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Motion, constant acceleration

  1. Jun 16, 2012 #1
    A car slows down from 23 m/s to rest in a distance of 85m. what was its acceleration, assumed constant?



    a=Δv/Δt x=1/2at^2



    i don't know where to start
     
  2. jcsd
  3. Jun 16, 2012 #2
    what is the velocity-displacement formula?
     
  4. Jun 16, 2012 #3
    v^2=2ax
     
  5. Jun 16, 2012 #4
    in that equation is v the change in velocity?
     
  6. Jun 16, 2012 #5
    Ignore this post, it was bad advice!
     
    Last edited: Jun 16, 2012
  7. Jun 16, 2012 #6
    v^2=2ax
     
  8. Jun 16, 2012 #7
    okay thanks i guess my teacher messed up
     
  9. Jun 16, 2012 #8
    I'm sorry, I just gave you some bad advice...ignore my first response.
     
  10. Jun 16, 2012 #9
    I don't know what I was thinking, but yeah you use (final velocity)^2 = (initial velocity)^2 + 2*a*x and just substitute the stuff you know and solve for a.
     
  11. Jun 17, 2012 #10
    Does anyone know how to derive V^2=V0^2+2as

    Ratch
     
  12. Jun 17, 2012 #11
    Take the velocity-time equation:
    [tex]
    v = v_{0} + a \, t
    [/tex]
    and the position time equation:
    [tex]
    x = v_{0} \, t + \frac{1}{2} \, a \, t^{2}
    [/tex]
    and eliminate time t.
     
  13. Jun 17, 2012 #12
    Work energy theorem
    KEv2-KEv0=mas
     
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