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Distance, velocity & Acceleration

  1. Feb 22, 2012 #1
    taking derivation of distance equation is easy as every body knows.
    differentiating distance we get velocity and differentiating velocity we get acceleration
    but if a bird fly from one tree to another of distance 50m in 3 sec. we can get it's velocity by
    v=s/t but how can we get it's acceleration ?
  2. jcsd
  3. Feb 22, 2012 #2


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    hi otomanb! :smile:
    if v = s/t, then the acceleration is zero :wink:
  4. Feb 22, 2012 #3
    But how ?
  5. Feb 22, 2012 #4


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    In calculating that velocity as "distance divided by time", you are calculating the average velocity, a constant. Since the velocity does not change, there is no acceleration.
  6. Feb 22, 2012 #5
    Only when the function x=f(t) is given, can we use differentiation
  7. Feb 22, 2012 #6
    Try thinking this way: there are many ways that the bird could have flown the 50 meters in 3 seconds. It could have rapidly accelerated to say 17 m/sec and flown the whole way there at that speed, and then quickly slowed down at the end. Or it could have gradually accelerated through the first 25 meters and then gradually slowed during the final 25 meters. Or it could have done a crazy flight speeding up and slowing down repeatedly. So as the others have said, the 50/3 = 16.7 m/sec is just the average, and to find the acceleration you need to know the actual location at each point along the way. (That's azureth's function; f(t)).
  8. Feb 22, 2012 #7
    Because v=s/t gives us the constant value and derivation of constant value gives us ZERO answer.
  9. Feb 22, 2012 #8
    but if i use equation of motion
    it give me some answer like 2.78m/s^2 :confused:
  10. Feb 22, 2012 #9


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    vi = 0, but you don't know vf

    you'll have to use another constant acceleration equation, s = vit + 1/2 at2 :wink:

    btw, is this a rocket-powered bird?

    what makes you think it flies with constant acceleration? :biggrin:
  11. Feb 22, 2012 #10
    not constant acceleration with constant "Velocity" and if velocity is constant acceleration is always ZERO
  12. Feb 22, 2012 #11


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    i'm confused …

    then why did you use a constant acceleration equation? :confused:
  13. Feb 22, 2012 #12
    sorry dont know that it's constant acceleration equation
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