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Homework Help: Vector Addition

  1. Sep 2, 2011 #1
    A woman walks 143m in the direction 55° east of north, then 178m directly east. Find the magnitude of the displacement vector.

    Answer: 306m

    Relevant equations:
    I will use vA as a shorthand to represent vector A and ||vA|| to represent the magnitude.

    Ax = ||vA||cos(theta)
    Ay = ||vA||sin(theta)
    ||vA|| = sqrt(Ax^2 + Ay^2)


    Since I know the magnitude of vA (assuming vA is the starting vector) and the angle in which she left the origin of the coordinate grid I can use the two equations stated above to find the values of Ax and Ay.

    Ax = 143m*cos(55) = 82.02m
    Ay = 143m*sin(55) = 117.14m

    Now I add vA + vB (<82.02m, 117.14m> + <178m, 0m>) to obtain vC, the vector displacement between her starting position to her final position. However, when I go and find the magnitude of the vector I always come out with the wrong answer, 285.19m.

    vC = <260.02m, 117.14m>
    ||vC|| = sqrt(260.02m^2 + 117.14m^2) = sqrt(81332.18m^2) = 285.19m.

    What did I do wrong? I thought this would be a simple problem but I keep coming out with the wrong answer. Can someone help me?
  2. jcsd
  3. Sep 2, 2011 #2
    The problem is that you're angle is wrong. Try drawing a small graph with the angle starting from the positive y-axis (north) and going towards the x-axis (east). You see that the 55° angle is made with the y-axis, and not the x-axis.
  4. Sep 2, 2011 #3
    Yes the angle is wrong. Since they say east of north (which I personally think is a stupid way to say) they mean 90-55=35 degrees. The rest is correct.
    Last edited: Sep 2, 2011
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