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

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An airplane flying due north at 220 km/h encounters a northeast wind of 55 km/h, prompting a discussion on calculating its ground speed and direction. The law of cosines is initially mentioned, but participants suggest using vector diagrams and component addition for clarity. The resultant speed is calculated to be approximately 261.79 km/h at an angle of 8.54 degrees east of north. Both contributors confirm their calculations align with an external verification tool, ensuring accuracy. The conversation emphasizes the importance of vector addition in solving such problems effectively.
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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
 
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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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