Question

A speedboat moving at 28 m/s approaches a no-wake buoy marker 91 m ahead. The pilot slows the boat with a constant acceleration of 4 m/s2 by reducing the throttle. What is the velocity (in m/s) of the boat when it reaches the buoy

Answer 1

Answer:

7.5 m/s

Explanation:

We can find its velocity when it reaches the buoy by applying one of Newton's equations of motion:

$v^2 = u^2 + 2as$

where v = final velocity

u = initial velocity

a = acceleration

s = distance traveled

From the question:

u = 28 m/s

a = -4 $m/s^2$

s = 91 m

Therefore:

$v^2 = 28^2 + 2 * (-4) * 91\\\\v^2 = 784 + -728 = 56\\\\v = \sqrt{56}\\ \\v = 7.5 m/s$

The velocity of the boat when it reaches the buoy is 7.5 m/s.