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21 July, 14:04

A speedboat moving at 31.0 m/s approaches a no-wake buoy marker 100 m ahead. the pilot slows the boat with a constant acceleration of - 4.00 m/s2 by reducing the throttle. (a) how long does it take the boat to reach the buoy? s (b) what is the velocity of the boat when it reaches the buoy? m/s

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  1. 21 July, 14:21
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    U = 31 m/s, the initial speed of the boat

    s = 100 m, distance to the buoy

    a = - 4 m/s², the acceleratin (actually deceleration)

    Part (a)

    Let t = time required to reach the buoy.

    Use the formula ut + (1/2) at² = s.

    (31 m/s) * (t s) - (1/2) * (4 m/s²) * (t s) ² = (100 m)

    2t² - 31t + 100 = 0

    Solve with the quadratic formula.

    t = (1/4) [31 + / - √ (31² - 800) ]

    = 10.92 s or 4.58 s

    Before selecting the answer, we should determine the velocity at the buoy.

    Part (b)

    When t = 10.92 s, the velocity at the buoy is

    v = (31 m/s) - (4 m/s²) * (10.92 s) = - 12.68 m/s

    Because of the negative value, this value of t should be rejected.

    When t = 4.58 s, the velocity at the buoy is

    v = (31 m/s) - (4 m/s²) * (4.58 s) = 12.68 m/s

    This value of t is acceptable.

    Answer (to nearest tenth):

    (a) The time is 4.6 s

    (b) The velocity is 12.7 m/s
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