Question

A U.S. Coast Guard boat leaves port and averages 35 knots traveling for 2 hours due east and then 3 hours due north. What is the boats distance and bearing from port?

Answer 1

The boat travels at an average speed of 35 knots and takes 5 hours (2 hours due east and then 3 hours due north). That means the speed is

`\begin{gathered} \text{speed}=\frac{\text{distance}}{\text{time}} \\ \text{speed}=(35)/(5) \\ \text{speed}=7\text{knots /hour} \end{gathered}`

The distance from port is calculated using the Pythagoras' theorem as follows;

`\begin{gathered} c^2=a^2+b^2 \\ c^2=2^2+3^2 \\ c^2=4+9 \\ c^2=13 \\ c=\sqrt[]{13} \\ c=3.6055\ldots \\ c=3.6 \end{gathered}`

The distance from port is 3.6 hours times 7 knots and that equals 25.2 knots

Next step, is to calculate the bearing from port using the cosine rule,

`\begin{gathered} a^2=b^2+c^2-2bc\cos A \\ 2^2=3^2+3.6^2-2(3*3.6)\cos A \\ 4=9+13-21.6\cos A \\ 4=22-21.6\cos A \\ 4-22=-21.6\cos A \\ -18=-21.6\cos A \\ (-18)/(-21.6)=\cos A \\ 0.8333=\cos A \\ 33.55=A \\ \text{The bearing therefore is,} \\ \text{Bearing}=90-33.55 \\ \text{Bearing}=56.45 \end{gathered}`

The distance therefore is,

28.98 miles from port

**(Note that 1 knot = 1.15 miles per hour)**

The bearing is 56.45 degrees north of east