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Average velocity and acceleration

  1. Sep 2, 2008 #1
    1. The problem statement, all variables and given/known data

    What is the average acceleration between time t = 2.2 s and time t = 9.6 s?
    [​IMG]
    I've already found the average velocity between the two points to be 1.09×10^-1 m/s


    2. Relevant equations

    a= "delta"V / "delta" T
    V= d/t

    3. The attempt at a solution

    v1= 9/2.2 = 4.09
    v2= 10/9.6 = 1.04

    a = (1.04-4.09) / (9.6-2.2) = -3.05 / 7.4 = -.412 m/s^2
     
  2. jcsd
  3. Sep 2, 2008 #2
  4. Sep 2, 2008 #3

    Redbelly98

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    Actually, v is the slope of the position-vs.-time graph.
     
  5. Sep 2, 2008 #4
    Ok, so I tried with the slope between 2.2 and 9.6

    V= (10-9)/(9.6-2.2) = 1/7.4 = .135

    a= .135/7.4 = .0183m/s^2

    still says I'm wrong.
     
  6. Sep 2, 2008 #5
    Got it, the main problem here was it involved calculus and my calc professor was sick the day she would've taught us that (that's a big thing)

    Anyway, I had to take the slopes as it approached 2.2 and 9.6 which I found to be 5 and -10 plug and chug, shoop dah woop and it came out to -2.03 m/s^2 which the lon-capa thing told me yes.

    and I only used up 18 of my 30 tries on it.
     
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