SOLUTION: a university flew its football team round trip for a game with a rival uiversty. the distance was 520 miles each way, and the round trip required a total flying time of 8.8 hours.

Question 183141: a university flew its football team round trip for a game with a rival uiversty. the distance was 520 miles each way, and the round trip required a total flying time of 8.8 hours. because of wind, the average speed of the plane in one directon was 30 miles per hour faster than it was in the opposite direction.what would be the speed of the plane in still air?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let s=speed of plane in still air
And let r=speed of the wind
s+r=plane speed with the wind
s-r=plane speed against the wind. We are told that:
s+r=(s-r)+30 subtract s from and add r to each side
2r=30
r=15 mph speed of wind
time against the wind=520/(s-15)
time with the wind=520/(s+15)
And we are told that time with wind + time against wind =8.8 hr, so:
520/(s-15) + 520/(s+15)=8.8 multiply each term by (s-15)(s+15)
520(s+15)+520(s-15)=8.8(s-15)(s+15) get rid of parens
520s+7800+520s-7800=8.8s^2-1980 simplify
1040s-8.8s^2+1980=0 or
8.8s^2-1040s-1980=0----------------quadratic in standard form. Solve using the quadratic formula:

Disregard negative value for s, speed is positive

mph---------------------speed of plane in still air
CK
520/(120.1+15) +520/(120.1-15)=8.8
3.849+4.95=8.8
~~~~8.8=8.8

Hope this helps---ptaylor

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