Answer:
The rate of deceleration is -0.14[tex]m/s^{2}[/tex]
Explanation:
Using one of the equations of motion;
v = u + at
where;
v = final velocity of the boat = 0m/s (since the boat decelerates to a stop)
u = initial velocity of the boat = 25m/s
a = acceleration of the boat
t = time taken for the boat to accelerate/decelerate from u to v = 3 minutes
Convert the time t = 3 minutes to seconds;
=> 3 minutes = 3 x 60 seconds = 180seconds.
Substitute the values of v, u, t into the equation above. We have;
v = u + at
=> 0 = 25 + a(180)
=> 0 = 25 + 180a
Make a the subject of the formula;
=> 180a = 0 - 25
=> 180a = -25
=> a = -25/180
=> a = -0.14[tex]m/s^{2}[/tex]
The negative value of a shows that the boat is decelerating.
Therefore, the rate of deceleration of the speed boat is 0.14[tex]m/s^{2}[/tex]
