# Determine the speed of an airplane?

An airplane makes a 990 km flight with a trailwind and returns, flying into the same wind. The total flying time is 3 hours 20 minutes, and the airplane's sped in still air is 600 km/h What is the speed of the wind?

5 stars to whoever can answer it with proof. Please.

An airplane makes a 990 km flight with a tailwind and returns, flying into the same wind. The total flying time is 3 hours 20 minutes, and the airplane's speed in still air is 600 km/h. What is the speed of the wind?

### 2 Answers

- Michael MLv 77 years agoFavorite Answer
Let v = speed of wind

speed of plane on the first part of the trip = 600 + v

speed of plane on the return trip = 600 - v

time for the first part of the trip = distance/speed = 990/( 600 + v)

time for the second part of the trip = 990 / (600 - v)

total time for the round trip is

990/( 600 + v) + 990 / (600 - v)

but this has to equal 3 hrs 20 min = 3 1/3 hrs = 10/3 hrs

990/( 600 + v) + 990 / (600 - v) = 10/3

99/( 600 + v) + 99 / (600 - v) = 1/3

1/( 600 + v) + 1 / (600 - v) = 1/ [ (99)(3) ]

(600 - v)/ {(600+v)(600 - v)} + (600 +v) / {(600 +v)(600 - v} = 1/297

1200/ { ( 600 + v)(600 - v) } = 1/297

(600 + v)( 600 - v) / 1200 = 297

(600 + v) ( 600 -v) = (297)(1200)

600^2 - v^2 = 356400

v^2 = 600^2 - 356400 = 3600

v = 60 km/hr

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- chenierLv 43 years ago
i'm no longer certain what you recommend through S seventy 8 E. Is that seventy 8 ranges east of south or seventy 8 ranges south of east? i will assume the former. If i have made the incorrect assumption, this may nonetheless inform you a thanks to figure it out. First, wreck each and each and every route right into a cardinal ingredient utilizing trigonometry. Then upload those to locate the route of the resultant vector. Use the Pythagorean theorem to locate the importance of the vector. 480 mph seventy 8 ranges east of south 480 * sin(seventy 8) = 469.fifty one mph east 480 * cos(seventy 8) = ninety 9.80 mph south fifty 8 mph 40 5 ranges west of south fifty 8 * sin(40 5) = 40-one.01 mph west fifty 8 * cos(40 5) = 40-one.01 mph south upload the souths and subtract the west from the east and also you get. 469.fifty one mph east - 40-one.01 mph west = 428.50 mph east ninety 9.80 mph south + 40-one.01 mph south = 100 and forty.80 one mph south The importance of the vector is sqrt(428.50^2 + 100 and forty.80 one^2) = 451.04 mph The heading is arctan(428.50/100 and forty.80 one) = seventy one.80 one ranges east of south

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