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My possible misunderstanding on a freely falling object

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  1. Sep 5, 2015 #1
    Consider the famous kinematics equations $$x_f=x_i+v_i\cdot t+\frac12at^2$$ My question is: does this equation work for every point A & B along the parabola formed when dropping an object?
     
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  3. Sep 5, 2015 #2

    Dale

    Staff: Mentor

    Yes. Each point is represented by a different t. To make that clear, I would write it as:

    $$x(t)=x_0+v_0 t+\frac{1}{2}a t^2$$
     
  4. Sep 5, 2015 #3
    Thanks for your answer. But I have something different in mind. Suppose that the point A has ##x_{A,0}## and ##v_{A,i}##, then the equation will have the form $$x_A(t)=x_{A,i}+v_{A,i}\cdot t+\frac12at^2$$ and for a point B which has ##x_{B,0}## and ##v_{B,i}## then the equation will have the form $$x_B(t)=x_{B,i}+v_{B,i}\cdot t+\frac12at^2$$ Will this yield the same values in that for instance if we use them to determine what will happen after ##t=5## as long as we know what ##x_{A,0}##, ##v_{A,i}##, ##x_{B,0}##, and ##v_{B,i}## are (and the time difference between the two)?
     
  5. Sep 5, 2015 #4

    Doc Al

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    Why don't you try it and see for yourself?
     
  6. Sep 5, 2015 #5
    I tried it with three points, always worked well (except the times when I made some mistakes -- of course), however I'm more interested into a rigorous proof.
     
  7. Sep 5, 2015 #6

    Doc Al

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    Should be doable with a bit of algebra. Go for it!
     
  8. Sep 5, 2015 #7
    I know it must be a simple thing I'm missing, but I can't figure it out, here's my try. Consider two points A & B such that ##x_{A,i}=R+x_{B,i}## and ##v_{A,i}\neq v_{B,i}## (for generality), then I should show that
    1. it implies that ##x_{A,f}=R+x_{B,f}##
    2. it implies that the final velocities wont change (how do I formulate that?)
    I arrived at the following equations $$v_{A,f}-v_{B,f}=v_{A,i}-v_{B,i}\\x_{A,f}-x_{B,f}-R=t(v_{A,i}-v_{B,i})$$
    And for that equality to happen we must have ##v_{A,i}-v_{B,i}=0## but this contradicts our assumption that ##v_{A,i}\neq v_{B,i}##, since clearly if -- for example -- you have an object falling onto the ground, if you take two distinct points in its trajectory, the velocity at each point wont be the same.

    So what am I missing?

    (here's a visualization of the situation)
     

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  9. Sep 5, 2015 #8

    Dale

    Staff: Mentor

    The given form of the equation only works for ##x_0## and ##v_0## defined at ##t=0##. At ##t=0## the object is only at one point. If you use the same formula for a point which is on the parabola at a different time then you are going to get a different parabola which has that point at ##t=0##.

    In other words, unless ##A=B## the equations that you have written represent two distinct parabolas. I.e. the first equation passes through ##A## at ##t=0## and the second equation passes through ##B## at ##t=0##, they are not the same parabola even if both parabolas pass through ##A## and ##B##.

    Also, I am not sure if you are thinking about parabolas through space or parabolas in spacetime. The equation you have written is for 1D motion, so it is only a parabola in spacetime.
     
  10. Sep 5, 2015 #9
    Ah, forgot about time, I should have added the following constraints ##t_B=T+t_A##, but again this doesn't yield much...
     
  11. Sep 5, 2015 #10

    Dale

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    So are you thinking of parabolas in space parametrized by time or are you thinking of parabolas in spacetime?
     
  12. Sep 5, 2015 #11
    I'm thinking of parabolas in space parametrized w.r.t time. (like the one I drawed)
     
    Last edited: Sep 5, 2015
  13. Sep 6, 2015 #12

    Doc Al

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    Why don't you treat each axis separately? I would think that would be simpler.
     
  14. Sep 6, 2015 #13
    I only treated the system w.r.t to a coordinate system in which the ##x##-axis is vertical.
     
  15. Sep 6, 2015 #14

    Doc Al

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    Ah, so you are only talking about the vertical axis (as I first thought) despite the mention of parabolas. OK.
     
  16. Sep 6, 2015 #15

    Dale

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    I am not quite sure why you want to put yourself through this, but OK. So a parabola in space can be defined by the parametric equations:
    ##x(\tau)=p \tau^2 + h##
    ##y(\tau)=2 p \tau + k##
    ##z(\tau)=0##
    Where the vertex of the parabola is at ##(x,y,z)=(h,k,0)##, and ##p## indicates how flat the parabola is. The x direction is vertical, the y direction is horizontal along the arc of the parabola and the z direction is perpendicular to the parabola.

    ##\tau## is not time, it is a somewhat arbitrary parameter. To use this approach then you would need to solve for the relationship between ##\tau## and ##t##. That would involve some more free parameters which would allow you to have the same spatial parabola represented by the same equations in ##\tau## independently of when the object passes through any given points on the parabola.

    I leave the details of figuring out the relationship between time and ##\tau## to you.
     
  17. Sep 6, 2015 #16
    How would this equation work for parabola? This is a straight line equation isn't it?
     
  18. Sep 6, 2015 #17

    Doc Al

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    The motion is along a straight line. But expressed as a function of time, the equation describes a parabola.
     
  19. Sep 6, 2015 #18
    The confusion here is that you have defined too many variables. Point i or 0, is a chosen initial point on the trajectory. A and B are two other points on the same trajectory. So there is no separate initial position for A and a separate one for B.
     
  20. Sep 7, 2015 #19
    E
    Equation is of straight line but it explains parabola?? How come?
     
  21. Sep 7, 2015 #20

    Doc Al

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    If you plot x as a function of time, it is not a straight line.
     
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