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Northern Component of velocity

  1. Sep 12, 2007 #1
    Components of velocity

    1. The problem statement, all variables and given/known data

    An airplane travels at 146 km=h toward the
    northeast.
    What is the northern component of its ve-
    locity? Answer in units of km=h.

    2. Relevant equations

    asquared + bsquared = csquared

    3. The attempt at a solution

    however i am not sure if u use the pythagorean formula to solve this?
     
    Last edited: Sep 12, 2007
  2. jcsd
  3. Sep 12, 2007 #2
    Yes, you're on the right track. (Assuming by "North East" they mean that 45 degree angle between north and east. That means NorthEast would be the hypotenuse/resultant of a two vectors: 1 north and 1 east of equal magnitudes.)
     
  4. Sep 12, 2007 #3

    berkeman

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    Staff: Mentor

    First, the plane speed is 146 km/h (not km=h).

    Next, you find the components in the north and east directions by multiplying the hypoteneuse by the sine or cosine of the appropriate angles....
     
  5. Sep 12, 2007 #4
    ooo i have to do sin/cosine of the angle
     
  6. Sep 12, 2007 #5
    so to find the horizontal velocity it would be the cos of 45 = vx/ 146? which equals 76.699 km/h

    and to get the vertical it would be sin of 45 = vy/146? which equals 124.231 km/h
     
    Last edited: Sep 12, 2007
  7. Sep 12, 2007 #6

    berkeman

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    Staff: Mentor

    Yes. That's an unusual way to write it, however. More like this (I'll use latex):

    [tex]v_y = 146 km/h * sin(45)[/tex]
     
  8. Sep 12, 2007 #7

    berkeman

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    Staff: Mentor

    Well, except for the math you did. The sin and cos of 45 degrees should be the same....
     
  9. Sep 12, 2007 #8
    ok thanks guys i got it i was doing some bad math... thank you
     
    Last edited: Sep 12, 2007
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