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Net Displacement (Probably very easy, but I don't get it)

  1. Jan 31, 2009 #1
    1. The problem statement, all variables and given/known data

    A person walks on level ground 0.6 km east, then 500 meters south, and then climbs to the top of a building (height 180 meters). What is the net displacement of the person?

    2. Relevant equations

    3. The attempt at a solution

    Really have no idea. I know that 600m east and then 500 m south equals a displacement of 1100 m. Or do you have to use Pythagorean theorem? And then for the extra 180 vertical meters, I have no idea.

    Some help please.
  2. jcsd
  3. Jan 31, 2009 #2


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    Homework Helper

    No, it does not. Yes, you must use the Pythagorean theorem. That will give you the distance in the horizontal plane. Then you consider the 180 m up, which gives you a triangle in a vertical plane which has a height of 180 and a base of about 780. Use the Pythagorean theorem to find the distance from beginning to end.

    The word "displacement" usually means the straight line distance AND the angles involved. You may have to find the angle in the horizontal triangle and the angle of elevation in the vertical triangle to complete the problem.
  4. Feb 1, 2009 #3
    i never really understood triangles in a vertical plane thoughhh
  5. Feb 1, 2009 #4
    Hey FAJ, use d=sqrt'sq 600 + sq 500 + sq 180'
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