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Instantaneous Acceleration?

  1. Jan 25, 2004 #1
    Is there such a thing as instantaneous acceleration or is it always zero? Problem gives: x(t)=8.5t^2-2t+6 and asks for instantaneous acceleration at t=3

    Any help would be greatly appreciated
     
  2. jcsd
  3. Jan 25, 2004 #2

    Doc Al

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    The instantaneous acceleration is just the acceleration at a particular time. In this case, it is certainly not zero. (Acceleration is the second derivative of distance with respect to time.)
     
  4. Jan 25, 2004 #3
    thank you so with the distance equation x(t)=8.5t^2-2t+6 the instantaneous acceleration is 17?
     
  5. Jan 25, 2004 #4

    Doc Al

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    Yep. And in this case, the acceleration is a constant: it doesn't even depend on the value of t.
     
  6. Jan 25, 2004 #5
    thank you so much
     
  7. Jan 28, 2004 #6
    how do you know it's a constant, and when would it not be constant?
     
  8. Jan 28, 2004 #7

    Doc Al

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    If you are given the position function x(t), take the second derivative and see what happens.
     
  9. Jan 28, 2004 #8
    i've learned today of a third derivative, which is a change or "jerk" in the acceleration. (a rate of change in a change of position) wouldn't that be changing always and never a constant?
     
  10. Jan 28, 2004 #9

    Doc Al

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    Not at all. It depends. In the example that started this thread, the "jerk" is zero at all times.

    Think of these simple examples:
    - a body at rest = v=0, a=0, jerk=0
    - a body moving with constant speed (in straight line): v=constant, a=0, jerk=0
    - a falling body (no air resistance, etc.): v= changing, a=constant, jerk=0
     
  11. Jan 28, 2004 #10
    Let us review first the definition of acceleration, and the question asked.

    given a position [tex] x [/tex] which varies with time.
    We know that the derivitve of that [tex] \frac{dx}{dt} [/tex] is the instantious time rate of change (derivitive) of the position, which is given the name velocity. From that we take the derivitive of that, so you have [tex] \frac{d^2x}{dt^2} [/tex] or the acceleration at an instant in time (given position)

    Now, we are asked what the acceleration at [tex] t = 3 [/tex] is. We are given the position function [tex] x(t)=8.5t^2-2t+6 [/tex]. Therefore taking the second derivitive [tex] \frac{d^2x(t)}{dt^2} [/tex] we find the derivitive to be [tex] 17 [/tex]

    Now consider if [tex] x(t) = t - 2 [/tex] in which [tex] \frac{d^2x(t)}{dt^2} [/tex] would be 0. As you can see in the position function, no acceleration occurs, it is simply a constant position change based on time (indicating a constant velocity).
     
    Last edited: Feb 11, 2004
  12. Feb 11, 2004 #11
    forstajh, the derivative is simply 17, not 17 m/s^2. No units were implied.
     
  13. Feb 11, 2004 #12
    Correct notmuch, I apologize for any confusion this may have caused to anyone. I am so used to using m/s^2 that I accidently put them even though no units were provided.

    Thanks for pointing out my error.

    -Jacob
     
  14. Feb 13, 2004 #13
    How does one find the rate of change of acceleration?

    i.e. two planets moving toward each other.
     
  15. Nov 7, 2008 #14
    Silverious, I am quite interested if someone has offered you any answers. please let me know.
    Thank you
     
  16. Nov 7, 2008 #15

    HallsofIvy

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    If you are given the position as a function of time for each object, say x1(t) and x2(t), then the distance between them is |x1(t)- x2(t)|. The rate at which the two objects are moving toward each other is the derivative of that. The acceleration of each object, relative to the other, is the second derivative.

    Now, if you are actually refering to two object, planets, you know that F= ma and that, for gravity, F= GmM/r2 so you can find the acceleration of either one directly from that.
     
  17. Nov 7, 2008 #16
    That is a good explanation. Thanks.
     
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