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 -3.9 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

Answer 1

A)
Find your givens
Vi = 33 m/s
a = -3.9 m/s^2
D = 100m
Vf = ?
t = ?
Now you would pick a eqution with only one unknown
Vf^2 = Vi^2 + 2ad
Plug in the knowns variables
Vf^2 = 33^2 + 2(-3.9)(100)
Vf^2 = 1089 - 780
Vf^2 = 309
Vf = 17.59
Therefor the velocity of the boat when it reaches the buoy is 24.82 m/s.
B)
its easy now to find time since you have every variable needed.
You may use a variety of equations here but i used
t = (vf - vi) / a
t = (17.59 - 33) / -3.9
t = 3.95
Therefor the amount of time it takes for the boat to reach the buoy is 3.95s