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Weight of vehicles when stopping

  1. Oct 5, 2007 #1
    1. The problem statement, all variables and given/known data

    Is it possible that a vehicle weighing 2000 kg and a vehicle weighing 13000 kg can both use the same skid to stop formula?? Does the difference in weight not come into account?


    2. Relevant equations

    Where :
    S = speed in km/h
    µ = .75
    d = skid distance

    S = 15.9 *( sgr ( µ d ))

    3. The attempt at a solution

    I know how to do the above equation. It is the derivation I do not know how to do.
    in order to prove or disprove the weight effect on the speed.
     
  2. jcsd
  3. Oct 6, 2007 #2
    solved myself

    Thanks, but I proved this one myself, with the use of kinetic energy. But if someone would like to check this over to make sure I am correct I would appreciate it.

    13000 kg vehicle leaves skid marks 21 meters long, the drag factor is .7

    13000 * 21 * .7 = 191100 joules

    V = sqr ((2gKE)/w)
    V = sqr ((2* 9.81*191100))/13000)
    V = sqr (3749382/13000)
    V = sqr 288.414
    V = 16.98 m/s
    S = 16.98 / .2777
    S = 61.14 km/h

    2000 kg vehicle leaves skid marks 21 meters long, the drag factor is .7

    2000 * 21 * .7 = 29400 joules

    V = sqr ((2gKE)/w)
    V = sqr ((2* 9.81*29400))/2000)
    V = sqr (576828/2000)
    V = sqr 288.414
    V = 16.98 m/s
    S = 16.98 / .2777
    S = 61.14 km/h

    Check skid to stop

    S = sqr (254 µ d ))
    S = sqr (254 * .7 * 21))
    S = sqr 3733.8
    S = 61.10 km/h
     
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