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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
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
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Physics
1 year
2021-08-26T04:57:32+00:00
2021-08-26T04:57:32+00:00 1 Answers
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Answers ( )
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:
[tex]v^2 = u^2 + 2as[/tex]
where v = final velocity
u = initial velocity
a = acceleration
s = distance traveled
From the question:
u = 28 m/s
a = -4 [tex]m/s^2[/tex]
s = 91 m
Therefore:
[tex]v^2 = 28^2 + 2 * (-4) * 91\\\\v^2 = 784 + -728 = 56\\\\v = \sqrt{56}\\ \\v = 7.5 m/s[/tex]
The velocity of the boat when it reaches the buoy is 7.5 m/s.