Chapter 8.3, Problem 39E

(a)

To determine

To find: The distance around the part to the nearest yard.

Answer to Problem 39E

The required distance is $191\text{yd}$ .

Explanation of Solution

Given information: The following graph is been given:

  

Formula used: $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation: To find the distance around the part to the nearest yard, you to find the value distance $PQ\&QR$ ,

Hence,

  $\begin{array}{l}PQ=\left|10-50\right|=40\\ QR=\left|80-10\right|=70\end{array}$

Now, the distance $PR$ ,

  $\begin{array}{l}PR=\sqrt{{\left(80-10\right)}^2+{\left(10-50\right)}^2}\\ =\sqrt{{\left(70\right)}^2+{\left(-40\right)}^2}\\ =\sqrt{4900+1600}\\ \approx 81\text{m}\end{array}$

Therefore, the total distance $=40+70+81=191\text{yd}$

(b)

To determine

To find: The distance $QM$ to the nearest yard.

Answer to Problem 39E

The required distance is $40\text{yd}$ .

Explanation of Solution

Given information: The following graph is been given:

  

Formula used: $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation: Get the midpoint of $\overline{PQ}$ first,

  $\begin{array}{l}\overline{PQ}=\left(\frac{80+10}{2},\frac{50+10}{2}\right)\\ =M\left(45,30\right)\end{array}$

Hence, $Q\left(10,10\right)$ and $M\left(45,30\right)$ ,

  $\begin{array}{l}QM=\sqrt{{\left(45-10\right)}^2+{\left(30-10\right)}^2}\\ =\sqrt{{35}^2+{20}^2}\\ =\sqrt{1625}\\ =40\text{yd}\end{array}$

(c)

To determine

To find: The time taken for jogs will be calculated.

Answer to Problem 39E

The required time is $1.5\min$ .

Explanation of Solution

Given information: The average speed of a man is $150\text{yards}\text{per minute}$ .

The following graph is been given:

  

Formula used: $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation: Since $PQ=40\approx QM=40$

Hence, $MR$ ,

  $\begin{array}{l}MR=\sqrt{{\left(80-45\right)}^2+{\left(10-30\right)}^2}\\ =\sqrt{{35}^2+{\left(-20\right)}^2}\\ =\sqrt{1625}\\ =40\text{yd}\end{array}$

And $RQ=70\&Qp=40$

Therefore, the total distance $=40+40+40+70+40=230\text{yd}$

Now, the required time will be:

  $\begin{array}{l}t=\frac{230\text{yd}}{150}\\ t=1.5\min \end{array}$