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Displacement and angle problem

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data

    A person going for a walk follows the path shown in the figure, where y1 = 290 m and θ = 61.0°. The total trip consists of four straight-line paths. At the end of the walk, what is the person's resultant displacement measured from the starting point?

    ____ m? ____ degrees? (from the + x axis)

    2. Relevant equations

    [​IMG]

    3. The attempt at a solution

    I've calculated and found the correct angle, which is 238.08 degrees but I don't know why I keep getting the wrong magnitude. I got 216.45 m and it's incorrect.

    here is my work:

    vector A = 100N @ 0 degrees ; A_x = 100 A_y = 0
    vector B = 290N @ 270 degrees; B_x = 0 B_y = -290
    vector C = 150N @ 210 degrees; C_x = -129.90 C_y = -75
    vector D = 200N @ 115 degrees; D_x = -84.52 D_y = 181.26
    ____________________________________________________
    F_x = -114.42; F_y = -183.74

    tan^-1 (-183.74/-114.42) = 58.1 degrees + 180 degrees = 238.08 degrees

    so for the magnitude I did

    -114.42^2 + -183.74^2 = c^2
    sq root of 46852.3 ... or 216.454 = c

    but that is wrong, can anyone tell me what c would be? thanks.
     

    Attached Files:

    Last edited: Sep 28, 2008
  2. jcsd
  3. Sep 28, 2008 #2
    sorry, I've attached the picture.

    The first vector is vector B in my solutions.
     
  4. Sep 28, 2008 #3

    Integral

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    Staff Emeritus
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    Gold Member

    Ok, I"ve got the pic. The only problem I see is the value of the angle of the 4th vector. You have 115deg, I get 119deg.
     
  5. Sep 28, 2008 #4
    thank youuuu! how do I mark this as solved now?
     
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