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How to Sketch a v-t Graph from a p-t Graph Using Tangents?

  1. Feb 17, 2015 #1
    Hi! I'm having troubles drawing a velocity-time graph from a position-time graph. I know parabolic p-t graphs have diagonal lines for their v-t graphs, but I'm not sure why. I also know tangents are important to use, but again, I don't understand why. Any clarification as to why this happens (in a way a high school student may understand :wink:) will be greatly appreciated! Thanks!
     
  2. jcsd
  3. Feb 17, 2015 #2

    BvU

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    Δ Hello C, welcome to PF :smile:

    You could look around here and see if you can pick up something to your liking.

    It all has to do with
    displacement = velocity x time.​

    If the velocity is constant, it's easy. Position-time graph is a straight line with slope (tangent!) velocity.

    If the velocity changes you have to consider smaller time intervals and you get something like
    displacement = average velocity x Δtime​
    and if you take Δtime small enough you get the velocity at that moment:
    velocity = displacement / Δtime
    on a position-time graph, displacement in a small time interval is Δp so to get v (for velocity), you take Δp and divide by Δtime
    v = Δp / Δtime​
     
  4. Feb 17, 2015 #3

    lightgrav

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    velocity is the slope of the position-time graph.
    technically, average velocity at the mid-time of the short line whose slope you're calculating.
    that is because vavg = Δx/Δt ... notice : it is rise/run .
    (tangent is just making the line between points so short, that it only seems to touch one point; )
    (it is nice for math, but in real lab experiments you have discrete data points at time intervals. )
    quadratic x(t) graphs have zero slope at some time (how fast is it there?) with + slope or - slope after then.
    these curves are really steep (is that fast, or slow?) at times that are very long time from the turn-around.
     
  5. Feb 17, 2015 #4
    Thanks for the help! The only thing I'm still struggling with is why a parabola (p-t graph) has a diagonal line as a v-t graph. A parabola goes both positively and negatively (positive and negative slope from the tangent lines) but in the v-t graph all that is shown with only a diagonal line (according to the way the parabola is opening). This is what's really tripping me up o_O.
     
  6. Feb 17, 2015 #5

    lightgrav

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    a parabola is really steep at long times from the turn-around.
    The slope on an x(t) graph is the value then on its v(t) graph.
    . . . the slope on a v(t) graph is the value on its a(t) graph ... a parabola x(t) means constant acceleration.
     
  7. Feb 18, 2015 #6

    BvU

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    Remember the v = Δp / Δtime ? Well, Δp can be negative as well as positive.

    If the p-t graph goes down, Δp is negative.
    If the p-t graph is horizontal, the position doesn't change and Δp is zero.
    And if the p-t graph goes up, Δp is positive.

    It can also go in the other direction: If you throw up a ball, its height first increases: v is positive. But this v reduces quickly until the ball is at its highest point, when v = 0 at some moment. After that it falls down, so height decreases and v is negative.
     
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