A sailboat sails from Marblehead Harbor directly east for 45 nautical miles, then \(60^{\circ}\) south of east for 20.0 nautical miles, returns to an easterly heading for 30.0 nautical miles, and sails \(30^{\circ}\) east of north for 10.0 nautical miles, then west for 62 nautical miles. At that time the boat becomes becalmed and the auxiliary engine fails to start. The crew decides to notify the Coast Guard of their position. Using graph paper, ruler, and protractor, sketch a graphical addition of the displacement vectors and estimate their position.

Each displacement can be represented as a vector with a magnitude (the distance traveled) and a direction. We can find the x and y components of each vector by multiplying the magnitude by the cosine and sine of the angle, respectively. For the first displacement (45 nautical miles east): - x-component: \(45 * \cos{0^{\circ}} = 45 \) - y-component: \(45 * \sin{0^{\circ}} = 0 \) For the second displacement (20 nautical miles, \(60^{\circ}\) south of east): - x-component: \(20 * \cos{60^{\circ}} = 10 \) - y-component: \(-20 * \sin{60^{\circ}} = -17.32\) For the third displacement (30 nautical miles east): - x-component: \(30 * \cos{0^{\circ}} = 30 \) - y-component: \(30 * \sin{0^{\circ}} = 0 \) For the fourth displacement (10 nautical miles, \(30^{\circ}\) east of north): - x-component: \(10 * \cos{60^{\circ}} = 5 \) - y-component: \(10 * \sin{60^{\circ}} = 8.66 \) For the fifth displacement (62 nautical miles west): - x-component: \(-62 * \cos{0^{\circ}} = -62\) - y-component: \(-62 * \sin{0^{\circ}} = 0 \)

Add the x-components and y-components separately to find the total displacement vector components: Total x-component: \(45 + 10 + 30 + 5 - 62 = 28\) Total y-component: \(0 - 17.32 + 0 + 8.66 + 0 = -8.66\)

Use the Pythagorean theorem to find the magnitude of the total displacement vector: Magnitude: \(\sqrt{(28)^2 + (-8.66)^2} =29.52 \text{ nautical miles}\) Find the angle relative to east by taking the inverse tangent of the y-component divided by the x-component: Angle: \(\tan^{-1}{\frac{-8.66}{28}} = -17.12^{\circ}\) So, the total displacement is 29.52 nautical miles at an angle of 17.12° south of east.

Draw the displacement vectors on graph paper using a ruler and protractor, making sure to label each vector and show their x and y components. Start by drawing the first vector, then draw the next vector starting from the endpoint of the previous one. Continue this process until all the vectors have been drawn. The endpoint of the last vector represents the final position of the sailboat. Since the total displacement is 29.52 nautical miles at an angle of 17.12° south of east, the final position can be estimated as approximately 29.52 nautical miles away from the starting point in a direction of 17.12° south of east.