Question

The Ferris Wheel In 1893, George Ferris engineered the Ferris Wheel. It was 250 feet in diameter. If the wheel make 1revolutionevery40seconds, then the function $h\left(t\right)=125\sin \left(0.157t-\frac{\mathrm{\pi }}{2}\right)+125$ 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 timet is an individual on the Ferris Wheel more than

125 feet above the ground?

Short answer

Part a. At time$10\mathrm{or}30\mathrm{seconds}$ an individual on the Ferris Wheel exactly 125 feet above

the ground.

Part b. At time$20\mathrm{or}60\mathrm{seconds}$ an individual on the Ferris Wheel exactly 250 feet above

the ground.

Part c. At the interval,$\left(10,30\right)$an individual on the Ferris Wheel is more than 125feet above the ground.

Step-by-step solution

Part (a) Step 1. Given Information

The given function is$h\left(t\right)=125\sin \left(0.157t-\frac{\mathrm{\pi }}{2}\right)+125$wherethe heighth, in feet, of a seat on the wheel as a function of timet, wheret is measured in seconds. We have to find time for an individual on the Ferris wheel exactly 125above the ground.

Part (a) Step 2. Finding the time

To find the time, substitute 125 to the given function.

$$\begin{gathered} h\left(t\right)=125\sin \left(0.157t-\frac{\mathrm{\pi }}{2}\right)+125 \\ 125=125\sin \left(0.157t-\frac{\mathrm{\pi }}{2}\right)+125 \\ 0=125\sin \left(0.157t-\frac{\mathrm{\pi }}{2}\right) \\ 0=\sin \left(0.157t-\frac{\mathrm{\pi }}{2}\right) \\ {\sin }^{-1}\left(0\right)=\left(0.157t-\frac{\mathrm{\pi }}{2}\right) \\ \mathrm{Sine}\mathrm{function}\mathrm{is}\mathrm{equal}\mathrm{to}0\mathrm{when}\mathrm{the}\mathrm{angle}\mathrm{are}0\mathrm{and}\mathrm{\pi }. \\ \left(0.157\mathrm{t}-\frac{\mathrm{\pi }}{2}\right)=0\mathrm{or}\left(0.157\mathrm{t}-\frac{\mathrm{\pi }}{2}\right)=\mathrm{\pi } \\ 0.157\mathrm{t}=\frac{\mathrm{\pi }}{2}\mathrm{or}0.157\mathrm{t}=\mathrm{\pi }+\frac{\mathrm{\pi }}{2} \\ 0.157\mathrm{t}=1.57\mathrm{or}0.157\mathrm{t}=4.17 \\ \mathrm{t}=10\mathrm{or}\mathrm{t}=30 \end{gathered}$$

Thus, the individual on the Ferris wheel exactly 125 feet above on the ground is at time$10\mathrm{or}30\mathrm{seconds}.$

Part (b) Step 1. Finding the time

To find the time, substitute 250 to the given function.

$$\begin{gathered} h\left(t\right)=125\sin \left(0.157t-\frac{\mathrm{\pi }}{2}\right)+125 \\ 250=125\sin \left(0.157t-\frac{\mathrm{\pi }}{2}\right)+125 \\ 250-125=\sin \left(0.157t-\frac{\mathrm{\pi }}{2}\right) \\ \frac{125}{125}=\sin \left(0.157t-\frac{\mathrm{\pi }}{2}\right) \\ 1=\sin \left(0.157t-\frac{\mathrm{\pi }}{2}\right) \\ {\sin }^{-1}\left(1\right)=\left(0.157t-\frac{\mathrm{\pi }}{2}\right) \end{gathered}$$

Now, during the first 80seconds of the ride, the wheel will complete two revolutions. Thus, the interval will be $[0,4\mathrm{\pi }).$

$$\begin{gathered} \mathrm{Sine}\mathrm{function}\mathrm{is}\mathrm{equal}\mathrm{to}1\mathrm{when}\mathrm{the}\mathrm{angle}\mathrm{are}\frac{\mathrm{\pi }}{2}\mathrm{and}\frac{5\mathrm{\pi }}{2}. \\ \left(0.157\mathrm{t}-\frac{\mathrm{\pi }}{2}\right)=\frac{\mathrm{\pi }}{2}\mathrm{or}\left(0.157\mathrm{t}-\frac{\mathrm{\pi }}{2}\right)=\frac{5\mathrm{\pi }}{2} \\ 0.157\mathrm{t}=\frac{2\mathrm{\pi }}{2}\mathrm{or}0.157\mathrm{t}=\frac{5\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{2} \\ 0.157\mathrm{t}=\mathrm{\pi }\mathrm{or}0.157\mathrm{t}=3\mathrm{\pi } \\ 0.157\mathrm{t}=3.14\mathrm{or}0.157\mathrm{t}=9.42 \\ \mathrm{t}=20\mathrm{or}\mathrm{t}=60 \end{gathered}$$

Thus, an individual on the Ferris wheel exactly250feet above the ground at$20\mathrm{or}60\mathrm{seconds}.$

Part (c) Step 1. Finding the interval

As we know that an individual on the Ferris wheel exactly 125feet above the ground at the time $10\mathrm{or}30\mathrm{seconds}.$

So, an individual on the Ferris Wheel more than 125feet above the ground at the time between intervals $10\mathrm{or}30\mathrm{seconds},10<\mathrm{t}<30.$

The interval will be$\left(10,30\right).$