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Simple question about velocity and acceleration

  1. Aug 27, 2005 #1
    I'm writing a lab report and came to the following conclusion, I was hoping someone might be able to verify it for me.

    The average velocity during the time interval [A,B] will be equal to the instantaneous velocity at the time (A+B)/2 if the acceleration during the time interval [A,B] is a constant.

  2. jcsd
  3. Aug 27, 2005 #2
    yeah, thats true
  4. Aug 27, 2005 #3

    Andrew Mason

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    The graph of velocity - time will have a slope a= v/t that is constant. In other words, it is a straight line with slope a.

    The definition of average speed is: [tex]v_{avg} = \frac{distance_{total}}{time_{total}}[/tex]

    What is the total distance covered (d = vt) in the time interval [A,B]? What is the total time?

    [Hint: think of distance as the area under the graph]

  5. Aug 28, 2005 #4
    Thanks Dave.

  6. Aug 28, 2005 #5
    The total distance covered would be the integral of velocity, and the total time would be d/v, right?

  7. Aug 28, 2005 #6

    Andrew Mason

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    The time would be B-A.

    [tex]t=\frac{distance_{total}}{v_{avg}}= B-A[/tex]

    [tex]d = \int_A^B vdt = v_A(B-A) + \frac{1}{2}(v_B - v_A)(B-A)[/tex]

    [tex]d/t = v_{avg} = d/(B-A) = v_A + \frac{1}{2}(v_B - v_A) = v_A + \frac{1}{2}(at) [/tex]
    [tex]v_{avg} = v_A + \frac{1}{2}(a(B-A)) = v_A + a\frac{(B-A)}{2}[/tex]

    which is the speed at time (B-A)/2.

  8. Aug 29, 2005 #7
    Thanks Andrew.

  9. Aug 29, 2005 #8


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    d/v certainly has UNITS of time,
    but you really NEED to specify what distance you mean
    (midpoint? location at mid-time? total distance?)
    and what velocity you mean
    (slowest=v_A ? fastest=v_B ? average? v at mid-point?)

    time-average of velocity = total displacement/total duration
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