A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 "'s."

What distance is traveled if the person is brought to rest at this rate from 115 ?

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  1. a = 30gs = 30 * 9.8m/s^2 = -294 m/s^2.


    V^2 = Vo^2 + 2a*d.
    d = (V^2-Vo^2)/2a.
    d = (0-1020.4)/-588 = 1.74 m.

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    To simplify what Henry said if you need clarification
    30g's is 30 times the acceleration of the earth within the first 10000 feet of the atmosphere, so in this case -294m/s^2 is the acceleration
    To find the distance convert your velocity to m/s and divide that by 2 times your acceleration
    Distance = (velocity^2)/-588)

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  3. kaon tae

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