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Simple Coordinate Problem

  1. Sep 25, 2009 #1
    "Simple" Coordinate Problem

    1. The problem statement, all variables and given/known data
    The following happens on a 2D X-Y Plane.

    A particle accelerates at {3t m/s2}i + 4t m/s2}j where t = seconds

    At t = 0, the position of the particle is {20.0 m}i + {40.0m}j

    At t = 0, the velocity of the particle is {5.00 m/s}i + {2.00 m/s}j

    At t = 4,

    (a) What is the position vector in unit-vector notation
    (b) What is the angle between its direction of travel and the positive direction of the x-axis


    2. Relevant equations
    dx = Vox*t + .5(a)t^2
    Dy = Voy*1 + .5(a)t^2


    3. The attempt at a solution

    dx = Vox*t + .5(a)t^2
    dx = 5(4) + .5(3(4)) (4)^2
    dx = 20 + 6(16)
    dx = 116

    Dy = Voy*1 + .5(a)t^2

    Dy = 2(4) + .5(4(4)) (4)^2
    Dy = 8 + 8(16)
    Dy = 136

    Therefore, it's position vector should be

    P = {136 m}i + {176 m}j

    If we remember to add on the initial displacement.

    The answer is {72.0 m)i + {90.7 m}j
     
  2. jcsd
  3. Sep 26, 2009 #2

    D H

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    Re: "Simple" Coordinate Problem

    The relevant equations you cited are for constant acceleration. You do not have a constant acceleration here.
     
  4. Sep 26, 2009 #3
    Re: "Simple" Coordinate Problem

    Huh... Right. True. However, I don't ever remembering having to deal with anything that isn't constant acceleration.

    Any hints on what to start with?
     
  5. Sep 26, 2009 #4

    D H

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    Re: "Simple" Coordinate Problem

    First you should do is check that you have the problem correct. If it is correct as written, I suspect that you have learned something about integration.
     
  6. Sep 26, 2009 #5
    Re: "Simple" Coordinate Problem

    Yes, the problem is correct.

    We have actually learned nothing about integration specifically, but I do know the basic calculus for integration and have a basic understanding of it.

    But I do not see how it applies.

    Basically, this question was not even assigned to us I don't think, I am simply going through all the questions my text gives me in the section, so... If you could give me a giant push or something, I'd appreciate it.
     
  7. Sep 26, 2009 #6

    D H

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    Re: "Simple" Coordinate Problem

    You know the acceleration as a function of time and the initial velocity. Acceleration is the derivative of velocity, so integrating acceleration with respect to time will yield velocity as a function of time. Integrating velocity will yield position.
     
  8. Sep 26, 2009 #7
    Re: "Simple" Coordinate Problem

    So in theory, wouldn't

    ax(t) = 3t
    vx(t) = (3/2)t^2 + c
    px(t) = (3/6)t^3 + ct + d

    If vx(0) = 5

    then

    vx(0) = (3/2)(0)^2 + c
    5 = 0 + c
    c = 5
    vx(t) = (3/2)t^2 + 5


    px(t) = (3/6)t^3 + ct + d
    px(t) = (3/6)t^3 + 5t + d
    px(0) = 20
    px(0) = (3/6)(0)^3 + 5(0) + d
    20 = 0 + 0 + d
    d = 20

    Therefore:

    p(x) = (3/6)t^3 + 5t + 20
    p(4) = (3/6)(64) + 20 + 20
    p(4) = 32 + 40 = {72 m}i

    Which... is the answer for the x coordinate. Thanks for the hint -- I knew it couldn't have been too hard once again just tired.
     
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