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Stopping distance and time of a vehicle

  1. Sep 22, 2012 #1
    1. The problem statement, all variables and given/known data

    The coefficient of friction is .48. The brakes are applied to a 1.05 x10^3 car travelling at 95km/hr. How far does the car move before it stops ? How long does it take for the car to stop?

    2. Relevant equations

    Fg = Fn, Fa= u x Fn, a=Fa/m, d=Vf^2-Vi^2/2a, t=d/v

    3. The attempt at a solution
    Fg = Fn =-9.81m/s^2 x 1050 kg = 10300 N
    Fa = u x Fn = .48 x 10300 N = 4944.2 N
    a = F/m = 4944.2 N / 1050 Kg = 4.71 m/s
    d = (Vf^2 - Vi^2)/2a =
    95 km/hr converts to 26.4 m/s
    so if Vf = 0 then -26.4 ^2 = 697.0 m/s^2
    697.0 m/s^2 / 2 x 4.71 m/s = 74.0 m
    then t = d/v so time is 74m/ 26.4 m/s = 2.80 secs

    Im just not convinced I have done this correct...?
     
  2. jcsd
  3. Sep 22, 2012 #2
    You have computed the distance correctly but your time is wrong. You should divide by the average velocity or use:

    dist = V0*t + .5*a*t^2, where a<0
     
  4. Sep 22, 2012 #3
    t=v/a
    ma=mgμ

    t=v/gμ
    Thus time taken for a body to stop depends only on initial velocity, coefficient of friction.
    Mass is not in the equation.
     
  5. Sep 22, 2012 #4
    Going back to my post, the average velocity is half the initial velocity which doubles your answer for the time. The time also is the root of:

    dist = V0*t + .5*a*t^2, where a<0 or -4.71 m/sec^2 which you computed above.
     
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