Final Answer:
The maximum acceleration magnitude of the horse is 0.6 m/s², and the minimum acceleration magnitude is -0.6 m/s².
Step-by-step explanation:
The maximum acceleration magnitude of the horse can be found using the formula a_max = rω² + rα, where r is the distance from the center of rotation, ω is the angular velocity, and α is the angular acceleration. Given that the merry-go-round is rotating at a constant 0.4 rad/s, we can calculate the angular acceleration using the formula α = dω/dt = 0, as there is no change in angular velocity over time. Therefore, the maximum acceleration magnitude is a_max = rω^2 = 7 * (0.4)² = 1.12 m/s².
To find the minimum acceleration magnitude, we use the same formula but with the opposite direction of motion, which gives us a_min = -rω² = -7 (0.4)² = -1.12 m/s². However, since the horse moves up and down according to y = 1.5 sin(θ) meters, we need to find the second derivative of y with respect to time to obtain the acceleration in terms of θ. The second derivative of y with respect to time gives us a = -1.5ω²sin(θ), and since sin(θ) ranges from -1 to 1, the minimum acceleration magnitude occurs when sin(θ) = -1, resulting in a_min = -1.5 (0.4)² * (-1) = -0.6 m/s².
This analysis shows that the maximum acceleration magnitude of the horse is 1.12 m/s² and the minimum acceleration magnitude is -0.6 m/s².