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Acceleration via graph

  1. Aug 4, 2009 #1
    This is a graph which shows time and displacement as function of time. I am told to find the average acceleration for the first ten and sixteen seconds. The answers are : 2.5 and 0.7 cm / s 2 . If the image is not showing below, download the attached image. http://img22.imageshack.us/img22/1050/acceleration.jpg [Broken]

    I found v(10) and v(0) and divided by ten which does not equal 2.5 cm/s^2 or .7cm/s^2. What i did try was however find v(10) and v(8) and divided by 2(1.6 something), which is much closer to 2.5 cm/s^2. Is there some method I am not using?
     

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  3. Aug 5, 2009 #2

    tiny-tim

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    Hi razored! :smile:
    I did that …

    did you remember to convert from m to cm? :wink:
     
  4. Aug 5, 2009 #3

    kuruman

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    Also, how did you get v(10) and v(0) from the graph? What assumptions did you make?
     
  5. Aug 5, 2009 #4
    I think you can't quite find the velocity as a function of time, since that would require a much better graph, anything you do would be merely an estimate. And I think that goes for both the thread starter and tiny-tim.
    I don't think the purpose of the exercise was for you to use a ruler to measure [tex]\frac{\Delta s}{\Delta t}[/tex] and then take the limit as best you can where [tex]\Delta t \rightarrow 0[/tex]
    Though I may be wrong, and that is the approach the question is asking for, this seems out of the scope of the question, and far from the ideal solution. Did you recently learn about the calculus-based definitions of displacement, velocity and acceleration? And were you taught how to derive (HA, PUN) each from a graph of one of them, even when the function isn't known?

    For at true AVERAGE acceleration, I suggest that you look at the graph of [tex]s(t)[/tex] as though it were the product of motion under constant acceleration. That is to say, that it is the product of motion under the constant average acceleration.

    If you were to travel for the same time under the constant, average, acceleration, you would achieve the same displacement as you would traveling under the varying acceleration, that is, to the best of my knowledge, the definition of the displacement-averaged acceleration.

    What formula do you know tying displacement with acceleration? How would you extrapolate the displacement-averaged acceleration from the graph, once you know that general formula?
     
    Last edited: Aug 5, 2009
  6. Aug 5, 2009 #5
    You can get v(10) quite accurately from the graph since it goes linear at 10s.Also, you can get a reasonable estimate of v(16)
     
  7. Aug 5, 2009 #6

    tiny-tim

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    I agree :smile:

    RoyalCat, don't be so negative! :rolleyes:
     
  8. Aug 5, 2009 #7
    Hehe, didn't even notice it turned linear at [tex]t=10[/tex]
    I suppose you could make that case. I wonder, how would the two results compare?

    The following is wrong.

    Would they be the same? It seems to me that your solution is just like mine, only it involves a bit more grunt-work, so to speak. :)

    You're finding [tex]\bar a[/tex] through [tex]v(t)=v_0+at \rightarrow a=\frac{v(t)-v_0}{t}[/tex]
    While I'm using the displacement as a function of time instead.
    Are the two equivalent, or am I missing something?

    EDIT:
    Bah, I was working outside of conventional definitions. Average acceleration is defined as [tex]\bar a\equiv \frac{\Delta v}{\Delta t}[/tex], while I chose to define it as the constant acceleration that would provide an equivalent displacement. Given an initial velocity of 0, the expression trims down to [tex]\bar a=\frac{2\Delta s}{(\Delta t)^2}[/tex]

    Really sorry if I confused the thread starter!

    Heh, I'm getting [tex]1.4 \tfrac{cm}{s^2}[/tex] and [tex]0.8 \tfrac{cm}{s^2}[/tex]

    My method is completely wrong as it doesn't take into consideration additional data, such as known velocities, so it produces an average acceleration that fits a displacement, but doesn't fit the velocities. While your method provides an average acceleration that fits the velocities, but doesn't fit the displacement, heh.
     
