Question 115488This question is from textbook Intermediate Algebra
: A plane filies 465 miles with the wind and 345 miles against the wind in the same length of time. If the speed of the wind is 20mph, find the speed of the plane in still air.
I am not sure where to begin. I have tried subtracting the wind speed and adding it, but I don not know where to go from there.
This question is from textbook Intermediate Algebra
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 r=rate (speed) of the plane in still air
(Note: with the wind, add wind speed; against the wind, subtract wind speed)
time flying with the wind=465/(r+20)
time flying against the wind=345/(r-20)
And we are told that the above times are equal, so:
465/(r+20)=345/(r-20) multiply each side by (r+20)(r-20){or cross-multiply}
465(r-20)=345(r+20) get rid of parens (distributive law)
465r-9300=345r+6900 subtract 345r from and add 9300 to both sides
465r-9300+9300-345r=345r-345r+6900+9300 collect like terms
120r=16200 divide both sides by 120
r=135mph--------------------------------speed of the plane in still air
CK
465/(135+20)=345/(135-20)
465/155=345/115
3=3
Hope this helps---ptaylor
|
|
|
