Question #12344

a speedboat moving at 30.0 m/s approaches a no-wake buoy marker 100 m ahead. the pilot slows the boat with a constant acceleration of -3.50 m/s^2 by reducing the throttle. (a) how long does it take the boat to reach the buoy? (b) what is the velocity of the boat when it reaches the buoy?

Expert's answer

(a) how long does it take the boat to reach the buoy?

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

-3.5T²/2 + 30T - 100 = 0, or

1.75T² - 30T + 100 = 0 ==>

T = (30 ± √(30² - 4*1.75*100))/(2*1.75) = (30 ± √200)/3.5 ≈ 4.5308s or 12.6120s.

We take the less time value. So, it takes near 4.5308s 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 = 30 - 3.5*4.5308 = 14.1422 m/s.

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

-3.5T²/2 + 30T - 100 = 0, or

1.75T² - 30T + 100 = 0 ==>

T = (30 ± √(30² - 4*1.75*100))/(2*1.75) = (30 ± √200)/3.5 ≈ 4.5308s or 12.6120s.

We take the less time value. So, it takes near 4.5308s 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 = 30 - 3.5*4.5308 = 14.1422 m/s.

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