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Instantaneous Velocity

  1. Mar 19, 2007 #1
    Hello everyone, my teacher taught me there are two methods of finding instantaneous velocity but I didn't understand her.

    One method was tangents and the other was average velocity at half-time. Could you please explain these methods to me? (in terms of graphical use)
  2. jcsd
  3. Mar 19, 2007 #2
    Well, that is incredibly situational and the way you explained it doesn't make a ton of sense.

    I'm guessing that what you mean is, when you plot displacement vs. time on a graph, the tangent at any point will have a slope that is the instantaneous velocity.

    I don't know what you mean by the other method. There are tons of ways to find a velocity, literally hundreds, it is all situational.
  4. Mar 19, 2007 #3
    Ya, I meant that when we plot a distance-time graph, we need to find the instantaneous velocity. I understand how to find the slope of the tangent, but the other method involves finding the midpoint. If you know what the midpoint method is, please do tell me. Thanks.
  5. Mar 19, 2007 #4
    I don't know of any systematic midpoint method that works under many conditions. Are you assuming anything to be constant?
  6. Mar 20, 2007 #5
    I feel I should point out that finding the average velocity of an object will not give you the instantaneous velocity of the object. In fact, this difference is precisely what makes the concept of instantaneous velocity so important. Unless you've taken calculus, one good way to find the instantaneous velocity of an object is to draw a tangent line to its displacement vs. time graph. Or, if an object is moving at a constant acceleration, you can use the following kinematic equations.

    [tex]s = s_{0} + v_{0}t + \frac{1}{2}at^2[/tex]

    [tex]v = v_{0}t + at[/tex]

    [tex]v^2 = v_{0}^2 +2a\left(s - s_{0}\right) [/tex]

    If an object experiences a constant acceleration, then these equations can be used to compute the instantaneous velocity for an object given the elapsed time or the displacement.

    Incidentally, there is a certain theorem in mathematics, the Mean Value theorem which says that given some time interval, a moving object's instantaneous velocity will be equal to its average velocity over the entire interval at at least one time in the interval. But the Mean Value Theorem doesn't tell you how many times this will happen, or at what time it will happen.
  7. Mar 20, 2007 #6
    The 2nd method can be applied only when the acceleration a is constant. Then the velocity is calculated as :


    and if you calculate the velocity at midpoint of the to=0 and t1, you have:
    vad = vo+(a*t1)/2= 1/2(2vo+a*t1) = 1/2[vo+(vo+a*t1)] =1/2(vo+v1)
    That is the average of vo and v1.
    vad : instant velocity at midpoint
    t1 : the time when you have velocity v1

    Or more simply, you can graph the v against time, there will be a triangle.
  8. Mar 20, 2007 #7


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    Ah, excellent! Haiha has it exactly. I was wondering myself what your teacher could be talking about but that has to be it.

    Although, however, it is going the other way. If an object has instantaneous velocity v0 at one point and instantaneous velocity v1 at another point and has constant acceleration then its average velocity between those two points is (v1+ v2)/2. Of course, if you know the velocity at one point and the average velocity, you can use that to find the instantneous velocity at any point (still with the assumption of constant acceleration).
    Last edited: Mar 20, 2007
  9. Mar 20, 2007 #8
    Thank you very much to all of you. I haven't taken calculus yet, but thanks to HallsofIvy it makes since because my teacher was talking about that too. I remember she talked about dividing the two points but it wasn't clear to me. I now understand that there must be constant acceleration to do the adding of velocity and dividing by 2. Thanks everyone.
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