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24 November, 04:12

How fast was a driver going if it left skid marks that were 63 feet long on wet concrete? (The coefficient of friction is 0.97) ?

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  1. 24 November, 06:53
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    Step-by-step explanation:

    Let m be the mass of the vehicle

    frictional force = μ mg

    μ is coefficient of friction,

    deceleration = μ g

    =.97 x 32

    a = - 31.04 ft / s

    v² = u² - 2 a s

    0 = u² - 2 a s

    u² = 2 a s

    = 2 x 31.04 x 63

    = 3911.04

    u = 62.54 ft / s
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