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Curious about average velocity (v bar)

  1. Oct 29, 2009 #1
    1. The problem statement, all variables and given/known data

    My question uses the following problem. I am confused as to why s=v(bar) * t works out the same as s=(v0-vfinal)/2 * t Enlighten me please!

    Problem: How long does it take to fall 1000ft with no drag? What is the final velocity?

    displacement: s=1000ft
    acceleration: a=9.8 (gravity)
    time: t=?

    2. Relevant equations

    s=v(bar) * t (distance = average rate * time)
    s=(1/2)at2 (distance = one half * acceleration * time squared)
    v=at

    3. The attempt at a solution

    Step 1) Convert 1000ft to 304.8 m
    Step 2) Rearrange s=(1/2)at2 to t=sqrt[2a/d]
    Step 3) Plugin known variables and solve for t giving 7.887 s
    Step 4) Plugin known variables into v=at giving v=77.29 m/s
    Step 5) Check answer with s=v(bar) * t
    304.8 m = (384.8-0)/(7.887-0) * 7.887 s

    I see that step 5 also works if you use this equation s = [(v0-vfinal)/2] * t

    Why is this the case?
     
  2. jcsd
  3. Oct 29, 2009 #2
    (Start - Final) /2 is just a way to compute the average. Take any 2 numbers and divide them by 2 and you get the average of those 2 numbers. V bar is the average velocity.
     
  4. Oct 30, 2009 #3

    Redbelly98

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    Staff Emeritus
    Science Advisor
    Homework Helper

    The expression in paranetheses should be

    (v0 + vfinal)​

    So your question becomes, why is

    vavg = (v0 + vfinal)/2 ?​

    If the acceleration is constant, then the velocity is a linear function of time. For linear functions, the average over some interval equals the average at the two endpoints.
     
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