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Another Vector Question

  1. Sep 10, 2008 #1
    1. The problem statement, all variables and given/known data

    A jetliner takes off from LA. after 5 min, the jet has an altitude of 5 km above the level of the airport and is 25 km west and 10 km north of the airport. What are the magnitude and the diretion of its displacement during this time interval? Specify the direction by stating two angles: the angle of the displacements direction west of north and the angle of the displacement's direction above the horizontal.

    2. Relevant equations

    for West of North:
    construct a triangle with adjacent side 10 km and hypotenuse 25 km
    then use cox=10/25 to find the angle

    For above the horizontal:
    construct a triangle with opposite side 5 km and hypotenuse 10
    then use tanx=5/10 to find angle

    For magnitude:

    use the magnitude formula for: [-25 km, 10 km, 5 km]


    3. The attempt at a solution

    Is horizontal always the Z component/up?

    west of north angle: 66*
    above horizontal angle: 27*
    direction of displacement: northwest
    magnitude: 27 km
     
  2. jcsd
  3. Sep 10, 2008 #2
    It's been awhile since I've done any of this so somebody correct me if otherwise, but for west of north and for above the horizontal, you should not be using any distance as the hypotenuse.

    You will need two different triangles for each situation which you did, but the hypotenuse is not known until you calculate it but you dont need it.
     
  4. Sep 10, 2008 #3
    Okay . .I understand now that I look at it again. Thank you.

    for north of west:
    construct triangle with opposite 25 and adjacent 10

    for above horizontal:
    construct triangle with opposite 5 and adjacent 10
     
  5. Sep 10, 2008 #4
    I believe so yes, that is what I had written down I think. I erased it but I think the angles I had were something like 68.198* and 26.565*
     
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