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Need help distinguishing between Postion, Speed, Velocity, and Acceleration

  1. Sep 19, 2005 #1
    I'm having a very hard time with these concepts. I do know that the derivative of Postion is Velocity and the differentiation of Velocity is Accleration.

    Most of which I have a hard time with is connecting the relationship between these terms, on graphs. I know that the graph of a postion will only increase/decrease when the Velocity is not equal to 0.


    Use the above Postion graph for example. How could you possibly find the instantenous velocity (or velocity for that matter) of the Postion graph at time = 3s? How can you tell that Velocity is increasing on a Position graph?

    I can't seem to define the Acceleration on a Velocity graph. What's funny is that I know that a Velocity with a linear line will have a Acceleration graph showing a constant line on 0; anything beyond this is confusing. Is it true that Acceleration will always be 0 unless the slope of the Velocity changes in a certain time interval? Maybe someone on here can help explain along with some graphs?

    Lets say that a ball has velocity of 3 m/s at time = 2s and acceleration is a constant 9.8 m/s/s. How would you figure out the Velocity of the ball at time = 3s? Do you just add 9.8 to 3 to find out your Velocity at time = 3s? What does m/s/s essentially mean?

    What is Speed technically? Speed is the magnitude of Velocity but Velocity is how fast it displaces over time. Is Velocity simply Speed + Direction = Velocity?

    In my textbook, the instantenous velocity is reffered to as "velocity" and the instantaneous acceleration "acceleration". Does this mean that an instantenous velocity and instantenous acceleration graph is no different than their respective "velocity" and "acceleration" graphs? Or is it that I'm not grasping the idea behind instanteous and instantenous velocity well enough?
    Last edited: Sep 19, 2005
  2. jcsd
  3. Sep 19, 2005 #2


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    you can think of velocity as spped in some direction,
    but I like to treat it as a list of 3 speeds, one in each direction.

    The velocity vs time graph is the key graph:
    the velocity at any instant is the slope of the location graph,
    and the slope of the velocity graph is the acceleration then.

    So, acceleration is zero unless the velocity graph is sloped.
    . . . the v(t) graph slope doesn't have to *change*.

    Experimentally we usually measure _average_ velocities
    and _average_ accelerations ... so your textbook just
    wanted to let you know that it will treat time intervals
    as "infinitessimal" - tiny - unless otherwise specified.
    In fact, the graph has velocities at every instant (right?)
    so the v(t) graph must be "instantaneous" velocities.

    For practical purposes, the average or instant is not a big deal.

    Learning to recognize that velocity produces displacement,
    and that slope of velocity implies aceleration - those are big.
  4. Sep 19, 2005 #3
  5. Sep 19, 2005 #4

    James R

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    If you're dealing with position-time, velocity-time and acceleration-time graphs the relationships are:

    The velocity at a particular time is the gradient of the tangent to the position-time curve at that time.
    The acceleration at a particular time is the gradient of the tangent to the velocity-time curve at that time.

    The change in velocity between two times is the area under the acceleration-time curve between the two times.
    The change in position between two times is the area under the velocity-time curve between the two times.
  6. Sep 20, 2005 #5


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    Velocity and acceleration at an "instant" is a tricky subject- in fact calculus was invented specifically to deal with it! To find the velocity "at t= 3" from the graph you give, you would like to replace the graph with a straight line- that's the situation where the velocity is a constant and its just "change in distance over change in time". The idea here is to replace the graph with its tangent line at t= 3. If you knew the formula, the derivative would allow you to do that (the derivative is the slope of the tangent line). Since all you are given is a graph, you have to "eyeball" it. It looks to me like if you put a straightedge on the graph at t= 3 and rotated it until it went "the same direction" as the graph there, it would go through (4, 2.5) (the graph crosses t= 4 lower than 2.5 but it "tails off" near t=4) and through (2, 0.5). If that's correct, distance changes from 0.5 to 2.5, a difference of 2.0 m, while time changes from 2 to 4, a change of 2 s. The speed at t= 3 must be about 2m/2s= 1 m/s. Now, as I said, that's "eyeballing"- it's only approximate and other people may see it differently- but hopefully close to 1 m/s
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