Question

0/2 points A fishing boat approaches a caution buoy 100 m ahead at a velocity of 28.0 m/s. The pilot reduces the throttle, skowing the bost with a constant acceleration of -3.20 ms (a) How long (in s) does it take the boat to reach the buoy? Signs are critical. Check the sign for the Initial velocity, the acceleration, and the change in position. Look for sign errors in your expression for the solution to the quadratic. Does a negotive time mak sense? s (b) What is the velocity (in my's) ofthe boat when it reaches the buoy? (Indicate the direction with the sign of your ansaer) m/s Need Help? ane

Answer 1

Using the equation of linear motion we have

S = ut + 0.5at^2

100 = 28t + 0.5*-3.2*t^2

1.6t^2 - 28t + 100 = 0

t = 28 - sqrt(28^2 - 4*1.6*100) / 3.2

= 5s

(B) v = u + at

v = 28 - 3.2*5

= 12 m/s