Question

A speedboat moving at 32.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of

−2.85 m/s²

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?

Answer 1

Initial velocity of the speedboat, u = 32.0 m/s

distance to be travelled, s = 100 m

acceleration, a = -2.85 m/s^2

(a) use the expression -

s = u*t + (1/2)*a*t^2

put the values -

100 = 32t + 0.5*(-2.85)*t^2

=> 1.42t^2 - 32t + 100 = 0

So, t = [ 32 + sqrt{32^2 - 4*1.42*100}] / (2*1.42) = [ 32 + sqrt{1024 - 568}] / (2.84)

= [32 + 21.3] / 2.84 = 18.8 sec

Other value of t is -

t = [ 32 - sqrt{32^2 - 4*1.42*100}] / (2*1.42) = [32 - 21.3] / 2.84 = 3.8 sec

First value of 't' is much larger because it is the time during the returning journey.

So, our answer is t = 3.8 sec.

(b) Velocity of the boat when it reaches the buoy is -

v = u + a*t

=> v = 32.0 + (-2.85)*3.8 = 32.0 - 10.83 = 21.17 m/s