Chapter 4.4, Problem 72E

Solution

To calculate: The time to reach 50 ft above the ground after reaching the low point if a Ferris wheel 50 ft in diameter makes one revolution every 40 sec. The center of the wheel is 30 ft above the ground.

The resultant time is 15.9 seconds.

Given information:

A Ferris wheel 50 ft in diameter makes one revolution every 40 sec. The center of the wheel is 30 ft above the ground.

Formula used: The period formula is $\text{Period}=\frac{2\pi }{\left|b\right|}$ .

Calculation:

The diameter of a Ferris wheel is 50 ft. It means its height varies from 0 ft to 50 ft. So the minimum value is 0 and maximum value is 50.

So the Amplitude is:

  $\begin{array}{c}\text{Amplitude}\left(a\right)=\frac{M-m}{2}\\ =\frac{50-0}{2}\\ =25\end{array}$

It takes 40 seconds for one revolution, means period of the function is 40 sec.

  $\begin{array}{c}\text{Period}=\frac{2\pi }{\left|b\right|}\\ 40=\frac{2\pi }{\left|b\right|}\\ \left|b\right|=\frac{\pi }{20}\end{array}$

The center of the wheel is 30 ft above the ground. It means the vertical translation is 30. Then $k=30$ .

As the function starts from the low point, means its minimum is at 0. Then $h=0$ .

Now substitute the values of $a,b,h,\text{and }k$ in the sinusoid function $y=-a\cos \left(b\left(t-h\right)\right)+k$ .

  $y=-25\cos \left(\frac{\pi }{20}\left(t-0\right)\right)+30$

Now find the when wheel is at 50 ft above the ground,

  $\begin{array}{c}50=-25\cos \left(\frac{\pi }{20}\left(t-0\right)\right)+30\\ 20=-25\cos \left(\frac{\pi }{20}t\right)\\ \cos \left(\frac{\pi }{20}t\right)=-\frac{20}{25}\\ \cos \left(\frac{\pi }{20}t\right)=-\frac{4}{5}\\ \frac{\pi }{20}t={\cos }^{-1}\left(-\frac{4}{5}\right)\\ \frac{\pi }{20}t=2.5\\ t\approx 15.9\end{array}$

Therefore, the time needed time to reach 50 ft above the ground after the low point is about 15.9 sec.