Solving Vector Problems: Finding an Airplane's Ground Speed and Direction

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Homework Help Overview

The discussion revolves around a vector problem involving an airplane's ground speed and direction, given its airspeed and the influence of wind. The subject area includes vector addition and trigonometry.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using the law of cosines and vector diagrams to analyze the problem. There are suggestions to break down the velocity vectors into their components and to visualize the scenario on a Cartesian plane.

Discussion Status

Some participants have provided guidance on using vector components for addition, while others have shared their calculations and results. There is a mix of confirmations and checks on the accuracy of the computed values, indicating an ongoing exploration of the problem.

Contextual Notes

Participants mention the use of a website for checking answers, which introduces an element of verification in their calculations. There are also references to potential typos in the results shared.

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An airplane flies due north at 220 km/h relative to the air. There is a wind blowing at 55 km/h to the northeast relative to the ground. What are the plane's speed and direction relative to the ground?

I started by using the law of Cosines.
a^2 = b^2 + c^2 - 2bccos(a)
That's the only thing I could think to try.
Help Please
 
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Have you tried drawing a vector diagram?
 
first, make a diagram of the thing on the cartesian plane...assume the wind is blowing 45 degrees east of north...draw in the magnitudes(the speed). figure out the other leg when you drop it down to the x axis...if you know what i am talking about. see where that takes you
 
ideasrule said:
Have you tried drawing a vector diagram?
Yes, I have. Which is where I got the idea for the Law of Cosines.
 
Oh, ok. You can use the law of cosines to figure out the magnitude of the resultant vector, but it's easier to add the x and y components of the two velocity vectors to get the x and y components of the resultant velocity vector.
 
Okay. I've solved it. It was 271.79 Km/h at 8.54 degrees east of north.
Thanks for your help.
:]
 
if you make the diagram...and add the 2 velocity vectors...you end up with...

220sin90 y(hat) + 220cos90 x(hat)
55sin45 y(hat) + 55cos45 x(hat)

add them up using the x and y components as variables...getting a total of...
258.9 y(hat) + 38.9 x(hat)

Your resultant vector is absolute value of R = square root of x^2 + y^2
this will get you your magnitude of 261.8. so this is the final speed.

THen to get the angle it is going...do arc tan of y/x...so 38.9/258.9
you get it will be heading 8.5 degrees east of north.

for some reason i don't like that answer all to much...maybe someone can check me on it but that is how you do it
 
Last edited:
Well, I know it's correct because I submit my answers onto a website that checks to see if it's right or not.
 
I sent in mine before i saw your answers...so. did you check your velocity? i double checked mine and it seems to come to the same every time...yours is right says your website?
 
  • #10
Hmm... That's odd because mine comes out right too.
 
  • #11
Ahh! Nevermind. Typo. Mine actually was 261.79. My bad. Ha.
 
  • #12
oh well...its been awhile since i got my lesson on vector addition so i would take yours hehe. feel free to help me with my current vector problem locatedo n the front page of the forums haha
 

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