A speedboat moving at 29 m/s approaches a no-wake buoy marker 98.3 m ahead. The pilot slows the boat with a constant acceleration of -3.61 m/s2 by reducing the throttle. How long does it take the boat to reach the buoy? (in s)

d = Vi t + (1/2) a t^2

98.3 = 29 t - 1.8 t^2

1.8 t^2 -29 t + 98.3 = 0
solve quadratic for t

[ 29 +/- sqrt ( 841 - 708) ] /3.6

[ 29 +/- 11.5 ] / 3.6

= 4.86 (ignore the big time, it is on the way back :)

To find the time it takes for the boat to reach the buoy, we can use the kinematic equation:

\(v_f = v_i + at\)

where:
\(v_f\) is the final velocity of the boat,
\(v_i\) is the initial velocity of the boat,
\(a\) is the acceleration of the boat, and
\(t\) is the time taken.

In this case:
\(v_f = 0\) m/s (since the boat comes to a stop when it reaches the buoy),
\(v_i = 29\) m/s,
\(a = -3.61\) m/s^2.

Substituting the values into the equation, we have:

\(0 = 29 + (-3.61)t\)

Rearranging the equation to solve for \(t\):

\(-3.61t = -29\)

Dividing both sides by \(-3.61\):

\(t = \frac{-29}{-3.61}\)

Using a calculator, we find that:

\(t \approx 8.03\) seconds

Therefore, it takes approximately 8.03 seconds for the boat to reach the buoy.