A student 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