Question 120519:  A plane flies 720 mi against a steady 30-mi/h headwind and then returns to the same point with the wind. If the entire trip takes 10 h, what is the plane’s speed in still air? 
 Found 2 solutions by  stanbon, solver91311: Answer by stanbon(75887)      (Show Source): 
You can  put this solution on YOUR website! A plane flies 720 miles against a steady 30 mph headwind and then returns to the same point with the wind, if the entire trip takes 10 hours what is the plane's speed in still air? 
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Let "p" be the plane's speed in still air. 
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Upwind DATA; 
Distance = 720 miles ; Rate = p-30; time = d/r = 720/(p-30) hrs. 
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Downwind DATa: 
Distance = 720 miles ; Rate = p+30: time = d/r = 720/(p+30) hrs 
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EQUATION: 
time up + time down = 10 hrs 
720/(p-30) + 720/(p+30) = 10 
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Divide thru by 10 to get: 
72/(p-30) + 72/(p+30) = 1 
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Multiply thru by (p-30)(p+30)=p^2-900 
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72(p+30) + 72(p-30) = p^2-900 
144p = p^2-900 
p^2-144p-900= 0 
(p-150)(p+6) = 0 
p = 150 mph (plane speed in still air is 150 mph) 
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Cheers, 
Stan H. 
 
 
 Answer by solver91311(24713)      (Show Source): 
You can  put this solution on YOUR website! Let r be the speed in still air, in accordance with rule one for word problems: Set the variable equal to the thing you want to know.
 
 
Air speed against the wind would then be   and airspeed with the wind would be  .  If the against the wind part of the trip takes t hours, then the return trip must take 10 - t hours, because the entire trip took 10 hours.
 
 
The formula for straight-line distance in terms of constant rate over a period of time is  .  So we can express time as a function of distance and rate by saying  .
 
 
In our case, the time for the outbound trip, t, is given by  .  The time for the return trip, 10 - t, is given by  
 
 
First solve this second equation for t:
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Now, note that we have two expressions for t in terms of r.  Set them equal to each other:
 
 
 
 
 
Multiply by  
 
 
 
 
 
Distribute and collect terms
 
 
  
 
 
 
Divide by 10
 
 
 
 
 
Note that   and  , so
 
 
  =>   or  
 
 
The solution set to the quadratic is therefore 150 or -6.  Since flying backwards doesn't make much sense, exclude the -6 value as an extraneous root.  The speed in still air is 150 miles per hour.
 
 
Check:
 
  and  , and finally, 6 + 4 = 10.  Answer checks.
 
 
Hope that helps, 
John 
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