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Uniform Accelerated Motion

  1. Feb 3, 2013 #1
    1. The problem statement, all variables and given/known data


    A car decelerates uniformly and comes to a stop after 10 s. The car's average velocity during deceleration was 50 km/h. What was the car's acceleration while slowing to a stop?

    {Since the acceleration is UNIFORM (constant), use vavg = (v0 + v)/2..... }



    2. Relevant equations



    3. The attempt at a solution

    V0 is not known, Final V = 0, t = 10, Vavg is 13.9 m/s and trying to find a

    But when I try to plug it in I feel I have to solve for v0
     
    Last edited: Feb 3, 2013
  2. jcsd
  3. Feb 3, 2013 #2

    tms

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    Start by writing down the equations that describe the car's motion.
     
  4. Feb 3, 2013 #3
    The equation I come up with was 13.9(10) = (v0 - 0/2)(10)
    But that equals to -27.8 m/s^2 which is not the right answer
     
  5. Feb 3, 2013 #4

    SammyS

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    Do it one step at a time.

    You have, vavg = (v0 + v)/2 , and you know vavg ≈ 13.9 m/s and v = 0 . So solve that for v0 .

    After that it's pretty straight forward to find the acceleration.
     
  6. Feb 3, 2013 #5

    tms

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    Write the equations using symbols, not numbers. It is easier to see what is happening with the physics that way.
     
  7. Feb 3, 2013 #6
    So 13.9m/s = (v0 + 0)/2 = 23.7 m/s
    V0≈23.7 m/s
    V≈0
    t≈10
    Then Δv/Δt..

    0-23.7m/s
    -----------
    10 s

    = -2.37 m/s2
     
  8. Feb 3, 2013 #7

    SammyS

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    Looks good !
     
  9. Feb 4, 2013 #8
    No, there still seems to be an arithmetic mistake :tongue:

    [itex]V_{0}[/itex][itex]\approx[/itex]27.8 m/s
     
  10. Feb 5, 2013 #9
    I'm not sure why I typed that but I know the answer is correct.
     
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