Part 1A ship leaves port on a bearing of 45.045.0degrees° north of east and travels 11.611.6 mi. The ship then turns due east and travels 3.33.3 mi. How far is the ship from port, and what is its bearing from port? Part 1The ship is [input]enter your response here miles from the port.(Do not round until the final answer. Then round to the nearest tenth as needed.)

Question

Part 1A ship leaves port on a bearing of 45.045.0degrees° north of east and travels 11.611.6 mi. The ship then turns due east and travels 3.33.3 mi. How far is the ship from​ port, and what is its bearing from​ port? Part 1The ship is [input]enter your response here miles from the port.​(Do not round until the final answer. Then round to the nearest tenth as​ needed.)

BlackTom AI · Accepted Answer

We start by parsing the ship's two legs. The first leg is 11.6 miles on a bearing of 45.0° north of east. That means the displacement components are: x1 = 11.6 cos(45.0°) and y1 = 11.6 sin(45.0°). Since cos and sin of 45° are equal, x1 and y1 are both 11.6 × √2/2 ≈ 8.21 miles.

The second leg is 3.3 miles due east, which adds to the eastward (x) component only: x2 = 3.3 and y2 = 0.

Combine the two legs: total eastward displacement x = x1 + x2 ≈ 8.21 + 3.30 = 11.51 miles, and total northward displacement y = y1 + y2 ≈ 8.21 + 0 = 8.21 miles.

Next we compute the straight-line distance from the port to the ship using the Pythagorean theorem: distance = √(x^2 + y^2) = √(11.51^2 + 8.21^2).
Compute the squares: 11.51^2 ≈ 132.38 and 8.21^2 ≈ 67.40. Sum ≈ 199.78. The square root gives a distance ≈ 14.12 miles.

Since the instruction says not to round until the final answer, we keep the intermediate value and then round at the end. Rounding to the nearest tenth yields about 14.1 miles.

For the bearing from port, we determine the angle north of east using the tangent of the angle: tan(θ) = y/x = 8.21 / 11.51 ≈ 0.713. Taking the arctangent gives θ ≈ 35.5°. This means the ship lies northeast of the port, with a bearing of approximately 35.5° east of north converted to standard bearing notation as 035.5°.

Summary of the displacements and results (without final rounding): total east displacement ≈ 11.51 miles, total north displacement ≈ 8.21 miles, distance ≈ 14.12 miles, bearing ≈ 035.5° from north. After rounding to the nearest tenth as required, the distance is about 14.1 miles, and the bearing would be about 035.5° (or 35.5°) clockwise from due north.

Homework:Module 11 Homework (Vectors)

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