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Simple finding linear acceleration

  1. Jun 3, 2014 #1
    1. The problem statement, all variables and given/known data

    In our lab, we times a cart going 1m. The time was 9.49 s. The cart was pulled by a string, over a pulley, attached to a free-falling mass. The purpose was to calculate the μ, the coefficient of friction.

    2. Relevant equations

    d = vit + 1/2at[itex]^{2}[/itex]

    a = (vf-vi)/t

    vi = 0 m/s
    d = 1.0 m
    t = 9.49 s

    3. The attempt at a solution

    Why does the acceleration come out differently in these two equations?

    Is it that the vf is not simply 1m/9.49 seconds?
    Is vf = 1/9.49 = 0.11 ?

    Because that is the average velocity, and we don't use average velocity because we have a constant acceleration?

    In the first equation I get a = 0.02m/s[itex]^{2}[/itex]
    and in the second equation I get a = 0.01 m/s[itex]^{2}[/itex]

    I think I know why, but I would like someone to say,"yes that is why you must use the first equation."
     
    Last edited: Jun 3, 2014
  2. jcsd
  3. Jun 3, 2014 #2

    Nathanael

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    Homework Helper

    [itex]V_{f}=9.8t[/itex]

    It's invalid to say [itex]V_{f}=\frac{d}{t}[/itex] because, as you said, that is the average velocity (which is less than the final velocity)



    Since initial velocity is zero, the expression [itex]\frac{d}{t}[/itex] will only give you half the final velocity. That is why your two accelerations were different by a factor of 2.
     
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