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Translating motion graphs. Dt to vt

  1. Mar 2, 2013 #1
    I am so confuse! It's regarding translating motion graphs.

    When translating a position time graph to a velocity time graph, if a curve line is followed by a straight diagonal line (both going up) do I use a step function?

    Cause in this video http://m.youtube.com/#/watch?v=EZXLkAYjmR0&desktop_uri=/watch?v=EZXLkAYjmR0

    At 3:04. He drew a diagnal line without calculating the slope... Isn't he suppose to use a step function?
    1. The problem statement, all variables and given/known data



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    3. The attempt at a solution
     
  2. jcsd
  3. Mar 2, 2013 #2

    PeterO

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    The two sloping sections before and after the curved line can be used to find the constant velocities before and after the acceleration section. The presenter has possibly assumed [for simlicity] that the acceleration between those two sections was constant. Does the curved section look like it may be parabolic? [I can't play the video].
     
  4. Mar 2, 2013 #3
    No, he souldn't use a step function in such a case.
    Curve line in d-t representation means accelerated motion (suppose the curve line is a parabola for simplicity, i.e. uniform acceleration)... then in the v-t representation it is a diagonal line, meaning that velocity is changing at constant rate.

    Then the diagonal line in d-t representation means constant velocity. In v-t representation this is instead a straight horizontal line.

    Therefore what you get is a diagonal line joined to an horizontal line. You would have a step function in v-t representation only in case you have two motions both at constant speed but with different speed (and at some point speed has to change instantaneously but you do not know why, it is not a realistic situation of course); in this case the d-t representation would be just two diagonal lines with different slopes.
     
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