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Vectors, Check my work.

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data

    A particle of mass 2.0 kg has position given by the vector r = (t^4 -3t^3) i + (t^3 - 9t) j + (2t -6) k
    (where i, j, and k are the unit direction vectors and r is given in meters and t in seconds. At what time in seconds is the particle located at the origin?

    For this I get t = 3(seconds).

    It's the second part I want you to check.

    1. what is the magnitude of the particle's velocity (in m/s) when the particle is at the origin?

    2. For the particle in question 1, what is the x/y/z-component of the net force (in Newtons) acting on it when it is at the origin? Do not write Newtons as part of your answer.

    2. Relevant equations

    V = Dr / Dt
    F= Ma

    3. The attempt at a solution

    1. Derive each component individually. Got:

    (4t^3-9t^2)i, (3t^2 - 9)j, (2)k
    Plug in 3 to get. 27 + 18 + 2 = Sqrt 47 = 6.86 m/s while at origin.

    2. Derived each component again to get the acceleration eq.
    I got (12t^2 - 18t)i, (6t)j, and (0)k

    Then did F = MA for each component.

    2.0(12(3)^2 - 18(3)) = 108 for x component.
    2.0(6(3)) = 36 for the y component.
    2.0(0) = 0 for the z component.

    Did I do this correctly?

    Thanks for helping :D.
  2. jcsd
  3. Feb 6, 2009 #2


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    Homework Helper
    Gold Member

    Hi Kaln0s, welcome to PF! :smile:

    Don't you have to square each component before summing them up and taking the square root?:wink:


    Looks good to me:approve:
  4. Feb 6, 2009 #3
    Oh god I feel dumb as hell lol. I've been doing Calculus and physics non-stop for the past day -_-.

    27^2 + 18^2 + 2^2 = 1057 sqrt = 32.5 m/s

    Thank you very much. :biggrin:
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