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Slope of position vs time squared graph?

  1. Oct 18, 2015 #1
    1. The problem statement, all variables and given/known data
    What would the slope of a position vs time squared graph represent?


    2. Relevant equations
    X=my^2+b


    3. The attempt at a solution
    I thought the slope of this graph would be velocity due to the fact that the normal position vs time graph slopes are velocity. However, I was told that the slope isn't velocity or acceleration. I'm not sure if it's correct or not?
     
    Last edited by a moderator: Oct 18, 2015
  2. jcsd
  3. Oct 18, 2015 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Your problem is in thinking that such a graph has a "slope". Strictly speaking "slope" is only defined for linear equations. If you mean "x= mt^2+ b" where t is time, measured in, say, seconds, while s is measured in meters, then the slope, not of the graph, but of the tangent line to the graph then we are approximating the quadratic function by a linear function, with x a distance and t time, then the slope of the tangent line is the speed at that instant.

    (I notice that you have used "y" instead of the standard "t" for time so it is possible that y refers to something other than time. If that is true, what does "y" represent?)
     
  4. Oct 18, 2015 #3

    gneill

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    Staff: Mentor

    Why don't you start with unit analysis? For each of the known quantities in the formula plug in a unit that would be associated with it. For example, if X is a distance, plug in the unit "meter". Then since for all physics equations the units must balance, determine what units the "m" in the formula must have in order for things to balance.
     
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