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9 February, 00:07

If such drivers are traveling on a straight road at 30 mph (44 ft/s) and their cars can decelerate at 27 ft/s², determine the shortest stopping distance d for a normal driver from the moment he or she see the pedestrians.

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  1. 9 February, 00:37
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    35.85 ft

    Explanation:

    initial velocity, u = 44 ft/s

    acceleration, a = - 27 ft/s²

    Let the stopping distance is d. the final velocity, v = 0 ft/s

    Us third equation of motion

    v² = u² + 2ad

    0 = 44 x 44 - 2 x 27 x d

    1936 = 54 d

    d = 35.85 ft

    Thus, the stopping distance is 35.85 ft.
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