Centripetal merry-go-round problem

[wegman14]

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.

 

[Kurdt]

Looks fine to me.

 

[wegman14]

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... "

 

[e(ho0n3]

"...centripetal acceleration of magnitude 1.87 m/s2" means that the magnitude of the centripetal acceleration vector is 1.87.

 

[Kurdt]

Similarly the speed is the magnitude of the velocity vector, so yes you use the speed for v in the equation.