The Ferris Wheel

• Aug 31st 2008, 06:31 AM
magentarita
The Ferris Wheel
In 1893, George Ferris engineered the Ferris Wheel. It was 250 feet in diameter. If the wheel makes 1 revolution every 40 seconds, then h(t) = 125sin[0.157t - (pi/2)] + 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 a person on the Ferris Wheel exactly 125 feet above the ground?

(B) During the first 40 seconds of the ride, OVER what interval of time t is a person on the Ferris Wheel more than 125 feet above the ground?

• Aug 31st 2008, 08:19 AM
Soroban
Hello, magentarita!

I will assume that: . $0.157 \:\approx\:\frac{\pi}{20}$

Quote:

In 1893, George Ferris engineered the Ferris Wheel. It was 250 feet in diameter.
If the wheel makes 1 revolution every 40 seconds, then: . $h(t) \:= \:125\sin\left(\frac{\pi}{20}t - \frac{\pi}{2}\right) + 125$
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 a person on the Ferris Wheel exactly 125 feet above the ground?

Can't you set up the equation?

The question is: When is $h = 125$?

We have: . $125\sin\left(\frac{\pi}{20}t - \frac{\pi}{2}\right) + 125 \:=\:125 \quad\Rightarrow\quad \sin\left(\frac{\pi}{20}t - \frac{\pi}{2}\right) \:=\:0$

Then: . $\frac{\pi}{20}t - \frac{\pi}{2} \;=\;\{0,\pi,2\pi,3\pi,\hdots\} \quad\Rightarrow\quad \frac{\pi}{20}t \;=\;\left\{\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2},\frac{7\pi}{2},\hdots\right\}$

Hence: . $t \;=\;\{10, 30, 50, 70, \hdots \}$

Answers: . $\boxed{t \;=\;10\text{ sec},\;30\text{ sec}}$

Quote:

(B) During the first 40 seconds of the ride, over what interval of time $t$
is a person on the Ferris Wheel more than 125 feet above the ground?

The rider is above 125 feet on the interval: $t \in (10,30)$

• Aug 31st 2008, 12:58 PM
magentarita
Soroban
Quote:

Originally Posted by Soroban
Hello, magentarita!

I will assume that: . $0.157 \:\approx\:\frac{\pi}{20}$

No, it's 0.157 minus pi/2...not approximately. How does knowing this now change the answers given?

Can't you set up the equation?

The question is: When is $h = 125$?

We have: . $125\sin\left(\frac{\pi}{20}t - \frac{\pi}{2}\right) + 125 \:=\:125 \quad\Rightarrow\quad \sin\left(\frac{\pi}{20}t - \frac{\pi}{2}\right) \:=\:0$

Then: . $\frac{\pi}{20}t - \frac{\pi}{2} \;=\;\{0,\pi,2\pi,3\pi,\hdots\} \quad\Rightarrow\quad \frac{\pi}{20}t \;=\;\left\{\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2},\frac{7\pi}{2},\hdots\right\}$

Hence: . $t \;=\;\{10, 30, 50, 70, \hdots \}$

Answers: . $\boxed{t \;=\;10\text{ sec},\;30\text{ sec}}$

The rider is above 125 feet on the interval: $t \in (10,30)$

• Aug 31st 2008, 01:33 PM
skeeter
Soroban knows what he's doing ...

$\frac{1 \, rev}{40 \, sec} = \frac{2\pi \, rad}{40 \, sec} = \frac{\pi}{20}$

nothing's changed.
• Aug 31st 2008, 05:17 PM
magentarita
skeeter...
Quote:

Originally Posted by skeeter
Soroban knows what he's doing ...

$\frac{1 \, rev}{40 \, sec} = \frac{2\pi \, rad}{40 \, sec} = \frac{\pi}{20}$

nothing's changed.

I know Soroban. He is found in most math forums. I know he is good.
He's a former math professor. I am not putting down his replies. However, your replies are starting to bug me. Please, let's make life easy for everyone. If you dislike my questions because they are too easy, please skip my posts and go to the next student. Understood?

Thanks,
Rita