Astudent sitting in a merry-go-round has an acceleration of 3.6 m/s2. if the tangential velocity of the student is 2.5 m/s, what is the distance of the student from the center of the merry-go-round?

The distance of the student from the center of the merry-go-round is, r = 1.74 m

Explanation:

Given,

The acceleration of the student in merry go round, a = 3.6 m/s²

The tangential velocity of the student is, v = 2.5 m/s

The acceleration of the merry go round is given by the formula,

                                a = v² / r

Therefore,

                                 r = v² / a

                                    = 2.5² / 3.6      

                                    = 1.74 m

Hence, the distance of the student from the center of the merry-go-round is, r = 1.74 m

there are no components of a physical goal mentioned on the list of choices that you've provided.