Chapter 7.3, Problem 104AYU

The Ferris Wheel In 1893, George Ferris engineered the Ferris wheel. It was 250 feet in diameter. If a Ferris wheel makes 1 revolution every 40 seconds, then the function

$h\left(t\right)=125\sin \left(0.157t-\frac{\pi }{2}\right)+125$

represents the height $h$, in feet, of a seat on the wheel as a function of time $t$, where $t$ is measured in seconds. The ride begins when $t=0$.

a. During the first 40 seconds of the ride, at what time $t$ is an individual on the Ferris wheel exactly 125 feet above the ground?

b. During the first 80 seconds of the ride, at what time $t$ is an individual on the Ferris wheel exactly 250 feet above the ground?

c. During the first 40 seconds of the ride, over what interval of time $t$ is an individual on the Ferris wheel more than 125 feet above the ground?

To determine

To find:

a. During the first 40 seconds of the ride, at what time $t$ is an individual on the Ferris wheel exactly 125 feet above the ground?

Answer to Problem 104AYU

a. $t=10s,30s$

Explanation of Solution

Given:

In 1893, George Ferris engineered the Ferris wheel. It was 250 feet in diameter. If a Ferris wheel makes 1 revolution every 40 seconds, then the function.

$h\left(t\right)=125\sin \left(0.157t-\frac{\pi }{2}\right)+125$ represents the height $h$, in feet, of a seat on the wheel as a function of time $t$, where $t$ is measured in seconds. The ride begins when $t=\text{ 0}$.

Calculation:

a. During the first 40 seconds of the ride, at what time $t$ is an individual on the Ferris wheel exactly 125 feet above the ground?

$h\left(t\right)=125\sin \left(0.157t-\frac{\pi }{2}\right)+125$

$125=125\sin \left(0.157t-\frac{\pi }{2}\right)+125$

$125\sin \left(0.157t-\frac{\pi }{2}\right)=0$

$\sin \left(0.157t-\frac{\pi }{2}\right)=0$

$0.157t-\frac{\pi }{2}={\sin }^{-1}\left(0\right)$

$0.157t-\frac{\pi }{2}=n\pi ,n=0,1,2\dots$

$0.157t=n\pi +\frac{\pi }{2}$

$t=\frac{\pi +n\pi }{0.314}$

When $n=0$, $t\approx 10.005s$

When $n=\text{ 1}$, $t\approx 30.015s$, So they are 125 feet about the ground at 10 and 30 seconds.

To determine

To find:

b. During the first 80 seconds of the ride, at what time $t$ is an individual on the Ferris wheel exactly 250 feet above the ground?

Answer to Problem 104AYU

b. $20s$

Explanation of Solution

Given:

In 1893, George Ferris engineered the Ferris wheel. It was 250 feet in diameter. If a Ferris wheel makes 1 revolution every 40 seconds, then the function.

$h\left(t\right)=125\sin \left(0.157t-\frac{\pi }{2}\right)+125$ represents the height $h$, in feet, of a seat on the wheel as a function of time $t$, where $t$ is measured in seconds. The ride begins when $t=\text{ 0}$.

Calculation:

b. During the first 80 seconds of the ride, at what time $t$ is an individual on the Ferris wheel exactly 250 feet above the ground?

$h\left(t\right)=125\sin \left(0.157t-\frac{\pi }{2}\right)+125$

$250=125\sin \left(0.157t-\frac{\pi }{2}\right)+125$

$125\sin \left(0.157t-\frac{\pi }{2}\right)=125$

$\sin \left(0.157t-\frac{\pi }{2}\right)=1$

$0.157t-\frac{\pi }{2}={\sin }^{-1}\left(1\right)$

$0.157t-\frac{\pi }{2}=\frac{\pi }{2}$

$0.157t=\pi$

$t=\frac{\pi }{0.157}\approx 20.010$

So they are 250 feet about the ground at $\text{20}\text{.010}$ seconds.

To determine

To find:

c. During the first 40 seconds of the ride, over what interval of time $t$ is an individual on the Ferris wheel more than 125 feet above the ground?

Answer to Problem 104AYU

c. $\left[0,0.03\right]\cup \left[0.39,0.43\right]\cup \left[0.86\cup 0.89\right]$

Explanation of Solution

Given:

In 1893, George Ferris engineered the Ferris wheel. It was 250 feet in diameter. If a Ferris wheel makes 1 revolution every 40 seconds, then the function.

$h\left(t\right)=125\sin \left(0.157t-\frac{\pi }{2}\right)+125$ represents the height $h$, in feet, of a seat on the wheel as a function of time $t$, where $t$ is measured in seconds. The ride begins when $t=\text{ 0}$.

Calculation:

c. During the first 40 seconds of the ride, over what interval of time $t$ is an individual on the Ferris wheel more than 125 feet above the ground?

$h\left(t\right)=125\sin \left(0.157t-\frac{\pi }{2}\right)+125$

$125>125\sin \left(0.157t-\frac{\pi }{2}\right)+125$

$0>125\sin \left(0.157t-\frac{\pi }{2}\right)$

$\sin \left(0.157t-\frac{\pi }{2}\right)<0$

$0.157t-\frac{\pi }{2}<{\sin }^{-1}\left(0\right)$

$0.157t-\frac{\pi }{2}<n\pi ,n=0,1,2\dots$

$0.157t<n\pi +\frac{\pi }{2}$

$t<\frac{\pi +n\pi }{0.314}$

When $n=\text{ 0}$, $t\approx \text{ 1}0.005s$

When $n=\text{ 1}$, $t\approx 30.015s$, Ferris wheel more than 125 feet above the ground is between 10 and 30 seconds.