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Direction of Displacement Vector [2D Kinematics]

  1. Sep 27, 2008 #1
    1. The problem statement, all variables and given/known data

    Question 1 [Correctly Solved]
    A pedestrian moves 2.1km east and then 20km north. Find the magnitude of the resultant displacement vector. Answer in units of km.

    Answer = 20.1km

    Question 2
    What is the direction of the displacement vector (using the counter-clockwise angular direction to be positive, within the limits of -180 degrees = 180 degrees)? Answer in units of degrees/

    3. The attempt at a solution

    [Theta] = (arctan) (Ay/Ax)
    [Theta] = (arctan) 20km/2.1km = 1.466

    ------

    I'm probably using the wrong equation. :x
    (using the counter-clockwise angular direction to be positive, within the limits of -180 degrees = 180 degrees) Also, I don't understand what that means. Will someone please explain? :blushing:
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 27, 2008 #2
    question #1: Draw it on a sheet of paper, make a Cartesian plane, use the upper y axis as north, right x-axis as east. You ll remember something!
    *hint* Pythagorean Theorem ;)

    Question #2: now im not sure if ill explain this right but! i think that this : "using the counter-clockwise angular direction to be positive, within the limits of -180 degrees = 180 degrees" means find the angle opposite to the 20km line (i think its the angle between the hypotenus and the 2.1km line)
    Let's say it is ;)
    now remember SOHCAHTOA and see if you can do it.

    Got more questions, just PM me. Goodluck!
     
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