Question

A speed boat moving at a velocity of 25 m/s runs out of gas and drifts to a stop 3 minutes later 100 meters away. What is its rate of deceleration?

Answer 1

Answer:

The rate of deceleration is -0.14$m/s^{2}$

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$m/s^{2}$

The negative value of a shows that the boat is decelerating.

Therefore, the rate of deceleration of the speed boat is 0.14$m/s^{2}$