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Non-constant acceleration

  1. Feb 23, 2012 #1
    What would be an example equation of a position function that has non-constant acceleration?

    I know y = y + vt + .5at^2 is constant acceleration, but why? is it because its to the second power?

    Does that mean that y = y + vt + .5at^3 describes a motion with a non constant acceleration?
     
  2. jcsd
  3. Feb 23, 2012 #2

    gneill

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    You should use other variable names or subscripts to differentiate initial value constants from the variables. Thus, for example:

    ## y = y_o + v_o t + \frac{1}{2} a t^2 ##

    Now, position y is a function of t. As such, how would you obtain the acceleration corresponding to such a function? (HINT: involves a bit of calculus)
     
  4. Feb 23, 2012 #3
    Have you taken calculus yet? It will probably answer your question once you start to work with derivatives.
     
  5. Feb 23, 2012 #4
    double prime gives a formula for acceleration. So are you saying that if i take the second derivative and end up with a constant, then it must be constant. And since the second derivative of a position equation of the third power will be y''(t)= at, then acceleration won't be constant...thus, y = y_0 + v_0t + at^3 is a position function of which acceleration is not constant? Is that what you are getting at?

    P.S. how do you do the latex reference? It looks different from using the physics forums reference list.
     
    Last edited: Feb 23, 2012
  6. Feb 23, 2012 #5

    gneill

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    Yup. That's it.
    The Latex interpreter allows several different tags to bracket formulas. '##' is one such, and is equivalent to '[itex]'.
     
  7. Feb 23, 2012 #6
    Thanks

    Latex interpreter? Is that a seperate program or source used to make the latex process a little more efficient/easy?
     
  8. Feb 23, 2012 #7

    gneill

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    Every PF post displayed by your browser is automatically interpreted for Latex. You don't need any separate programs or tools. Anything you place between the appropriate tags will be so interpreted. Of course you'll have to learn a bit of Latex syntax to make use of it...
     
  9. Feb 23, 2012 #8
    Yea, I have to get better/faster with the Latex syntax.

    Btw, another question about constant acceleration as it would relate to angle of a vector:

    If I have a position function with an x component of zero acceleration, and a y component of a position function is of constant acceleration (raised to the second power), and then I graph the acceleration vector on an acceleration vs acceleration graph, since the acceleration vector in the y direction will be constant and the acceleration vector in the x direction is zero, does that mean that my acceleration vector angle will always be 90 degrees?
     
    Last edited: Feb 23, 2012
  10. Feb 23, 2012 #9

    gneill

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    When you plot y position vs x position of, say, a particle in motion, you usually do it as y(t) vs x(t), yielding a curve that represents the trajectory of the particle over some time period. A position vector at any given instant would be drawn from the origin to a point on that trajectory.

    If the accelerations in the x and y directions are both constant for all time (and zero acceleration is just a constant acceleration that happens to be zero), then the entire "trajectory" for such a plot would be a single point on the ay axis, and only one acceleration vector could be drawn: from the origin to that point. So yes, such an acceleration vector would always have an angle of 90 degrees (w.r.t. to the ax axis).
     
  11. Feb 23, 2012 #10
    Okay, going on a tangent here (no pun intended), i'm not sure I understand what you mean by "(and zero acceleration is just a constant acceleration that happens to be zero)" If acceleration is zero, as in x = x[itex]_{0}[/itex] + v[itex]_{x}[/itex]t + 0, then how can it be considered a "constant acceleration" if it doesn't exist? Is it because although it has no value (a value of 0), it consistently has a value of zero throughout the entire trajectory, thus can be referred to as being "constant?"
     
  12. Feb 23, 2012 #11

    gneill

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    Don't confuse language conventions with mathematical equations :smile: Zero is a perfectly good value for a variable to have. If it's a constant zero then you have the option of dropping its term from the equation. An object with constant velocity has acceleration = 0.
     
  13. Feb 23, 2012 #12
    constant just means that it's always the same

    if it's always zero, then it's always the same, and thus it is constant
     
  14. Feb 24, 2012 #13

    Thanks. So if i'm understanding how these equations should work, if I happen to have a position function with a non zero constant acceleration on both the x and y axis, then i'm going to have an acceleration vector that is neither vertical nor horizontal, but at some angle, right?
     
  15. Feb 24, 2012 #14

    gneill

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    Yes. Acceleration is a vector quantity. As such it has magnitude and direction just like any other vector, and can be decomposed into x and y components.
     
  16. Feb 24, 2012 #15
    Okay, now that I am understanding the acceleration vector better, I am confused or unsure about ALL of an acceleration vector's components.

    So, as in the example with a position function of zero constant acceleration for the x component and a position function of some positive constant acceleration, the acceleration vector will be vertical with no x components.

    However, it does have a component of centripital acceleration and some other acceleration, right?
     
  17. Feb 24, 2012 #16

    gneill

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    The components that you extract from a vector depend upon your choice of coordinate system. Rotating a Cartesian coordinate system about the origin through some angle does not change a vector in any way (magnitude and direction in space), but it does change the sizes of the components that are 'projected' onto its axes.

    Centripetal acceleration assumes a polar coordinate system with a center of rotation. In certain circumstances it is convenient to erect a 'temporary' polar coordinate system to analyze a particular point along the trajectory of an object that is not otherwise engaged in circular motion. In such a case a small (differential element size) section of the trajectory about the point in question can be interpreted as having a curvature corresponding to a radius. This is known as the 'instantaneous radius of curvature' for that point on the trajectory. If the instantaneous velocity of the particle is known for that point, then along with the radius of curvature an instantaneous centripetal acceleration can be calculated.

    Such an acceleration is not an additional acceleration affecting the object; it is simply another interpretation, via a different coordinate system, of the net forces acting on the object.
     
  18. Feb 24, 2012 #17
    Okay, thanks, I understood that very well. Never covered the term polar coordinates in trig, but if thats analagous to the unit circle, then i'm comfortable with the polar coordinate system.

    So, being that I feel we are on the same page thus far, could we switch to a new page..in otherwords, there is a thread that I started in which the topic seems to be evolving towards the same details we are discussing here in this thread. However, the discussion here seems to be progressing more productively than the other thread. thread

    So along the same lines of what we are discussing here, could you take a look at post # 3 of this thread, https://www.physicsforums.com/showthread.php?t=580771 and let me know if you identify any confusion i'm having? If all sounds correct, then I'll feel more comfortable moving on to maybe angular acceleration topics.
     
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