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Solving distance in this problem. Easy

  1. Feb 27, 2013 #1
    During the braking test of. 1988 car, the car came to rest from an initial velocity of 96km/h (w) in 3.0 seconds. Assuming that deceleration remains constant.

    How far did the car travel after brakes were applied?

    I calculated the deceleration. It is 8.9m/s (e)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 27, 2013 #2

    berkeman

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    Staff: Mentor

    Keep going...
     
  4. Feb 27, 2013 #3
    I used this equation to solve for Distance.

    Vf^2 - vi^2/ 2 aav

    VHF is final velocity and vi is initial velocity, aav is average velocity

    =0^2 - 96^2/ 2 (8.9)

    =-517.75

    ...
     
  5. Feb 27, 2013 #4
    There
     
  6. Feb 27, 2013 #5

    SammyS

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    Gold Member

    This equation is missing something.[/QUOTE]

    One thing is an equal sign (so it's not an equation).

    It's missing something else also.
     
  7. Feb 27, 2013 #6
    Are you sure? This seems wrong to me (units?)
     
  8. Feb 27, 2013 #7
    8.9 is the deceleration.

    My question is

    How far did the car travel after brakes were applied?
     
  9. Feb 27, 2013 #8

    CWatters

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    Nope. As sjb-2812 has suggested the units are wrong. It should be 8.9 m/s2

    I would use a different equation of motion...

    http://en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations

    eg..

    V2 = U2 + 2as

    so

    s = (V2 - U2)/2a
     
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