A speedboat moving at 33.0 m/s approaches a no-wake buoy marker 100 m ahead. The

Question

A speedboat moving at 33.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of ?3.20 m/s2 by reducing the throttle. (a) How long does it take the boat to reach the buoy? (b) Find the total time the bullet is in contact with the board.

Explanation / Answer

Let D be the distance, a the acceleration, V the initial velocity and T the time. Then D = VT + aT²/2, or aT²/2 + VT - D = 0. Substituting given values we obtain -3.2T²/2 + 33T - 100 = 0, or 1.6T² - 33T + 100 = 0 ==> T = (33 ± v(33² - 4*1.6*100))/(2*1.6) = ˜ 3.718s or 16.93s. We take the less time value. So, it takes near 3.718s the boat to reach the buoy. (b) what is the velocity of the boat when it reaches the buoy? Let Vf be the velocity ob a boat when it reaches the buoy: Vf = V + aT = 33 - 3.2*3.712= 21.1 m/s.

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