5. mamadou takes a trip to an amusement park a ferris wheel. the ferris wheel has a maximum heigh 440 and a…

5. mamadou takes a trip to an amusement park a ferris wheel. the ferris wheel has a maximum heigh 440 and a minimum height of 20. at time ( t = 0 ) min mamadous car first reaches the bottom of the ferris wheel. the next time it reaches the bottom of the ferris wheel occurs at a time of ( t = 10 ) minutes. the graph below shows two full revolutions on the ferris wheel, as well as a dashed midline, starting at time ( t = 0 ). write an equation relating ( y ), mamadous height in feet above the ground, and ( t ), time in minutes, to represent the given context.

5. mamadou takes a trip to an amusement park a ferris wheel. the ferris wheel has a maximum heigh 440 and a minimum height of 20. at time ( t = 0 ) min mamadous car first reaches the bottom of the ferris wheel. the next time it reaches the bottom of the ferris wheel occurs at a time of ( t = 10 ) minutes. the graph below shows two full revolutions on the ferris wheel, as well as a dashed midline, starting at time ( t = 0 ). write an equation relating ( y ), mamadous height in feet above the ground, and ( t ), time in minutes, to represent the given context.

Answer

Explanation:

Step1: Determine the midline ( M )

The midline ( M=\frac{\text{Max}+\text{Min}}{2}=\frac{440 + 20}{2}=\frac{460}{2}=230).

Step2: Determine the amplitude ( A )

The amplitude ( A=\frac{\text{Max}-\text{Min}}{2}=\frac{440-20}{2}=\frac{420}{2} = 210).

Step3: Determine the period ( T )

The period ( T = 10) (since the time between two consecutive bottom - points is (10) minutes). The formula for the angular frequency ( \omega=\frac{2\pi}{T}), so ( \omega=\frac{2\pi}{10}=\frac{\pi}{5}).

Step4: Determine the phase shift ( \varphi )

Since at (t = 0), the function is at its minimum. For a cosine function (y = A\cos(\omega t-\varphi)+M), when (y=\text{Min}) at (t = 0). The general form of a cosine function (y=A\cos(\omega t-\varphi)+M). We know that (y = A\cos(\omega t-\varphi)+M). When (t = 0), (y=20). Substituting (A = 210), (M = 230), (\omega=\frac{\pi}{5}) into (y=A\cos(\omega t-\varphi)+M), we get (20=210\cos(-\varphi)+230). Then (\cos(-\varphi)=\cos(\varphi)=\frac{20 - 230}{210}=- 1). So (\varphi=\pi).

The equation of the cosine function is (y = A\cos(\omega t-\varphi)+M). Substituting (A = 210), (\omega=\frac{\pi}{5}), (\varphi=\pi), and (M = 230) into the formula, we get (y=210\cos\left(\frac{\pi}{5}t-\pi\right)+230). Using the identity (\cos(a - b)=\cos a\cos b+\sin a\sin b), (\cos\left(\frac{\pi}{5}t-\pi\right)=-\cos\left(\frac{\pi}{5}t\right)). So (y=-210\cos\left(\frac{\pi}{5}t\right)+230).

Answer:

(y=-210\cos\left(\frac{\pi}{5}t\right)+230)