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Clarification of 3D Vectors

  1. Sep 25, 2007 #1
    An object has an initial velocity given by V = vi + vj, where v=10m/s. If an acceleration of A = ak, where a = 10 m/s^2 is applied for 10 seconds determine the final velocity for the object.

    How far did the object travel in 10s?

    How far from the z axis is the object?

    Well to start. I used the equation:
    Vf = Vi + at
    Vf = (10i + 10j) + (10k)(10s)
    Vf = 10i + 10j + 100k

    Then to determine distance, I used:
    D^2 = (change in x)^2 + (change in y)^2 + (change in z)^2

    D = sqrt.(10,200) m?
    I am not sure if I calculated this right.

    Then for the last part of the question, I am not really sure how to determine this. I know that the object moved 100m in the x direction, and 100m in the y direction, and I think it moved 500m in the z direction given the Vf= 100m/s in z direction and Vi = 0m/s in z direction.

    So I am assuming we have to use pythagorean theorem again? I am not sure. Please help, and comment on any of the previous answers please.
  2. jcsd
  3. Sep 25, 2007 #2
    for the second part, i think you should use
    s = ut + half*at^2 to find its position vector and then use the distance formula to find its distance from the origin.

    for the third part, that will be the z-component of the above position vector that you find.
  4. Sep 25, 2007 #3
    so if I use that equation:

    i get:

    s(t) = 100i + 100j + 500K

    D = sqrt. (270,000) m?

    Then for distance from z axis is simply sqrt. (500) m?
  5. Sep 25, 2007 #4
    i did a mistake here.

    for the second part, i guess so...

    for the third part, it is square root of (x-component^2+y-component^2) = sqrt(10000+10000). if you draw the final coordinate of the particle, i think you will see why.
  6. Sep 25, 2007 #5
    i am not really understanding how you determined the 3rd part...

    The way you did it, it simply looks like the displacement between the x and y components...not the distance from the z axis.
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