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21 October, 05:00

During a car collision, the knee, thighbone, and hip can sustain a force no greater than 4000 N. Forces that exceed this amount could cause dislocations or fractures. Assume that in a collision a knee stops when it hits the car's dashboard. Also assume that the mass of the body parts stopped by the knee is about 20% of the total body mass. The person with a mass of 65 kg is initially traveling at 18 m/s (40 mi/h).

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  1. 21 October, 05:28
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    a) t = 0.0585 s

    b) s = 0.5265 m

    Explanation:

    Given:

    - maximum possible force F = 4000 N

    - mass of person = 65 kg

    - mass of knees and other m = 0.2*65 = 13 kg

    - initial velocity u = 18 m/s

    - Final velocity v = 0

    Find:

    a) Minimum stopping time

    b) Minimum stopping distance.

    Solution:

    Minimum stopping time can be evaluated by Newton's second law of motion. Where change in momentum per unit time is Average force experienced by the body:

    F = m * (v - u) / dt

    dt = 13 * (0 - (-18)) / 4000

    Hence, dt = 0.0585

    Minimum stopping distance can be evaluated by using equation of motions and required deceleration is as follows:

    a = - 4000 / 13

    a = - 307.692 m/s^2

    And,

    s = (v^2 - u^2) / 2*a

    s = (0 - 18^2) / 2 * (-307.692)

    s = 0.5265 m

    Answer: The minimum stopping time and distance are t = 0.0585 s and s = 0.5265 m, respectively.
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