a ferris wheel is 10 meters in diameter and boarded from a platform that is 3 meters above the ground. the…

Explanation:

Step1: Determine the amplitude

The diameter of the ferris wheel is 10 meters, so the radius (amplitude (A)) is (\frac{10}{2} = 5) meters.

Step2: Determine the vertical shift

The platform is 3 meters above the ground, and the six - o'clock position is level with the platform. The center of the ferris wheel is at a height of (3 + 5=8) meters above the ground (since the radius is 5 and the platform is 3 meters high). So the vertical shift (k = 8).

Step3: Determine the period and angular frequency

The wheel completes 1 full revolution in 6 minutes. The period (T = 6) minutes. The formula for the angular frequency (\omega=\frac{2\pi}{T}), so (\omega=\frac{2\pi}{6}=\frac{\pi}{3}). Since the ferris wheel starts at the six - o'clock position (the lowest point), the function is a cosine function with a negative sign (because at (t = 0), we are at the minimum of the cosine function). The general form of a cosine function for height is (h = A\cos(\omega t)+k), but since we start at the minimum, it will be (h=-A\cos(\omega t)+k). Substituting (A = 5), (\omega=\frac{\pi}{3}) and (k = 8) into the formula, we get (f(t)=- 5\cos\left(\frac{\pi}{3}t\right)+8).

Answer:

(f(t)=-5\cos\left(\frac{\pi}{3}t\right)+8)