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

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