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Question

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1. –/2 points Notes Question: SerCP8 2.P.056.
A speedboat moving at 34.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of -3.65 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

Explanation / Answer

Given that a speed of the boat is u = 34.0m/s distance s = 100m   The pilot slows the boat with a constant acceleration is a = - 3.65 m/s^2 b ) the velocity of the boat when it reaches the buoy                 v^2 - u^2 = 2as                v^2 = ( 34.0m/s)^2 - 2 * 3.65 m/s^2 * 100m                v = 20.6 m/s a ) time  it take the boat to reach the buoy is        v = u + at      20.6 m/s = 34.0 m/s - 3.65m/s^2 * t       t = 3.67 s

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