SOLUTION: 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?

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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) About Me  (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) About Me  (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 r-30 and airspeed with the wind would be r%2B30. 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 d=rt. So we can express time as a function of distance and rate by saying t=d%2Fr.

In our case, the time for the outbound trip, t, is given by t=720%2F%28r-30%29. The time for the return trip, 10 - t, is given by 10-t=720%2F%28r%2B30%29

First solve this second equation for t:

-t=%28720%2F%28r%2B30%29%29-10

t=10-%28720%2F%28r%2B30%29%29

t=%2810%28r%2B30%29-720%29%2F%28r%2B30%29

t=%2810r%2B300-720%29%2F%28r%2B30%29

t=%2810r-420%29%2F%28r%2B30%29

Now, note that we have two expressions for t in terms of r. Set them equal to each other:

720%2F%28r-30%29=%2810r-420%29%2F%28r%2B30%29

Multiply by %28r%2B30%29%28r-30%29

720%28r%2B30%29=%2810r-420%29%28r-30%29

Distribute and collect terms

720r%2B21600=10r%5E2-300r-420r%2B12600
10r%5E2-1440r-9000=0

Divide by 10

r%5E2-144r-900=0

Note that -150%2A6=-900 and -150%2B6=-144, so

%28r-150%29%28r%2B6%29=0 => r=150 or r=-6

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:
720%2F%28150-30%29=720%2F120=6 and 720%2F%28150%2B30%29=720%2F180=4, and finally, 6 + 4 = 10. Answer checks.

Hope that helps,
John