14. A buoy oscillates in simple harmonic motion $y = A \cos(\omega t)$ as waves move past it. The buoy moves a total of 4.5 feet (vertically) between its low point and its high point. It returns to its high point every 8 seconds. Write an equation describing the motion of the buoy if it is at its high point at $t = 0$. Find the rate of change of the vertical motion of the buoy at the instant $t = 3$, and explain how you know the direction (up or down or neither) of the buoy at that instant?
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5 feet between its low point and its high point, so the amplitude of the motion is $A = 4.5/2 = 2.25$ feet. Show more…
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