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Creating the Graph of a Position Function

  1. Dec 16, 2009 #1
    1. I need to create a graph of a position function where a particle at point pi, has zero velocity, and positive acceleration.

    2. Relevant Equations f(x)=sin(x); f'(x)=cos(x); f''(x)=-sin(x)

    3. Attempts at a Solution If I choose sin(x) as my function, then velocity is zero at pi. I am unclear if I have this next part right. Since the slope of the velocity function is increasing at pi, acceleration is therefore positive. I think I have that right. But I am not sure.
  2. jcsd
  3. Dec 16, 2009 #2
    You're answer is right infront of you. You defined f'(x). We know f'(x) describes the slope of a curve at any given point. You want to know if the slope at the point pi is positive or negative, so compute f'(pi).
  4. Dec 17, 2009 #3
    Sorry for being dense, but I am just not real sure what you are getting at. If f(x) = sin(x), the value of pi at f'(x) is -1, but the slope is 0, so there is zero velocity. (Do I have it right so far?)

    I am not understanding what it gets me to compute the value of f'(pi).

    The value of f''(x) = 0, but function has a positive slope at that point, so the acceleration is increasing. (Do I have that right?)
  5. Dec 17, 2009 #4
    If you are plotting f'(x): the value at f'(pi) is the velocity and the acceleration is either the slope which you can evaluate from your plot, or which can be solved explicitly by finding f''(x) and evaluating at that value. In other words what is -sin(pi)?

    The equation you have listed comes close to satisfying the required properties but not quite.
  6. Dec 17, 2009 #5
    Ok, so since the value of the second derivative at pi is a negative value, then the acceleration is negative.? So If I start with f(x)=cos(x), then f''(x)=1 so then the acceleration is positive. Right?
  7. Dec 17, 2009 #6
    check your work: what is sin (pi). (Make sure your calculator is in radians).

    Let me save you some time by suggesting this: plot sin(x), cos(x),-sin(x) which are
    f(x), f'(x), and f''(x) in this case. You can do the same with f(x)=cos(x) but the results will be the same.

    With sines and cosines functions the first derivative will be 90 degrees phase shifted and the second derivative 180.

    What this means is that the acceleration will always look like the mirror image of the position function.

    There are many, many solutions to this problem, but you will have to make your initial function more complicated. One possibility would be set it up with two terms whose derivatives add up to a value of zero when x=pi, but have at least one non zero second derivatives at x=pi.
  8. Dec 17, 2009 #7
    By x you mean position in space?
    Then the velocity is not the derivative in respect to x but in respect to time.
    Maybe you mean something like this
    y(t) - position function (gives position versus time)
    y'(t) - velocity
    y''(t) - acceleration.
    Then you require that when y=pi, y'(t)=0 and y''(t)>0

    Is this what you need?
  9. Dec 17, 2009 #8
    What we want is that f'(x) = -sin(x) so f''(x) = -cos(x). If we have this then the velocity is zero and the acceleration is positive.(It's at a maximum specifically)

    What f(x) can we use to obtain this result?
  10. Dec 17, 2009 #9
    You don't have anything like this. There is no speed or acceleration unless x varies in time.
    If you actually mean "find a function of x so that the first derivative is zero and the second derivative is positive" then it makes sense. Not any derivative is a velocity.
    If the question is about derivatives only, then f(x)=cos(x) satisfies the conditions;
    f'(x)=-sin(x) = 0 at x=pi
    f''(x)=-cos(x) =1 >0 at x=pi
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