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Kinematics - How fast does the car stop?

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  1. Feb 11, 2014 #1
    Kinematics -- How fast does the car stop?

    1. The problem statement, all variables and given/known data
    A car is moving with a speed of 32.0 m/s. The driver sees an accident ahead and slams on the brakes, giving the car an acceleration of -3.50 m/s2. How far does the car travel after the driver put on the brakes before it comes to a stop?

    vi=32 m/s

    vf=

    t=

    a=-3.5 m/s2

    delta-x=



    2. Relevant equations



    3. The attempt at a solution
    How would this problem be solved? Does it require two steps?

    Is Vf=0m/s and Vi= 32 m/s?

    Because if that's the case, wouldn't I use the formula a=vf-vi/t

    So:

    -3.5 m/s2=(0 m/s - 32 m/s)/t

    -3.5 m/s2 = (-32 m/s)/t

    t= -32 m/s/-3.5 m/s2

    t= 9.14 s

    Then to find distance, use s=d/t, rearrange as d=s*t

    d=32 m/s*9.14 s

    d=292 m
     
  2. jcsd
  3. Feb 11, 2014 #2

    lewando

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    Gold Member

    Your "t" is correct, but your "d" is not. d=s*t is valid only when you have a constant s (rate), which is not the case here-- s is decreasing. Pick a relevant kinematic equation and you will be home free.
     
  4. Feb 11, 2014 #3
    d=0.5(vi+vf)*t
    d=0.5(0 m/s + 32 m/s)*9.14 s
    d=146.24 m
     
  5. Feb 12, 2014 #4

    lewando

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    Gold Member

    Looks good. Watch your significant digits.
     
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