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Mechanics General Motion in a Straight Line

  1. May 12, 2014 #1
    1. The problem statement, all variables and given/known data
    A particle P starts from rest at a point O and moves in a straight line. P has acceleration 0.6t m s−2
    at time t seconds after leaving O, until t = 10.
    (i) Find the velocity and displacement from O of P when t = 10.
    After t = 10, P has acceleration −0.4t m/s^2 until it comes to rest at a point A.
    (ii) Find the distance OA.


    2. Relevant equations
    v = ∫a dt & s = ∫v dt


    3. The attempt at a solution
    For (i) v = ∫a dt = 0.3t^2 +c (c=0) = 0.3 x 10^2 = 30 m/s
    s = ∫v dt = 0.1t^3 +c (c=0) = .1x10^3 = 100 m

    But for (ii) I did the same thing...
    a = -0.4t m/s^2
    → v = ∫a dt = -0.2 t^2 + c = -0.2 t^2 +30 because the initial velocity (when t=10s) is 30 m/s. However the answer said it should be 50 m/s? Could anyone explain me that please?

    Also s=∫v dt = (-0.2t^3)/3 + 30t +100 (but the answer said -1000/3????)
     
  2. jcsd
  3. May 12, 2014 #2

    BiGyElLoWhAt

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    1. are your answers to (i) correct?
    2. What are your limits of integration for (ii) and how did you get them?
     
  4. May 12, 2014 #3

    haruspex

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    The answer said what should be 50 m/s? The velocity in (i)?
     
  5. May 12, 2014 #4
    this is the answer sheet:
    Screenshot (186).png
     
  6. May 13, 2014 #5

    haruspex

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    That's clearly an error on the answer sheet. The text on the lower right box says taking the initial speed as 30 m/s.
    If you care to post your answers to (ii) someone will check them.
     
  7. May 13, 2014 #6

    tms

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    These should be definite integrals, not indefinite ones. You got away with it here, because one of the limits of integration is 0. However, this is what trips you up on the second part of the problem.
     
  8. May 13, 2014 #7

    tms

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    It seems okay to me. The initial speed is not 30 m/s, but the speed at [itex]t = 10[/itex], the start of the second part of the problem, is 30 m/s.
     
  9. May 13, 2014 #8

    haruspex

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    Yes, the text in the lower right box is correct, and agrees with Giiang's result. But the text in the lower left box says 50 m/s.
     
  10. May 13, 2014 #9

    tms

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    That 50 comes from the integration in the first part of question ii.
    [tex]v(t_F) = v(10) + \int_{10}^{t_F} (-0.4t) dt = 0.[/tex]
    Solving that for [itex]t_F[/itex], along the way you get
    [tex]v(t_F) = v(10) - 0.2t_F^2 + 20 = 0.[/tex]
     
  11. May 13, 2014 #10

    BiGyElLoWhAt

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    Alright look, the books answers are right. Your initial conditions and method of solving for (ii) is wrong. Plain and simple. The constant of integration doesn't always equal your initial condition. You need to actually solve for it. that's where the 50 comes in from. It doesn't say the initial velocity is 50, it comes from the fact that v(10)=30 as your initial condition. Plug it in and solve.
     
  12. May 13, 2014 #11

    BiGyElLoWhAt

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    once again...
     
  13. May 17, 2014 #12
    I got it. Thanks!

    So,
    a = -0.4t m/s^2
    → v = ∫a dt = -0.2 t^2 + c

    and it was found in (i) that v(10) = 30
    → 30 = - 0.2 10^2 +c → c = 50

    The same goes with displacement.
     
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