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Kinematics for car in 2D

  1. May 9, 2009 #1
    1. The problem statement, all variables and given/known data

    The speed limit in a school zone is 40 km/h (about 25 mi/h). A driver sees a child run onto the road 13 m ahead of his car. He applies the breaks, and the car decelerates at a uniform rate of 8.0 m/s^2. If the driver's reaction time is 0.25 s, will the car stop before hitting the child?



    2. Relevant equations

    d = Vot + 1/2at^2
    V^2 = Vo^2 + 2ad



    3. The attempt at a solution

    40 km/h x [1000 m/1 km] x [1 h/3600 s] = 11 m/s

    Vo = 11 m/s
    X = ?
    Xo = 0
    a = -8.0 m/s^2
    t = 0.25 s

    X - Xo = Vot + 1/2at^2
    X = (11 m/s)(0.25 s) + 1/2(-8.0 m/s^2)(0.25 s)^2
    = 2.5 m

    Vo = 11 m/s
    V = 0
    X = ?
    Xo = 2.5 m
    a = -8.0 m/s^2

    V^2 = Vo^2 + 2a(X - Xo)
    X = Xo + [V^2 - Vo^2/2a]
    = 2.5 + [0 - (11 m/s)^2/2(-8.0 m/s^2)
    = 10 m

    Therefore, 2.5 m + 10 m = 12.5 m, and the car will stop in time.

    Seems right, can anybody please verify this outcome?
     
  2. jcsd
  3. May 9, 2009 #2
    Dunno if I'm right, but this is how I see this problem.

    First, I calculate the distance car has traveled in that 0.25 seconds

    [tex]d=v*t=\frac{100m}{9s}*\frac{1s}{4}=\frac{25m}{9}[/tex]

    After that, the car slows down

    [tex]v^{2}=v_{0}^{2}+2ad[/tex]
    [tex]0=\frac{100}{9}^{2}-2*8d[/tex]
    [tex]d=\frac{100*100}{81*2*8}[/tex]

    Now to sum up those 2 distances

    [tex]d=\frac{25}{9}+\frac{10000}{1296}=\frac{13600}{1296}\approx10.5m[/tex]

    But once again, I don't know if I'm rigth (Although I agree the boy will live :biggrin:)
     
  4. May 9, 2009 #3
    Oh, I see where I went wrong. No deceleration during the first part, and a constant velocity. Thank-you for your help!
     
  5. May 9, 2009 #4
    You're welcome :D
     
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