Chapter 6.2, Problem 55E

(a)

To determine

To Find: The equation that relates s and the distance d between the two ships at noon.

Answer to Problem 55E

The required equation is $\sqrt{9s^2-108s+1296}$ .

Explanation of Solution

Given:

The two ships leave the port at nine am one travels at the bearing of N $53°\text{W}$ at 12 miles per hour and other travels at a bearing of $67°$ to the west at s km per hour.

Calculation:

The visual representation of the problem is shown in Figure 1

  

Figure 1

The value of the angle C is calculated as,

  $\begin{array}{c}C=180°-\left(53°+67°\right)\\ =60°\end{array}$

The distance travelled by the first ship from 9 am to noon is obtained as,

  $\begin{array}{c}d=12\left(3\right)\\ =36\text{km}\end{array}$

The distance travelled by the first ship from 9 am to noon is obtained as,

  $\begin{array}{c}d=s\left(3\right)\\ =\text{3s}\end{array}$

Then, by the law of cosines the side c and d is obtained as,

  $\begin{array}{c}{d}^2=a^2+b^2-2ab\cos D\\ =\sqrt{{36}^2+9s^2-2\left(36\right)\left(3s\right)\cos 60°}\\ =\sqrt{9s^2-108s+1296}\end{array}$

(b)

To determine

To Find: The speed of s that the second ship must travels so that the ship is 43 miles apart at the noon.

Answer to Problem 55E

The second ship must travel at the speed of $15.87$ km per second.

Explanation of Solution

Consider the given function is,

  $d^2=\sqrt{9s^2-108s+1296}$

Then,

  $\begin{array}{l}43=\sqrt{9s^2-108s+1296}\\ {43}^2=9s^2-108s+1296\\ s=-3.87,15.87\end{array}$

Thus, the second ship must travel at the speed of $15.87$ km per second.