Question 64140This question is from textbook Beginning Algebra
: A plane flies 720 miles against a steady headwind and then returns to the same point with the wind. If the entire trip takes 10 hours what is the planes speed in still air?
I have attempted to solve this by using the formula:
720/(x+30)=720/(x-30) and needless to say do not get the correct answer. The 10 hours is throwing me off. I think the answer is 144mph but cannot get the formula to produce the result. What am I missing? Thanks.
This question is from textbook Beginning Algebra
Answer by josmiceli(19441)   (Show Source):
You can put this solution on YOUR website! The plane travels 2*720 mi or 1440 mi in 10 hours
d / t = r , so 1440 / 10 = 144 m i/hr is the planes
speed in still air.
Whatever the windspeed is, it's effect in one direction
will cancel it's effect in the opposite direction
because the plane goes the same distance with
and against the wind.
Actually, I believe this problem, as stated, is physically
impossible. The plane can't reverse direction without
slowing down and maintain it's speed in the opposite di-
rection.
Even if you change the problem and say that the plane keeps
going in the same direction for 1440 mi and it is only the
windspeed that reverses diretion at the halfway point,
any value that you select for the windspeed (other than 0)
will make the planes speed in still air turn out to be
greater than 144 mi/hr.
w = windspeed
p =planes speed instill air
The equation I end up with is
that's using
The only way you can get p = 144 is if w = 0
If I say w = 18 mi/hr, I get p = 146.21 mi/hr
If I say w = 9 mi/hr, I get p = 144.5 mi/hr
There is a simple answer to this, but you just couldn't
do it in real life. A lot of problems are like that
and my answer is- if a simple answer is what they
want, give it to them, but watch out for the
"trick" questions when they want you to think a lot.
I could be all wrong about this, but I don't think so.
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