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Deriving the rate time equation for a Jerk (J)

  1. Sep 24, 2007 #1

    ~christina~

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    1. The problem statement, all variables and given/known data

    Automotive engineers refer to the time rate change of acceleration as the Jerk.
    Assuem an object moves in 1 dimention such that it's Jerk is constant.

    a) determine expressions for it's
    1. acceleration ax(t)
    2. velocity vx(t)
    3. position x(t)

    given initial acceleration, velocity, and positio are axi, vxi, xi respectively

    b) show that ax^2= axi^2 + 2J(vx-vxi)

    2. Relevant equations

    not sure because time rate of acceleration = Jerk...but what is the time unit..is it just sec?

    I know that the time rate change for velocity is acceleration though..

    a= v/t

    Not sure about the position




    3. The attempt at a solution

    including the equation attempt above..the whole problem is to determine the equation..I'm just not sure how to put it together

    I know that the problem is under deriving equations from calculus though just I know that if you get the derivative of the position then you get velocity eqzn and if you derive that you get the acceleration.
    And if you get the integral of the acceleration then you get the velocity equation and if you get the integral of that you get the position function once again..

    other than that I just don't know how to relate equation deriving to a "Jerk"

    I have problems with deriving eqzns..


    for b.) I'd don't know what they want me to do with it...exactly...

    Thanks:uhh:
     
  2. jcsd
  3. Sep 24, 2007 #2

    cepheid

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    First, of course the standard units of time are seconds, no matter which unit system you are using.

    Saying that the jerk is the rate of change of acceleration with time is the same as saying that the jerk is the derivative of the acceleration with respect to time. In other words:

    velocity is the first derivative of position
    acceleration is the second derivative of position
    jerk is the third derivative of position.

    Based on this, I think you can see that:

    [tex] J(t) = \textrm{const.} = J_x [/tex]

    [tex] a_x(t) = \int J_x \, dt [/tex]
    [tex] v_x(t) = \int a_x(t) \, dt [/tex]
    [tex] x(t) = \int v_x(t) \, dt [/tex]

    Now it's just a matter of computation.
     
  4. Sep 24, 2007 #3

    ~christina~

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    check

    But what do I compute?? and how do I prove the eqzn

    b) show that ax^2= axi^2 + 2J(vx-vxi)

    how do I show that??


    When You say compute don't you have to have a original equation to compute from?

    Is it...since Jx= constant

    ax(t)=[tex]\int Jx dt[/tex] = Jt

    then vx(t)=[tex]\int ax(t)dt [/tex]= [tex]\int Jtdt [/tex] = J/2*t^2

    then x(t)= [tex]\int vx(t)dt [/tex]= [tex]\int J/2* t^2 [/tex]= J/6*t^3


    Is this fine?
     
    Last edited: Sep 24, 2007
  5. Sep 24, 2007 #4

    learningphysics

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    add constants when you take the integrals.
     
  6. Sep 24, 2007 #5

    ~christina~

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    Okay Thank You. But what about the part B?

    b) show that ax^2= axi^2 + 2J(vx-vxi)

    how do I show that??
     
  7. Sep 24, 2007 #6

    learningphysics

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    When you add the constants, this will make sense.

    eg:

    ax = Jt + axi

    vx = (J/2)t^2 + axi*t + vxi etc...
     
  8. Sep 24, 2007 #7

    ~christina~

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    hm I thought you meant the constant c at the end of the integral...ex usually X+C or etc.

    So do you mean that I add...it like this...

    ax(t)=[tex] \int Jxdt[/tex] = Jt + axi

    then vx(t)=[tex] \int ax(t)dt[/tex]=[tex] \int Jt + axi[/tex]= J/2*t^2 + axi + vxi

    then x(t)=[tex] \int vx(t)dt[/tex]=[tex] \int J/2*t^2 + axi +vxi [/tex]= J/6*t^3 +axi + vxi + xi

    b) show that ax^2= axi^2 + 2J(vx-vxi)

    now..what do I add???

    You said that it was:
    When you add the constants, this will make sense.

    eg:

    ax = Jt + axi

    vx = (J/2)t^2 + axi*t + vxi etc...



    Am I going in the right direction with this??
    b/c would I use the end equation for x or add them all together?
     
  9. Sep 24, 2007 #8

    learningphysics

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    that's not right. it should be J/2*t^2 + axi*t + vxi
     
  10. Sep 24, 2007 #9

    ~christina~

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    see below
     
    Last edited: Sep 24, 2007
  11. Sep 24, 2007 #10

    ~christina~

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    Oh..I forgot that..well is right now?

    x(t)=[tex]\int vx(t)dt [/tex]= J/6*t^3+ axi/2*t^2+ vxi*t + xi

    but how do I get what they want me to get?? since they have ax and axi and etc in the equation?:bugeye:
     
  12. Sep 24, 2007 #11

    learningphysics

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    Yeah, that looks right.

    you should be able to do part b) using your vx and ax formulas.
     
  13. Sep 24, 2007 #12

    learningphysics

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    work out ax^2 using your formula for ax.
     
  14. Sep 24, 2007 #13

    ~christina~

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    I'm stumped.....I really don't know where I get ax^2 from that equation....
    do I use originally what I got for the indef interals of each?
    thus

    ax= Jt +axi
    vx= J/2*t^2 + axi*t + vxi
    x= J/6*t^3+ axi/2*t^2+ vxi*t + xi

    I'm not sure do I add them together...all of the equations??
     
  15. Sep 24, 2007 #14

    learningphysics

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    ax = Jt + axi

    ax^2 = J^2t^2 + 2Jt(axi) +axi^2 = 2J(J/2*t^2 + axi*t) + axi^2

    Do you see how you can get vx-vxi in the above?
     
  16. Sep 24, 2007 #15

    ~christina~

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    Yes since you can plug in

    vx= J/2*t^2 + axi*t + vxi
    vx-vxi= J/2*t^2 + axi*t so you substitute that into

    ax^2 = J^2t^2 + 2Jt(axi) +axi^2 = 2J(J/2*t^2 + axi*t) + axi^2

    and get...

    ax^2 = J^2t^2 + 2Jt(axi) +axi^2 = 2J(vx-vxi) + axi^2


    I get that now...I got confused from your previous post that said I had to add constants..so it was simply square the original ax...
     
  17. Sep 24, 2007 #16

    learningphysics

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    Yeah, sorry... I just meant to include constants (axi, vxi, xi).

    Anyway, that looks right. good job.
     
  18. Sep 24, 2007 #17

    ~christina~

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    yay!

    thanks for all your help
     
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