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    mustafahalil's Avatar
    mustafahalil Posts: 1, Reputation: 1
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    #1

    Jun 4, 2009, 02:49 PM
    Accident reconstruction
    A car skids on a dr , level road surface for distance of 17.5 it.s final speed was 24kmh
    Skid testing with the same car gives a coefficient of friction 0.73


    (I) how much further would the car have skidded had it contiued on until it stopped?

    (ii) Had the car been travelling at initial speed of 72kmh what distance would it have skidded (a) to 24kmh and (b) to stop:confused:
    carcrashexpert's Avatar
    carcrashexpert Posts: 25, Reputation: 2
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    #2

    Jun 4, 2009, 05:37 PM

    For the following, I assume you mean 17.5 m (you didn't specify).

    Additional skid distance:
    V^2 = Vo^2 + to a S
    so, S = (V^2 - Vo^2) / (2 a)
    Where V, final speed, is zero, and Vo, initial speed, is 24km/h (divide by 3.6 to convert to m/s). Multiply friction coefficient f by gravity (9.81 m/s^2) to get acceleration, a, in m/s^2.
    S = ((24/3.6)^2) / (2*9.81*.73) = 2.3 m

    Skidding from 72 to 24, use the same equation:
    V^2 = Vo^2 - to a S
    (24/3.6)^2 = (72/3.6)^2 - 2*9.81*.73*S
    S = 24.8 m

    Total skid distance without impact is straightforward, just add the two distances. (27.1 m)

    Hope that helps. More than that, you have to hire me! ;)

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