anonymous
  • anonymous
I have a distance problem and I know the equation is d=rt. I'm just not sure how to solve the problem: A flight crew flew 420 km in 3 h with a tailwind. Flying against the wind, the flight crew flew 440 km in 4 h. Find the rate of the flight crew in calm air and the rate of the wind.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
the total distance traveled divide by total amount time should give you an average of how much traveled per hour
anonymous
  • anonymous
so it would be 420/3? and 440/4 to get the rate?
anonymous
  • anonymous
since you are looking for a rate of travel in calm air, i think (420+440)/(3+4) should do it

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anonymous
  • anonymous
ok Thank you so much!!
anonymous
  • anonymous
How would I find the rate of the wind?
anonymous
  • anonymous
is the rate of the wind considered same for both cases?
anonymous
  • anonymous
I guess it would be....the answer I get for the first part is 122.8 km/h. Do I use that answer to determine the rate of the wind?
anonymous
  • anonymous
one way to find out is - since you have the rate of travel at calm wind, you can subtract rate of travel at either tail wind or against the wind from calm wind
anonymous
  • anonymous
wind=15 km/hr, aircraft speed = 125 km/hr
anonymous
  • anonymous
i don't think you can just add the rate of travel since the time at which air plane travels is different
anonymous
  • anonymous
i think the rate of wind might be different for both cases
anonymous
  • anonymous
@robtobey, how did you arrive at 125 km/h = aircraft speed and 15 km/hr = wind? Was the original equation wrong or was mymath wrong? :)
anonymous
  • anonymous
msayer3, There is no indication from the problem statement that the wind velocity changed from my point of view.
anonymous
  • anonymous
yeah, that's what i'm wondering.
anonymous
  • anonymous
I will post the solution in a few minutes.
anonymous
  • anonymous
ok...thank yo very much
anonymous
  • anonymous
r*t = d r = d/t Let x = the wind speed First flight leg: \[r+x==\frac{d}{t}\text{/.} \{d\to 420,t\to 3\} \]r + x = 140 Second flight leg:\[r-x=\frac{d}{t}\text{/.} \{d\to 440,t\to 4\} \]r - x = 110 2 r = 140 + 110 r = 125 r + x =140 125 + x = 140 x = 15
anonymous
  • anonymous
I use Mathematica for the calculations /. means sort or, "replace the symbols as follows"
anonymous
  • anonymous
Ok...I think I got it now. Thank you again for your help.
anonymous
  • anonymous
Glad I could help you out.

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