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Change of acceleration

  1. Sep 22, 2004 #1
    Automotive engineers refer to the time rate of change of acceleration as the "jerk." If an object moves in one dimension such that its jerk "J" is constant. How would I determine expressions for its acceleartion Ax, velocity Vx, and position X, given that its inital acceleration, speed, and position are Axi, Vxi, and Xi, respectively.

    How would I show that Ax^2=Axi^2+2J(Vx-Vxi)?

    I'm not sure where to start on this one.
     
  2. jcsd
  3. Sep 22, 2004 #2

    Pyrrhus

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    You don't know how to integrate, i don't believe you can do this problem.

    It's something like

    [tex] \frac{da_{x}}{dt} = J [/tex]

    [tex] a_{x} = \int Jdt [/tex]

    [tex] a_{x} = Jt + C[/tex]
    and then you need to find a value for C, and on and on...

    or

    [tex] \int^{a_{x}}_{a_{xo}}da_{x} = \int^t_{0} Jdt [/tex]

    [tex] a_{x} ]^{a_{x}}_{a_{xo}}= Jt]^t_{0} [/tex]

    [tex] a_{x} - a_{xo}= Jt - 0 [/tex]

    Edit: To make it clearer.
     
    Last edited: Sep 22, 2004
  4. Sep 22, 2004 #3

    robphy

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    You may recall from the "constant acceleration" case that
    [itex]v^2=v_0{}^2+2a(x-x_0)[/itex].
    Where did this equation come from? How is it derived?


    Now, your problem is to derive
    [itex]a^2=a_0{}^2+2J(v-v_0)[/itex]
    for the "constant jerk" case.


    More hints:

    "constant velocity" case: [itex]x=x_0+vt[/itex]
    "constant acceleration" case: [itex]v=v_0+at[/itex] and [itex]x=x_0+ \ldots[/itex]
    "constant jerk" case: [itex]a=\ldots[/itex] and [itex]v=\ldots[/itex] and [itex]x=\ldots[/itex]
     
  5. Sep 23, 2004 #4
    I'm actually confused by what the question is asking. Can you possibily refrase the question so a high schooler can understand what it's asking? About the "How would I determine expressions for its acceleartion Ax, velocity Vx, and position X, given that its inital acceleration, speed, and position are Axi, Vxi, and Xi, respectively."

    What is the question trying to get at? What in the world is a Jerk? Is it when acceleration has a slope on an acc. vs time graph?

    Thanks
     
  6. Sep 23, 2004 #5

    robphy

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    Yes! Jerk is the slope of the acc. vs. time graph.

    So, you should think back...
    when velocity had a constant nonzero slope on a velocity vs time graph, what expressions did you write for the velocity and the position?

    Apply the same reasoning your problem.
     
  7. Sep 23, 2004 #6
    I dont think I ever wrote any expression for velocity and position.
    It's probably what you wrote...X=Xo+vt

    I'm not sure how to apply this to my problem since I dont know what the question is asking. Probably a rephrase of the question would be most helpful.

    Thanks
     
  8. Sep 24, 2004 #7

    Pyrrhus

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    Urban, i don't know how you can do this problem without Riemann's Integrals, but Rophy's idea is quite good, the kinematic equations you know for constant acceleration happens when dv/dt = a, where a is a constant. Now for this problem it will be da/dt = j, where j is a constant. Imagine how can you use the kinematic equations you know for this case.
     
  9. Sep 24, 2004 #8

    Pyrrhus

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  10. Sep 24, 2004 #9
    Thanks a lot. Even if I did learn to do integral, how would this help me in this problem? What does getting the area under a position v time, velocity v time, and acceleration v time, graph tell me?
     
  11. Sep 24, 2004 #10

    Pyrrhus

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    dx = vdt (Velocity vs time graph area it will give displacement)
    dv = adt (Acceleration vs time graph area it will give velocity)
     
  12. Sep 24, 2004 #11
    sooo... da/dt = j ... da=jdt will give me acceleration?

    When you say "velocity" is it average velocity?
     
  13. Sep 24, 2004 #12

    Pyrrhus

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    In a Jerk vs Times graph yes :smile:

    Yes average velocity.
     
  14. Sep 24, 2004 #13
    When the question asks "How would I determine expressions for its acceleartion Ax, velocity Vx, and position X" is it asking for me to derive a formula with variables Axi, Vxi, and Xi...so it would be X=blah*J^2+blah*J+blah...and Vx would be the dirivative of X...so on?

    Also, how must I use Integral to obtain this fomula for Jerk?
     
  15. Sep 24, 2004 #14

    Pyrrhus

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    Look at my first post here.
     
  16. Sep 24, 2004 #15
    where did the variable C come from?
     
  17. Sep 24, 2004 #16
    Also, when you write "dx/dt = v" are you saying the integral of the relationship between displacement and time is the average velocity?
     
  18. Sep 24, 2004 #17

    Pyrrhus

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    Well if you have y = x^2 + 4 or y = x^2 o y= x^2 + 53, the derivatives will be the same, 2x, so you put a C when you integrate representing that unknown constant.
     
  19. Sep 24, 2004 #18

    Pyrrhus

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    [tex] \frac{dx}{dt} [/tex] is Instantenous Velocity while [tex] \frac{\Delta x}{\Delta t} [/tex] is average velocity, so if you integrate and get the whole area down a curve of Acceleration vs Time, it will be the average velocity, it depends on what you're getting.
     
  20. Sep 24, 2004 #19
    therefore, Instantenous Acceleration is the Jerk.

    to determine expressions for its acceleartion Ax, velocity Vx, and position X, all I have to do is find the integral of the Jerk for Ax, the integral of Ax for Vx, then the integral of Vx for X.
     
  21. Sep 24, 2004 #20

    Pyrrhus

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    Exactly, all depending of course on time differential.
     
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