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19 August, 09:00

The combination of crumple zones and air bags and seat belts might increase the distance over which a person stops in a collision to as much as 1.00 m. what is the force exerted on a 65.0-kg driver who decelerates from 18.0 m/s to rest over a distance of 1.00 m?

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  1. 19 August, 11:49
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    From Newton's second law of motion,

    F = ma

    We know m = 65 kg. Next, we have to find a through this formula:

    v² = v₀² + 2ad

    0² = (18 m/s) ² + 2 (a) (1 m)

    Solving for a,

    a = - 162 m/s²

    Thus,

    F = (65 kg) (-162 m/s²) = - 10,530 Newtons
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