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Finding distance with known constant acceleration

  1. Sep 3, 2008 #1
    1. How far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g?



    2. X=Xo+ VoT and V1=Vo +at



    3. I converted 60g into 588.6 m/s^2 and then plugged it into the V1 equation to Find V1 which turns out to be 21.1896 m/s (after converting 36 ms into 0.036 seconds). I then entered the 21 m/s into the X1 equation and X=0+21.1896 * 0.036. I keep getting x = 0.76m. I am not confident that I am doing this question right or what I should be doing but that is my logic.
     
  2. jcsd
  3. Sep 3, 2008 #2

    DaveC426913

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    Gold Member

    60g? Six, zero, g's?
     
  4. Sep 3, 2008 #3
    yup 60 g's.
     
    Last edited: Sep 3, 2008
  5. Sep 4, 2008 #4

    DaveC426913

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    Gold Member

    You've got v1 and v0 mixed up.

    You are given v1; it is 0. You need to solve for v0. (and then make sure you've got your signs correct for both v and a.)
     
    Last edited: Sep 4, 2008
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