Average Velocity vs Instaneous Velocity

  1. I just need some clarification.

    In a lab that i did, a metal ball goes down a slanted ramp, onto a horizontal ramp of d distance, and then goes airborne (off the table). The instant the ball enters the horizontal ramp a timer is started, and - when it goes airborne - the timer is stopped. Thus, i have t time.

    When i calculate using the formula [tex]v = d/t[/tex], i am calculating instantaneous velocity of the ball the instant it goes airborne correct?

    When i consider where the ball goes onto the of horizontal ramp to be the zero point, i can say that the ball has displaced d distance, right? And using the same logic with the timer, i can say that the change in time is t time, correct? Thus i have Δd and Δt.

    Using the formula [tex]v_{avg} = \Delta d/ \Delta t[/tex], yields me the average velocity correct?

    Thanks in advance.
     
  2. jcsd
  3. tiny-tim

    tiny-tim 26,053
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    hi rainstom07! :wink:
    yes :smile:

    v = d/t is the instantanteous velocity from point 0 to point d

    this only works because the velocity is (we assume) constant …

    it would not work if, for example, there was friction
    yes :smile:

    (is there something that was bothering you about this?)
     
  4. Can you explain why a non-constant velocity doesn't work? For instance, if the ball goes down the slanted ramp, i start the timer when it goes airborne and then stopped the timer when it hits the floor; thus, i obtain t seconds. I then measure the distance between the ramp and the floor; thus, i obtain the vertical d distance.

    can i say that [tex]v_y = d_y/t[/tex] is the instantaneous velocity (y-component) of the ball the instant it hits the ground?

    I thought that if you consider the distance as d, then you cannot consider the distance as the Δd. The same goes with time.

    Thanks for your help!
     
  5. tiny-tim

    tiny-tim 26,053
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    (write "itex" rather than "tex", and it won't keep starting a new line :wink:)
    i'm confused :redface:

    what's dy ? :confused:
     
  6. The height (the distance between the ramp and the floor).
     
  7. tiny-tim

    tiny-tim 26,053
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    no, the instantaneous velocity increases (ie the ball gets faster as it falls further)

    dy/t is the average velocity, which is less than the final velocity :smile:
    you can if the speed is constant

    (but not otherwise)
     
  8. So let me get this straight.

    If you consider the distance as the Δd, then you cannot consider the distance as d for a non-constant velocity (i.e. there is acceleration)?

    I guess this leads to a very interesting question (and probably the source of my confusion). In the equation [tex]v = d/t[/tex], what is the d stand for, physically? It surely cannot be the height; the height is Δd, right?
     
  9. tiny-tim

    tiny-tim 26,053
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    it's really only a dimensional equation …

    it reminds you that speed is distance per time :smile:

    (as a real, useful, equation it only works when speed is constant :redface:)
     
  10. :confused::confused::confused::confused::confused:

    so there's no physical manifestation of the d in the equation for a non-constant velocity? huh?
     
  11. tiny-tim

    tiny-tim 26,053
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    d = ut + 1/2 at2 (constant acceleration)
     
  12. interesting,..but yea d stands for displacement which is different from distance..and the instantaneous time would be a derivative. if time wasnt constant

    i got a question about velocity but i dont know if itll be alright to put it here so ill put it anyways
    ok first imagine you are in a car as passenger moving at constant velocity relative to the ground. up ahead you see a cup sitiing on a table stationary ralative to the ground. as you get closer to the cup you stick your hand out to get it..when you pass by it you grab it but dont stop...then as soon as you get it you let it go so it falls as a projectile,..
    my question is what is the velocity of the cup as soon as you grab it? myguess is that the cup is now in your reference frame and as soon as you grab it its moving at your speed. so the time it takes the cup to go from zero to my speed will be zero?or some small number..my other question is ,so it falls a greater distance from the point of letting go, will it make a difference how long you hang on to the cup ignoring air resistance or will it fall the same distance from the point of letting go? and will my speed stay constant or will i decelerate a litle?you know because of momentum? would it be like running over a fly i dont feel it but i slowed down a bit...many thanks for your time.
     
  13. tiny-tim

    tiny-tim 26,053
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    hi uknew! :smile:
    some small number …

    your hand is soft, so it will deform slightly, which takes time (and even the cup itself will deform very slightly … there's no such thing as a perfectly rigid object :wink:)
    same distance
    conservation of momentum applies in every collision …

    the cup gains momentum, so you-and-the-car lose momentum :smile:
     
  14. thanks tiny tim.
    so say for example that a wheel is moving at speed of sound or some high number ( just pretend) and me being unbreakble and at rest grab onto the outer rim of the wheel, my body will accelerate to the speed of the wheel in a time t..if the wheel is moving at constant velocity when i grab it and continues at constant speed (if at all possible).. how can i calculate the time my body will reach the same speed as the wheel?will i use kinematics, impuse, momentum or change in momentum or change i energy? or all? how wouldi use impulse decipline?since ft=mv, what force would i use? if my only givens are my mass and speed of the wheel and mass of the wheeel.
     
  15. tiny-tim

    tiny-tim 26,053
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    you would have to use the compressibility of the wheel and of your body
     
  16. interesting thanks
     
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