    Last edited: Aug 5, 2009
  9. Aug 5, 2009 #8

    kuruman

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    I think RoyalCat's initial suggestion, "For at true AVERAGE acceleration, I suggest that you look at the graph of as though it were the product of motion under constant acceleration. That is to say, that it is the product of motion under the constant average acceleration" is to the point. For any interval starting at 0 s when the initial velocity is zero to final time tf, the displacement can be written as
    [tex]\Delta x=\frac{1}{2}a_{ave} (\Delta t)^{2}[/tex]

    Proof
    Let vf = velocity at the end of the interval. Since the initial velocity is zero at t=0,

    [tex]a _{ave}= \frac{v_{f}}{\Delta t} \rightarrow v_{f}=a_{ave}*\Delta t[/tex]

    and from the kinematic equation

    [tex]\Delta x = \frac{1}{2}(0+v_{f})\Delta t = \frac{1}{2}a_{ave}(\Delta t)^{2}[/tex]

    If you put in this expression, the numbers that razored reports as being the answers, you don't get the displacements in the graph. Either the above definition for the average acceleration is not what the problem-maker had in mind, or the answers have been incorrectly reported.
     
  10. Aug 5, 2009 #9

    tiny-tim

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    You two are making this really complicated! :cry:

    I think the question just intended average acceleration = (change in speed)/time …

    especially since it's to be read straight off the graph. :smile:
     
  11. Aug 5, 2009 #10

    kuruman

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    I agree, but what is "it" that can be read straight off a position vs. time graph?
     
  12. Aug 5, 2009 #11

    tiny-tim

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    The speed

    it's just the slope, which you can get by counting the squares (up vs. along).​
     
  13. Aug 5, 2009 #12

    kuruman

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    For me it is the position, especially when the graph crosses intersection points on the grid. Vive la difference.:smile:
     
  14. Aug 5, 2009 #13
    Haha, I'm afraid you haven't read my latest post on the subject. I came to the conclusion I was operating under an incorrect definition.

    [tex]\bar a\equiv \frac{\Delta v}{\Delta t}[/tex] for most intents and purposes.

    I think the only reason we're quarreling over whether to average the acceleration with respect to displacement, or with respect to velocity, is because the velocity at [tex]t=16 sec[/tex] is not immediately apparent. If it were, it would be rather clear that the exercise was meant to find velocity as the slope of the graph and then average the acceleration with respect to velocity, but since the best approximation we can get for [tex]v(t=16 sec)[/tex] we can get is "a reasonable estimate," as dadface put it, I think the accurate displacement-averaged acceleration might have its place here too.

    Well, we are just making it overly complicated, but the instructions should be a it more clear, or the available data a bit more unambiguous, :rofl:

    The problem with [tex]v(t=16 sec)[/tex] is that in order to measure it, you'd need to measure [tex]\lim_{\Delta t\rightarrow 0}\frac{\Delta s}{\Delta t}[/tex], and that course of action is FAR from obvious.

    Ah well, the textbook's answers make it quite clear that they were asking the students to use the [tex]\bar a\equiv\frac{\Delta v}{\Delta t}[/tex] definition.

    On the bright side, I think this might be an interesting point for the thread starter to bring up in class, to show the difference between the different kinds of averages you can use on the graph.
     
  15. Aug 5, 2009 #14

    kuruman

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    I don't think anyone is quarreling. We are just trying to figure out this thing to help razored who is probably sitting back enjoying this exchange. I cannot claim to be able to read the mind of the problem-maker, but let me ask you this. How would you solve the following two problems if only numbers are given but no graph?

    An object moving in a straight line starts from rest and reaches position 40 cm in 10 s. What is the object's average acceleration? Answer: 0.8 cm/s2

    As opposed to

    An object moving in a straight line starts from rest and reaches velocity 23 cm/s (as best as I can read it) in 10 s. What is the object's average acceleration? Answer: 2.3 cm/s2

    My point here is that one can only guess what the problem-maker thinks is given by showing the graph.
     
  16. Aug 5, 2009 #15
    I was just using colorful language, not to be taken seriously in the least bit. :tongue:

    The thing is that the graph gives both of these things, and the leading definition for average acceleration is [tex]\bar a\equiv\frac{\Delta v}{\Delta t}[/tex]

    So yes, this problem is ambiguous. ^^; But the answers make it pretty obvious which solution the question was aiming for.
     
  17. Aug 5, 2009 #16

    kuruman

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    Not at all taken seriously.

    It is better to give a hint for getting to the answer as opposed to the answer itself. Working backwards from an answer is not conducive to acquiring good habits.
     
  18. Aug 5, 2009 #17
    Well, unfortunately, that's the only way of separating the wheat from the chaff in this case. :x
    Working by strict definitions aside, that is.
     
  19. Aug 5, 2009 #18
    I understand what you mean by TRUE acceleration. I guess the author was intending to find instantaneous velocities,which can be pretty accurately approximated, over a time interval.


    Thank you for the advice/help.
     
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