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Suppose a Ferris wheel is 250 feet in diameter. If a Ferris wheel makes 1 revolution every 40 seconds, then the function [tex]h(t)=125 \sin \left(0.157 t-\frac{\pi}{2}\right)+125[/tex] represents the height [tex]h[/tex], in feet, of a seat on the wheel as a function of time [tex]t[/tex], where [tex]t[/tex] is measured in seconds. For the particular seat under discussion, the ride begins when [tex]t=0[/tex]. Complete parts (a) through (c) below.

(a) During the first 40 seconds of the ride, at what time(s) [tex]t[/tex] is an individual riding in the seat on the Ferris wheel exactly 125 feet above the ground?

[tex]t=\square[/tex]
(Round to the nearest integer as needed. Use a comma to separate answers as needed.)

(b) During the first 80 seconds of the ride, at what time(s) [tex]t[/tex] is an individual riding in the seat on the Ferris wheel exactly 250 feet above the ground?

[tex]t=\square[/tex]
(Round to the nearest integer as needed. Use a comma to separate answers as needed.)

(c) During the first 40 seconds of the ride, over what interval of time [tex]t[/tex] is an individual riding in the seat on the Ferris wheel more than 125 feet above the ground?



Answer :

GinnyAnswer
Let's break down each part of the problem step-by-step.

### (a) Times when the height is exactly 125 feet:

We have the height function [tex]\( h(t) = 125 \sin \left(0.157t - \frac{\pi}{2}\right) + 125 \)[/tex].

To find when the seat is exactly 125 feet above the ground, we set [tex]\( h(t) = 125 \)[/tex] and solve for [tex]\( t \)[/tex]:

[tex]\[ 125 \sin \left(0.157t - \frac{\pi}{2}\right) + 125 = 125 \][/tex]

Subtract 125 from both sides:

[tex]\[ 125 \sin \left(0.157t - \frac{\pi}{2}\right) = 0 \][/tex]

Divide by 125:

[tex]\[ \sin \left(0.157t - \frac{\pi}{2}\right) = 0 \][/tex]

The sine function equals zero at integer multiples of [tex]\(\pi\)[/tex]:

[tex]\[ 0.157t - \frac{\pi}{2} = n\pi \][/tex]

Solving for [tex]\( t \)[/tex]:

[tex]\[ 0.157t = n\pi + \frac{\pi}{2} \][/tex]

[tex]\[ t = \frac{n\pi + \frac{\pi}{2}}{0.157} \][/tex]
[tex]\[ t = \frac{\pi(2n + 1)/2}{0.157} \][/tex]

For the first 40 seconds, the solutions are:

[tex]\[ t = 10 \text{ seconds, and } t = 30 \text{ seconds} \][/tex]

These solutions correspond to when the seat is exactly 125 feet above the ground.

Therefore:
[tex]\[ t = 10, 30 \][/tex]

### (b) Times when the height is exactly 250 feet:

Set [tex]\( h(t) = 250 \)[/tex]:

[tex]\[ 125 \sin \left(0.157t - \frac{\pi}{2}\right) + 125 = 250 \][/tex]

Subtract 125 from both sides:

[tex]\[ 125 \sin \left(0.157t - \frac{\pi}{2}\right) = 125 \][/tex]

Divide by 125:

[tex]\[ \sin \left(0.157t - \frac{\pi}{2}\right) = 1 \][/tex]

The sine function equals 1 at:

[tex]\[ 0.157t - \frac{\pi}{2} = \frac{\pi}{2} + 2k\pi \][/tex]

Solving for [tex]\( t \)[/tex]:

[tex]\[ 0.157t = \pi + 2k\pi \][/tex]

[tex]\[ t = \frac{\pi(1 + 2k)}{0.157} \][/tex]

For the first 80 seconds, the solution corresponding to these values is:

[tex]\[ t = 20 \text{ seconds} \][/tex]

Thus:
[tex]\[ t = 20 \][/tex]

### (c) Time interval when height is more than 125 feet:

To find when [tex]\( h(t) > 125 \)[/tex]:

[tex]\[ 125 \sin \left(0.157t - \frac{\pi}{2}\right) + 125 > 125 \][/tex]

Subtract 125 from both sides:

[tex]\[ 125 \sin \left(0.157t - \frac{\pi}{2}\right) > 0 \][/tex]

Divide by 125:

[tex]\[ \sin \left(0.157t - \frac{\pi}{2}\right) > 0 \][/tex]

This is true when:

[tex]\[ 0.157t - \frac{\pi}{2} > 0 \][/tex]

Solving for [tex]\( t \)[/tex]:

[tex]\[ 0.157t > \frac{\pi}{2} \][/tex]

[tex]\[ t > \frac{\pi}{(2 \cdot 0.157)} \][/tex]

For the condition [tex]\( h(t) > 125 \)[/tex]:

[tex]\[ t > 10 \][/tex]

Thus, the interval when the height is more than 125 feet during the first 40 seconds is:
[tex]\[ 10 < t < 40 \][/tex]

So, the interval of [tex]\( t \)[/tex]:

[tex]\[ (10, 40) \][/tex]