Centripetal merry-go-round problem

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The problem involves a man on a rotating merry-go-round with a speed of 3.43 m/s and a centripetal acceleration of 1.87 m/s². To find the radius from the center, the equation Ac = v²/r is applied. The calculated radius is 6.29 meters using the formula (3.43²/1.87). Clarification is provided that the terms "centripetal acceleration of magnitude" and "speed" refer to the magnitudes of their respective vectors. The solution process confirms that the speed can indeed be used as the velocity in the equation.
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Homework Statement


A carnival merry-go-round rotates about a vertical axis at a constant rate. A man standing on the edge has a constant speed of 3.43 m/s and a centripetal acceleration of magnitude 1.87 m/s2. How far is the man from the center of the merry-go-round?


Homework Equations


Ac= v^2/r


The Attempt at a Solution


I understand I am looking for the radius in the equation, I am just not sure about the other aspects of it. I get a radius of 6.29 the way i do it by: (3.43^2/1.87). Not confident though.
 
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Looks fine to me.
 
alright, so the constant speed of 3.43 m/s can be plugged in as the velocity? and I am confused on the wording "centripetal acceleration of magnitude... "
 
"...centripetal acceleration of magnitude 1.87 m/s2" means that the magnitude of the centripetal acceleration vector is 1.87.
 
Similarly the speed is the magnitude of the velocity vector, so yes you use the speed for v in the equation.
 

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