IDNLearn.com makes it easy to get reliable answers from experts and enthusiasts alike. Get thorough and trustworthy answers to your queries from our extensive network of knowledgeable professionals.

A Ferris wheel rotates around in 30 seconds. The maximum height above the ground is 55 feet, and the minimum height above the ground is 5 feet. Which function models the height as a function of [tex]t[/tex] in seconds?

[tex]
\begin{array}{l}
f(t) = 30 \sin \left( \frac{\pi}{15} t \right) + 25 \\
f(t) = 25 \sin (60 \pi t) + 30 \\
f(t) = 25 \sin (2 \pi t) + 30 \\
f(t) = 30
\end{array}
[/tex]


Sagot :

To model the height of a Ferris wheel as a function of time [tex]\( t \)[/tex] (in seconds), we need to develop a sinusoidal function that represents the vertical position of a point on the Ferris wheel over time.

Given:
- The rotation period [tex]\( T \)[/tex] is 30 seconds.
- The maximum height is 55 feet.
- The minimum height is 5 feet.

Step-by-Step Solution:

1. Determine the amplitude:
The amplitude [tex]\( A \)[/tex] of the function is half the difference between the maximum and minimum heights.
[tex]\[ A = \frac{\text{max height} - \text{min height}}{2} = \frac{55 - 5}{2} = 25 \text{ feet} \][/tex]

2. Determine the vertical shift:
The vertical shift [tex]\( D \)[/tex] (also known as the midline of the sinusoidal function) is the average of the maximum and minimum heights.
[tex]\[ D = \frac{\text{max height} + \text{min height}}{2} = \frac{55 + 5}{2} = 30 \text{ feet} \][/tex]

3. Determine the angular frequency:
The angular frequency [tex]\( \omega \)[/tex] is computed using the rotation period. Since the period of a sine wave is [tex]\( 2\pi \)[/tex] radians and the Ferris wheel completes a rotation every 30 seconds, we have:
[tex]\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{30} = \frac{\pi}{15} \text{ radians per second} \][/tex]

With these values, we can write the height [tex]\( f(t) \)[/tex] as a function of time [tex]\( t \)[/tex]:

[tex]\[ f(t) = A \sin(\omega t) + D \][/tex]

Substituting the values:
[tex]\[ f(t) = 25 \sin\left(\frac{\pi}{15} t \right) + 30 \][/tex]

We can now identify the correct function from the given options:

- [tex]\( f(t) = 30 \sin \frac{\pi}{15} t + 25 \)[/tex]
- [tex]\( f(t) = 25 \sin 60 \pi t + 30 \)[/tex]
- [tex]\( f(t) = 25 \sin 2 \pi t + 30 \)[/tex]
- [tex]\( f(t) 30 \)[/tex]

The correct function is:

[tex]\[ f(t) = 25 \sin\left(\frac{\pi}{15} t\right) + 30 \][/tex]