Question

A speedboat moving at 32.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot...

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?

Homework Answers

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

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