Chapter 9.3, Problem 5MRPS

Solution

To find: The number of feet covered.

The restaurant revolves $568.73\text{ feet}$ in $1\text{ hour }55\text{ minutes}\text{.}$

Given:

It is given that the restaurant has radius of $47.25\text{ feet}$ and makes one complete revolution in about an hour.

Calculation:

The total distance covered in a circular motion is $2\pi r.$ Now, the restaurant completed one revolution per hour, so the distance covered in one revolution is:

  $\begin{array}{l}D=2\pi r\\ D=2\times 3.14\times 47.25\\ D=296.73\text{ feet}\end{array}$

Next, the distance covered in $1\text{ hour }55\text{ minutes}$ is:

  $\begin{array}{l}D=296.73\times \frac{115}{60}\\ D=\frac{34123.95}{60}\\ D=568.73\text{ feet}\end{array}$

Hence, the distance covered in $1\text{ hour }55\text{ minutes}$ is $568.73\text{ feet}\text{.}$

To find: the distance covered when the person seated $5\text{ feet}$ away from the window.

The distance covered when the person seated $5\text{ feet}$ away is $508.55\text{ feet}\text{.}$

Given:

It is given that the restaurant has radius of $47.25\text{ feet}$ and makes one complete revolution in about an hour.

Calculation:

The customer seated $5\text{ feet}$ away from the window, then the distance from the centre of the circular restaurant is:

  $\begin{array}{l}r=47.25-5\\ r=\text{42}\text{.25 feet}\end{array}$

Now, the distance covered by the person in $\frac{115}{60}\text{ hr}$ will be:

  $\begin{array}{l}\text{D}=2\pi r\left(\frac{115}{60}\right)\\ \text{D}=\frac{2\times 3.14\times 42.25\times 115}{60}\\ \text{D }=\frac{30512.95}{60}\\ \text{D}=508.55\text{ feet}\end{array}$

Hence, the distance covered when the person seated $5\text{ feet}$ away is $508.55\text{ feet}\text{.}$