Chapter 6.1, Problem 53E

Solution

To calculate: The speed of the ship and current.

Speed of ship = $13.66\text{ mph}$

Speed of current = $7.07\text{ mph}$

Given:

Distance covered by ship $=20\text{ mi}$

Direction of ship $=$

  ${30}^{\circ }$

Time $=2\text{ hours}$

Explanation:

Consider the following figure:

  

Suppose, the distance covered in south is $A$ and the distance covered in the west of south is $B$ . So, by the law of sines,

  $\frac{A}{\sin {105}^{\circ }}=\frac{B}{\sin {30}^{\circ }}=\frac{20}{\sin {45}^{\circ }}$ ------(1)

From the equation (1), one can write as follows:

  $\begin{array}{c}\frac{A}{\sin {105}^{\circ }}=\frac{20}{\sin {45}^{\circ }}\\ A=\frac{\sin {105}^{\circ }}{\sin {45}^{\circ }}\times 20\\ A\approx 27.3205\end{array}$

From the equation (1),

  $\begin{array}{c}\frac{B}{\sin {30}^{\circ }}=\frac{20}{\sin {45}^{\circ }}\\ B=\frac{\sin {30}^{\circ }}{\sin {45}^{\circ }}\times 20\\ B\approx 14.14\text{ miles}\end{array}$

So, the speed of the ship is,

  $\begin{array}{c}{s}_1=\frac{A}{t}\\ =\frac{27.3205}{2}\\ =13.66\text{ mph}\end{array}$

And the speed of current is,

  $\begin{array}{c}{s}_2=\frac{B}{t}\\ =\frac{14.14}{2}\\ =7.07\text{ mph}\end{array}$

Conclusion:

Hence, the speed of ship and current is $13.66\text{ mph}$ , $7.07\text{ mph}$ respectively.