**Sailboat Race Course Problem**A sailboat race course consists of four legs, defined by the displacement vectors **A**, **B**, **C**, and **D**, as the drawing indicates. The magnitudes of the first three vectors are as follows:- \( A = 2.90 \, \text{km} \)- \( B = 4.90 \, \text{km} \)- \( C = 5.30 \, \text{km} \)The finish line of the course coincides with the starting line. Using the data in the drawing, find:(a) The distance of the fourth leg.(b) The angle \( \theta \).**Diagram Explanation**: The diagram is a quadrilateral with vectors labeled as follows:- Vector **A** is at the bottom, starting from the point labeled "Start" and finishing at point "Finish", oriented with an angle of \( 40.0^\circ \) to the horizontal line.- Vector **B** is along the right side, oriented at an angle of \( 35.0^\circ \) to the horizontal line.- Vector **C** is at the top, oriented at an angle \( \theta \).- Vector **D** closes the loop and represents the fourth leg, which brings the path back to the starting point.**Solution Boxes**:(a) Distance of the fourth leg:- Number: 6.67- Units: km(b) Angle \( \theta \):- Number: 21.29- Units: degrees

Answered: gnitudes of the first three vectors are…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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**Sailboat Race Course Problem**

A sailboat race course consists of four legs, defined by the displacement vectors **A**, **B**, **C**, and **D**, as the drawing indicates. The magnitudes of the first three vectors are as follows:

- \( A = 2.90 \, \text{km} \)
- \( B = 4.90 \, \text{km} \)
- \( C = 5.30 \, \text{km} \)

The finish line of the course coincides with the starting line. Using the data in the drawing, find:

(a) The distance of the fourth leg.

(b) The angle \( \theta \).

**Diagram Explanation**:

The diagram is a quadrilateral with vectors labeled as follows:

- Vector **A** is at the bottom, starting from the point labeled "Start" and finishing at point "Finish", oriented with an angle of \( 40.0^\circ \) to the horizontal line.
- Vector **B** is along the right side, oriented at an angle of \( 35.0^\circ \) to the horizontal line.
- Vector **C** is at the top, oriented at an angle \( \theta \).
- Vector **D** closes the loop and represents the fourth leg, which brings the path back to the starting point.

**Solution Boxes**:

(a) Distance of the fourth leg:

- Number: 6.67
- Units: km

(b) Angle \( \theta \):

- Number: 21.29
- Units: degrees

Expert Solution

Step 1

a. First, lets get the sum of the horizontal components of each vectors ;
Dx = -AcosθA + BcosθB + CcosθC
= -2.9cos40 + 4.9cos35 + 5.3cos23
= -2.22152888505 + 4.01384501702 + 4.8786757233
Dx = 6.67099185527 km

Then solve the sum of vertical components of each vectors ;
Dy = AsinθA + BsinθB - CsinθC
= 2.9sin40 + 4.9sin35 - 5.3sin23
= 1.86408406809 + 2.81052453812 - 2.07087498099
Dy = 2.60373362522 km

Now solve for the distance of the fourth leg ;
D = Dx2 + Dy2

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