5. A Ferris wheel at an amusement park measures 16m in diameter. It makes 3 rotations
every minute. The bottom of the Ferris wheel is 1m above the ground. Riders board the
Ferris wheel at the minimum point.
a) Determine the equation that models Emily's height (m) with respect to time (in seconds)
above ground. [3A]
b) A 12m tree stands near the Ferris wheel. For how long (in seconds) is Emily higher than
the tree during the first rotation? Round to 2 decimal places. [4A]

Respuesta :

Following are the responses to the given points:

For point a:

[tex]Diameter\ (d)= 16\ m\\\\[/tex]

Calculating the 3 rotations for every minute:

Calculating time for completing 1 rotation:

[tex]1\ rotation=\frac{60}{3}= 20\ second\\\\period=20 \ second\\\\[/tex]

The standard form of the equation of the sine and cosine function is:

[tex]y=A \sin \{ B(x-c)\} +D\\\\y=A \cos \{ B(x-c)\} +D\\\\[/tex]

Calculating the Amplitue:

[tex]A=\frac{max-min}{2}=\frac{17-1}{2}=\frac{16}{2}=8\\\\Period=\frac{2\pi}{B}\\\\20=\frac{2\pi}{B}\\\\B=\frac{2\pi}{20}\\\\B=\frac{\pi}{10}\\\\[/tex]

Calculating the phase shift:

for [tex]\sin[/tex] function: [tex]c=5[/tex]

for [tex]\cos[/tex] function: [tex]c=10[/tex]

Calculating the vertical shift:

[tex]\to D=\frac{max+ min }{2}=\frac{17+ 1}{2}=\frac{18}{2}=9\\\\y=8 \sin \{ \frac{\pi}{10}(t-5)\} +9\\\\y=8 \cos \{ \frac{\pi}{10}(t-10)\} +9\\\\[/tex]

For point b:

[tex]y> 12\ m\\\\12=8 \sin \{ \frac{\pi}{10}(t-5)\} +9\\\\12-9=8 \sin \{ \frac{\pi}{10}(t-5)\} \\\\3=8 \sin \{ \frac{\pi}{10}(t-5)\} \\\\\frac{3}{8}=\sin \{ \frac{\pi}{10}(t-5)\} \\\\\sin^{-1}\frac{3}{8}=\frac{\pi}{10}(t-5) \\\\\frac{\pi}{10} (0.38439677)=(t-5) \\\\1.22357+5=t \\\\t=6.22357\ second\\\\t=6.22\ second\\\\\sin^{-1}\frac{3}{8}=\frac{\pi}{10}(t-5) \\\\\frac{\pi}{10} (2.7571961)=(t-5) \\\\t=8.7764+5\\\\t=13.78\ second\\\\t_2-t_1=13.7764-6.22357= 7.55283\approx 7.55\ second \\\\[/tex]

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Rotation: brainly.in/question/39626227

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