A Ferris wheel

A Ferris wheel with a diameter of 100 feet makes five revolutions every 8 minutes. The base of the wheel is 4 feet above the ground. Your friend gets on at 3 PM sharp.

a) Write an equation in seconds to express your friend's height in feet at any given time.

b) What are your friend's heights after one minute and 2 minutes?

c). Find the first time and the second time in seconds. Is your friend at 90 feet high?

Final Answer:

a = 54-50*cos(t*pi/48)
h₁ =  89.3553 ft
h₂ =  54 ft
t₃ =  36.2812 s
t₄ =  59.7188 s

Step-by-step explanation:

$$\begin{gathered} d=100\text{ft} \\ r=d/2=100/2=50\text{ft} \\ {T}_5=8\text{min}\to \text{s}=8\cdot 60\text{s}=480\text{s} \\ n=5 \\ T=T_5/n=480/5=96\text{s} \\ h=4\text{ft} \\ f=1/T=1/96=\frac{1}{96}\doteq 0.0104\text{Hz} \\ \omega =2\pi \cdot f=\frac{2}{96}\cdot pi \\ \omega =\pi /48=3.1416/48\doteq 0.0654 \\ a=h+r-r\cdot \cos (\omega \cdot t) \\ a=54-50\cdot \cos (t\cdot \pi /48) \end{gathered}$$

$$\begin{gathered} t_1=1\text{min}\to \text{s}=1\cdot 60\text{s}=60\text{s} \\ {h}_1=h+r-r\cdot \cos (\omega \cdot t_1)=4+50-50\cdot \cos (0.0654\cdot 60)=89.3553\text{ft} \end{gathered}$$

$$\begin{gathered} t_2=2\text{min}\to \text{s}=2\cdot 60\text{s}=120\text{s} \\ {h}_2=h+r-r\cdot \cos (\omega \cdot t_2)=4+50-50\cdot \cos (0.0654\cdot 120)=54\text{ft} \end{gathered}$$

$$\begin{gathered} h_3=90\text{ft} \\ {h}_3=54-50\cdot \cos (t_3\pi /48) \\ 90=54-50\cdot \cos (t_3\pi /48) \\ -0.72=\cos (t_3\pi /48) \\ {t}_3=36.281194783384=36.2812\text{s} \end{gathered}$$

$t_4=59.718805216616=59.7188\text{s}$

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