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Simple kinematics question

  1. Sep 28, 2010 #1
    Hi, sorry for not using the template, but my question doesn't really fit.

    Basically, I'm trying to move between displacement-time graphs, velocity-time graphs, and acceleration-time graphs.

    Ex: http://img832.imageshack.us/img832/3391/10113039.jpg [Broken]

    I have a sheet filled out, but it only has some of the different types of accelerated motion (like those in the example), and I'm trying to understand why it moves like it does, rather than just the answer (I wouldn't be able to do it with any examples that aren't on my sheet if I don't know how it works).

    I've been thinking trying to do this for over an hour, but the differences between how slow to fast and fast to slow curves affect the vt/at graphs is so huge, and I can't figure out how exactly it changes it (only the middle one in the screenshot is slow to fast). I've tried other resources (such as google) as well.

    I know that you want people to show an effort, but this isn't really something you can show your work for, and I feel I've done all that I can do; either you know it or you don't, so I was hoping someone could explain (for other examples besides the image as well i.e. fast to slow curves, slow to fast curves, with both positive and negative slope, and both above and below the x-axis).

    PS. I have a test tomorrow. I'd ask my teacher, but I've been sick the last week and he isn't giving me any time to ask questions so I'm on my own..

    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 29, 2010 #2
    Consider the derivatives of the functions on your graphs, that may help
  4. Sep 29, 2010 #3


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    For the first one,

    -Notice that as t increases, the slope of d decreases. Since velocity is the slope of the position graph, what does a zero slope on a distance graph mean in terms of the velocity? Once you have the velocity graph, what would the derivative of the velocity (acceleration) look like?
  5. Sep 30, 2010 #4
    Well if there was a slope of zero on the position graph, you would not be moving
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