What is the acceleration at point A on a rotating merry go round?

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The acceleration of point A on a rotating merry-go-round can be determined by considering both its radial and tangential components. Since point A is moving outward with a constant velocity V while the merry-go-round rotates with a constant angular velocity ω, the total acceleration is a combination of centripetal and tangential acceleration. The centripetal acceleration is given by the formula a_c = ω²a, where 'a' is the distance from the center O. Additionally, the tangential acceleration is zero if the speed V is constant. Therefore, the total acceleration at point A is primarily centripetal, directed towards the center O.
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Ok so I have this circular merry go round and its rotating with a constant angular velocity \omega. There is a point A on the merry go round moving radially outward with a constant velocity V. Let the center of the circular merry go round be O. I want to find the acceleration of point A with respect to the inertial frame O.

anyone have any ideas?
 
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