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Vectors and Components-Part One

  1. Dec 29, 2013 #1
    1. The problem statement, all variables and given/known data

    An aircraft is traveling at 200 mi/h on a heading of 38 degrees. Find the northerly and easterly components of its velocity.



    2. Relevant equations
    Sine=opposite/hypotenuse
    Cosine=adjacent/hypotenuse

    3. The attempt at a solution

    The problem states that the angle given is a heading. In order to use our equations, it is best to convert to our non-heading angle. I don't know what that's called, but it's 52 degrees. So to find the northerly component, I did 200cos(52) and got 123.1 mi/h. To get the easterly component, I did 200sin(52) and got 157.6 mi/h. Are these the right answers so far? Do I need to convert back to a "heading" answer? How would I do that?


    Thanks so much for a helpful and descriptive answer! :smile:
     
  2. jcsd
  3. Dec 29, 2013 #2

    PhanthomJay

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    A heading in aviation terms is measured clockwise frpm magnetic north. If the airplane is heading 38 degrees, it means it is moving in a direction 38 degrees east of north, that is, in a generally northeast direction. The problem should more specifically state or sketch the heading, often denoted for this case as N 38 E. in which case , the northerly component of the velocity is 200 cos 38 or 200 sin 52, whichever triangle you choose(draw a sketch!).
     
  4. Dec 29, 2013 #3
    Okay, I think I got all that. :-) I forgot to mention that I already drew a diagram. The problem isn't any more specific than I indicated, unfortunately. Are you stating that my answer for the northerly component is correct?

    Thank you! :smile:
     
  5. Dec 29, 2013 #4

    PhanthomJay

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    Your answer is incorrect. The plane is heading 038 degrees per its compass reading at a speed of 200 mph. Its northerly component is 200 cos 38, not 200 cos 52. Please look at your sketch again to confirm this result.
     
  6. Dec 29, 2013 #5
    I think I see what you're saying. I should have made my sketch larger and neater. I got it backwards, right? It would be 200 sin 52 if I were to use the this angle, correct?
     
  7. Dec 29, 2013 #6

    PhanthomJay

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    Correct!
     
  8. Dec 29, 2013 #7
    So the northerly component is 157.6 mi./h. Thank you! Was the easterly component correct?
     
  9. Dec 29, 2013 #8

    PhanthomJay

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    Why no the easterly component is incorrect...what should it be ?
     
  10. Dec 29, 2013 #9
    Is is 123.1 mi./h? Just for clarity, I got the right answer for the northerly component, correct?

    Thanks!
     
  11. Dec 30, 2013 #10

    PhanthomJay

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    Yes both values for the northerly and easterly components are now correct. ( You might want to round them off however at least to the nearest whole number with no decimal point. ).
     
  12. Dec 30, 2013 #11
    So just 158 mi/h and 123 mi/h. Thank you!
     
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