(214+w) = 12
819(214+w) + 1505(214-w)= 12(214^2-w^2)
(2324)(214)+(819-1505)w = 12(45796)-12w^2
12w^2-686w-52216=0
6w^2-343w-26108=0
w=[343+-sqrt(343^2-4*6*-26108]/12
w=[1205.69]/12
w=100.47 mph (speed of the wind)
Cheers,
Stan H.

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