Question 184013: A plane in calm air travels 150 MPH. In wind, it travels 180 miles, turns around and travel back 180 miles with a total travel time of 2.5 hours. What is the rate of travel for with wind and against wind?
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 of the wind
Then 150+r=plane's rate of travel with the wind
And 150-r=planes rate of travel against the wind
Time travelling with the wind=180/(150+r)
Time travelling against the wind =180/(150-r)
And we are told that the above two times add up to 2.5 hr, so:
180/(150+r)+180/(150-r)=2.5 multiply each term by (150+r)(150-r)
180(150-r)+180(150+r)=2.5(22,500-r^2) divide each term by 2.5
72(150-r)+72(150+r)=22,500-r^2 get rid of parens
10,800-72r+10,800+72r=22,500-r^2 or
21,600-22,500=-r^2
-r^2=-900 multiply each side by -1
r^2=900 take sqrt of each side
r=+or- 30 mi/hr wind speed is positive,so:
r=30 mph-----------------------rate of the wind
Then 150+r=150+30=180 mph-----plane's rate of travel with the wind
And 150-r=150-30=120 mph ----planes rate of travel against the wind
CK
180/180+180/120=2.5
1+1.5=2.5
2.5=2.5
Hope this helps---ptaylor
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