A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 rad/s².The child is standing 4.65 m from the center of the merry-go-round.What is the magnitude of the linear acceleration of the child? A)8.05 m/s² B)7.27 m/s² C)2.58 m/s² D)3.46 m/s² E)4.10 m/s²

Question

Quizplus · Accepted Answer

The linear acceleration is the vector sum of the tangential acceleration (a_t = r * α) and the centripetal acceleration (a_c = r * ω^2). Calculating these: a_t = 4.65 m * 0.745 rad/s² = 3.46 m/s² and a_c = 4.65 m * (1.25 rad/s)² = 7.27 m/s². The magnitude of the total linear acceleration is √(a_t² + a_c²) = √(3.46² + 7.27²) ≈ 8.05 m/s².

A Child Is Riding a Merry-Go-Round That Has an Instantaneous