SOLUTION: An airplane flies between two cities 450 kilometers apart, traveling with a wind of 45bmiles per hour going, and against the same wind when returning. The trip out takes one less h

Question 88582: An airplane flies between two cities 450 kilometers apart, traveling with a wind of 45bmiles per hour going, and against the same wind when returning. The trip out takes one less hour that the return trip. What is the speed of the plan in still air?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Distance(d)=Rate(r)times Time(t) or d=rt; r=d/t and t=d/r
let r=rate (speed) of plane in still air
rate out with wind=r+45
rate returning against wind=r-45
also
let t=time it takes returning against the wind
then t-1=time it takes going out with the wind
Now we know that:
time going out with the wind=time returning against the wind minus 1 hour
So----
450/(r+45)=(450/(r-45))-1 multiply both sides by (r+45)(r-45) to get rid of fractions
450(r-45)=450(r+45)-(r+45)(r-45) get rid of parens
450r-20250=450r+20250-r^2+2025 subtract 20250, 2025 and also 450r from both sides
450r-450r-20250-20250-2025=450r-450r+20250-20250-r^2+2025-2025 collect like terms
-42525=-r^2 0r
r^2=42525 take square root of both sides
r= plus or minus 206.2
r=206.2 mph-------------------------speed in still air
CK
time out=450/(206.2+45)=450/(251.2)=1.791 hours
time returning=450(206.2-45)=450/(161.2)=2.791 hr
2.791-1.791=1 hour

Hope this helps----ptaylor

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