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Exercise: position vector of a particle

  1. Jan 5, 2016 #1
    1. The problem statement, all variables and given/known data
    The acceleration of a particle moving only on a horizontal xy plane is given by a = 3ti + 4tj, where a is in meters per second squared and t is in seconds. At t = 0, the position vector r = (20.0 m)i + (40.0 m)j locates the particle, which then has the velocity vector v = (5.00m/s)i + (2.00m/s)j. At t = 4.00s, what are
    (a) its position vector in unit-vector notation and
    (b) the angle between its direction of travel and the positive direction of the x axis?

    2. Relevant equations
    x = x0 + vox * t + 1/2 * ax * t2
    y = y0 + voy * t + 1/2 * ay * t2

    3. The attempt at a solution
    So, since ax = 3t and ay = 4t, I got the following values for the position vector:
    (136m)i + (176m)j

    The problem is that the values aren't the solution given by the book (which is (72m)i + (90.7m)j ).
    I keep reading but nothing comes to my mind besides the fact that probably I did something wrong with the acceleration.
     
  2. jcsd
  3. Jan 5, 2016 #2

    SteamKing

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    The equations you cited in Section 2 are valid only for constant acceleration. Your acceleration is changing with time.
     
  4. Jan 5, 2016 #3
    The book doesn't give any example with a non-constant acceleration. I don't know how to proceed in this case.
    Could you guide me to the solution?
     
  5. Jan 5, 2016 #4

    SteamKing

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    Are you sure?

    Have you seen these expressions:

    v = ds / dt

    a = dv /dt
     
  6. Jan 5, 2016 #5
    Yes, I know them.
    The last one is an instantaneous acceleration, right?
     
  7. Jan 5, 2016 #6

    SteamKing

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    Yes, it is.

    When acceleration is changing with time, you must integrate acceleration w.r.t. time to find the instantaneous velocity, and then integrate velocity w.r.t. time to find position.

    v(t) = ∫ a(t) dt + v0

    s(t) = ∫ v(t) dt + v0 ⋅ (t) + s0

    where v0 and s0 represent the initial velocity and position, respectively.
     
  8. Jan 5, 2016 #7
    If from the second equation I simply write
    s(t) = ∫ v(t) dt + s0 I have the solution.
    Is it okay? Because if I do the second equation as you wrote it, it has a +20 on the first component and a +8 on the second component. I guess it was a typo from you.
     
  9. Jan 5, 2016 #8

    SteamKing

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    s0 can be written in terms of a position vector with i and j components.

    All of the integrations can be performed separately to obtain the i and j components of the velocity and the position of the particle as a function of time.
     
  10. Jan 5, 2016 #9
    Yeah, I know. The last equation you wrote would have these numbers:

    s(t) = (3 * t3) / 6 i + (4 * t3) / 6 j + 5 * t i + 2 * t j + 5 * t i + 2 * t j + 20 i + 40 j

    Am I right?
     
  11. Jan 5, 2016 #10

    SteamKing

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    I think you have included v0 ⋅ t twice in the expression for s(t). You'll need to evaluate s(t) for t = 4 sec.
     
  12. Jan 5, 2016 #11
    That's exactly what I wanted to make you notice. In your equations I have two of them:
    With:
    v(t) = (3 * t2) / 2 i + (4 * t2) / 2 j + 5 i + 2 j
    and
    s(t) = (3 * t3) / 6 i + (4 * t3) / 6 j + 5 * t i + 2 * t j + (5 i + 2 j) * t + 20 i + 40 j
     
  13. Jan 5, 2016 #12

    SteamKing

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    v(t) = ∫ a(t) dt + v0

    ∫ v(t) dt = s(t)

    ∫ v(t) dt = ∫ [∫ a(t) dt + v0] dt = ∫∫ a(t) dt dt + ∫ v0 dt = s(t) + v0 ⋅ t + s0

    Since:
    v0 = 5 i + 2 j

    If you multiply v0 by t, then you have 5 ⋅ t i + 2 ⋅ t i

    You are doing only one integration where v0 is present, not two.
     
  14. Jan 5, 2016 #13
    Wait, I'm getting confused.
    First we do this integral:
    v(t) = ∫ a(t) dt + v0 = (3 * t2) / 2 i + (4 * t2) / 2 j + 5 i + 2 j
    after this, we do this other integral:
    s(t) = ∫ v(t) dt + v0 ⋅ (t) + s0 = ∫ (3 * t2) / 2 i + (4 * t2) / 2 j + 5 i + 2 j dt + v0 ⋅ (t) + s0
    Am I right? Because this is what comes out from your equations.
     
  15. Jan 5, 2016 #14

    SteamKing

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    No, it's not.

    v0 is a constant vector, namely v0 = 5i + 2j. It does not depend on t.

    When integrating a(t), v0 is added after the integration is done, which is why v(t) = ∫ a(t) dt + v0

    When integrating v(t), then another constant of integration, which is called s0, is added after the integration of v(t) is done.
    But because v(t) includes v0, when v(t) is integrated, the v0 term must be integrated at well. Since v0 is a constant, its integral is just v0 ⋅ t.

    s(t) = ∫ (v(t) + v0) dt = ∫ v(t) dt + ∫ v0 dt = ∫ v(t) dt + v0 ⋅ t

    I think your confusion stems from the fact that the v(t) inside the integral sign involves only the part of v(t) which varies with time.

    I think these equations will be more clear:

    v(t) = ∫ a(t) dt = v1(t) + v0

    s(t) = ∫ v (t) dt = ∫ (v1(t) + v0) dt = ∫ v1(t) dt + ∫ v0 dt = ∫ v1(t) dt + v0 ⋅ t + s0
     
  16. Jan 5, 2016 #15
    Oooh! So the second equation was simply "s(t) = ∫ v (t) dt".
    Okay, I got it now.
    Thank you!
     
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