Question

A sailboat under auxiliary power is proceeding on a bearing 25° north of west at 6,25km/h in still water.Then a tail wind blowing 15km/h in the direction 35° south of west, alters the course of the sailboat.What is the resultant speed and direction of the saiboat (correct to 2 decimal places)?

Answer 1

$\textsf{Calculuating­ the components of vector}\\ A_1 A_2, A_3 A_2 \,\textsf{and}\, A_3 A_1\,\textsf{in the system of rectangular axes}\\\textsf{and}\,A_2\, \textsf{is a point}\, 6.25\,\textsf{units away from}\\ A \,\textsf{at a bearing of}\, 25\degree, A_3 \,\textsf{is a point}\\ 15 \,\textsf{units away from}\, A \\\textsf{at a bearing of}\, 215\degree \\ \textsf{We first construct a diagram including}\\\textsf{­point}\, A_1, A_2, A_3 \,\textsf{at the given bearing}\\\textsf{an­d determine the magnitude of}\, A_3 A_2. \\\textsf{Since the magnitude of velocity}\\\textsf{i­s speed (see the figure).}\\ \textsf{Hence, the}\, x \, \textsf{and}\, y \,\textsf{components­ of vector}\\ \overrightarrow{A_1 A_2},­ \,\textsf{and}\,\overrightarrow{A_3 A_1}\, \textsf{are respectively given by} \\ {A_1 A_2}_x = -6.25\cos(25\degree)­ = -5.66\\ {A_1 A_2}_y = 6 \sin(25\degree) = 2.64\\ \overrightarrow{A_1 A_2}\,\textsf{can be written as}\\ \overrightarrow{A_1 A_2­} = {A_1 A_2}_x \textbf{i} + {A_1 A_2}_y\textbf{j}, \\\textsf{where}\, \textbf{i} \,\textsf{and}\, \textbf{j} \,\textsf{are unit vectors.}\\ \therefore \overrightarrow{A_1 A_2} = -5.66\textbf{i} + 2.64\textbf{j}\\ \textsf{Similarly,} \\ {A_3 A_1}_x = 15\cos(215\degree) = -12.29\\ {A_3 A_1}_y = 15 \sin(215\degree) = -8.60\\ \overrightarrow{A_3 A_1} = -12.29\textbf{i} - 8.60\textbf{j}\\ \textsf{By the vector law of addition,}\\ \overrightarrow{A_3 A_1} + \overrightarrow{A_1 A_2} = \overrightarrow{A_3 A_2} \\ \therefore \overrightarrow{A_3 A_2} = -17.95\textbf{i} - 5.96\textbf{j}\\ |A_3 A_2| \, \textsf{is the resultant speed and}\\\textsf{direct­ion of the sail boat.}\\ \therefore |A_3 A_2| = \sqrt{17.95^2 + 5.96^2} = 18.91\\ \textsf{Therefore, the resultant speed is}\, 18.91 \,\textit{km/­h,}\,\textsf{to}\,2\textsf{d.p.}\\\textsf{Also, the direction of the sail boat is given by}\\ \theta = \arctan\left(\frac{5.96}{17.95}\right) = 18.37\degree\\ \therefore \textsf{ The direction of the sail boat is}\, W\,18.37\degree S$