the blades of a wind - mill turn on an axis that is 30 feet from the ground. the blades are 10 feet long and…

the blades of a wind - mill turn on an axis that is 30 feet from the ground. the blades are 10 feet long and complete 2 rotations every minute.\nwrite a sine model, (y = asin(bt)+k), for the height (in feet) of the end of one blade as a function of time (t) (in seconds). assume the blade is pointing to the right when (t = 0) and that the wind - mill turns counterclockwise at a constant rate.\n(y = 30sin(\frac{pi}{15}t)+10)\n(y = 30sin(\frac{pi}{15}t)+30)\n(y = 10sin(\frac{pi}{15}t)+10)\n(y = 10sin(\frac{pi}{15}t)+30)

the blades of a wind - mill turn on an axis that is 30 feet from the ground. the blades are 10 feet long and complete 2 rotations every minute.\nwrite a sine model, (y = asin(bt)+k), for the height (in feet) of the end of one blade as a function of time (t) (in seconds). assume the blade is pointing to the right when (t = 0) and that the wind - mill turns counterclockwise at a constant rate.\n(y = 30sin(\frac{pi}{15}t)+10)\n(y = 30sin(\frac{pi}{15}t)+30)\n(y = 10sin(\frac{pi}{15}t)+10)\n(y = 10sin(\frac{pi}{15}t)+30)

Answer

Answer:

D. $y = 10\sin(\frac{\pi}{15}t)+30$

Explanation:

Step1: Determine the amplitude $a$

The length of the blade is the amplitude, so $a = 10$.

Step2: Calculate the angular - frequency $b$

The wind - mill makes 2 rotations per minute. One rotation is $2\pi$ radians and there are 60 seconds in a minute. So the angular speed $\omega=2\times2\pi$ radians per 60 seconds. Then $b=\frac{2\times2\pi}{60}=\frac{\pi}{15}$.

Step3: Find the vertical shift $k$

The axis of the wind - mill is 30 feet from the ground, so the vertical shift $k = 30$. The sine model is $y=a\sin(bt)+k$, substituting $a = 10$, $b=\frac{\pi}{15}$, and $k = 30$ gives $y = 10\sin(\frac{\pi}{15}t)+30$.