A speedboat moving at 28 m/s approaches a no-wake buoy marker 91 m ahead. The pilot slows the boat with a constant acceleration of 4 m/s2 by

Question

A speedboat moving at 28 m/s approaches a no-wake buoy marker 91 m ahead. The pilot slows the boat with a constant acceleration of 4 m/s2 by reducing the throttle. What is the velocity (in m/s) of the boat when it reaches the buoy

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Trung Dũng 3 weeks 2021-08-26T04:57:32+00:00 1 Answers 0 views 0

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    2021-08-26T04:58:44+00:00

    Answer:

    7.5 m/s

    Explanation:

    We can find its velocity when it reaches the buoy by applying one of Newton’s equations of motion:

    v^2 = u^2 + 2as

    where v = final velocity

    u = initial velocity

    a = acceleration

    s = distance traveled

    From the question:

    u = 28 m/s

    a = -4 m/s^2

    s = 91 m

    Therefore:

    v^2 = 28^2 + 2 * (-4) * 91\\\\v^2 = 784 + -728 = 56\\\\v = \sqrt{56}\\ \\v = 7.5 m/s

    The velocity of the boat when it reaches the buoy is 7.5 m/s.

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )