/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 A ship leaves the island of Guam... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 70

A ship leaves the island of Guam and sails 285 km at 62.0\(^{\circ}\) north of west. In which direction must it now head and how far must it sail so that its resultant displacement will be 115 km directly east of Guam?

Short Answer

Expert verified
The ship must travel 353 km at 45.1掳 south of east.

Step by step solution

01

Breakdown of Initial Movement

First, we need to understand the vector movement of the ship's initial journey. The ship sails 285 km at an angle of 62.0掳 north of west. We'll break this movement into its northward and westward components. The westward (x) component is given by \(285 \times \cos(62.0掳)\) and the northward (y) component by \(285 \times \sin(62.0掳)\).
02

Calculate Components of Initial Displacement

Calculate the actual numerical values for the components using trigonometry.\[\text{x-component (westward)} = 285 \times \cos(62.0掳) \approx 133.0 \text{ km (West)}\]\[\text{y-component (northward)} = 285 \times \sin(62.0掳) \approx 250.5 \text{ km (North)}\]
03

Understanding Desired Resultant Displacement

The problem states that the resultant displacement should be 115 km directly east. This means that the final x-component of the resultant vector is +115 km (east direction), and the final y-component should be 0 km.
04

Set Up Equations for Resultant Displacement

Let's assume the ship needs to sail \(d\) km at an angle \(\theta\) south of east. The eastward (x) component of this trip is \(d \times \cos(\theta)\) and the southward (y) component is \(d \times \sin(\theta)\). From the previous steps, the eastward component compensates the -133 km, and the northward component needs to completely negate the 250.5 km northward to achieve 0 km net in the north-south direction.
05

Solve for Distance and Angle

First, for x-component, \(d \times \cos(\theta) = 133 + 115 = 248 \text{ km (East)}\), and for y-component, \(d \times \sin(\theta) = 250.5 \text{ km (South)}\). Using these:- \(d^2 \times \cos^2(\theta) + d^2 \times \sin^2(\theta) = 248^2 + 250.5^2\)Solve for \(d\), then use \(\tan^{-1}\left(\frac{250.5}{248}\right)\) to find \(\theta\). \[d \approx \sqrt{248^2 + 250.5^2} \approx 353 \text{ km}\]\[\theta \approx \tan^{-1}\left(\frac{250.5}{248}\right) \approx 45.1掳\]
06

Conclusion

To achieve the desired resultant displacement of 115 km east, the ship must travel approximately 353 km at an angle of 45.1掳 south of east.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Components
When dealing with vector displacement, breaking down vectors into their components is a fundamental step. Each vector can be decomposed into two parts or components - the horizontal (x-component) and the vertical (y-component). These components allow us to analyze movements in different directions separately, making calculations more manageable.

In the case of the ship, the initial journey at 62.0掳 north of west forms a vector. To find the westward (x) component, we use the cosine of the angle:
  • Westward component = 285 km 脳 cos(62.0掳).
For the northward (y) component, we use the sine:
  • Northward component = 285 km 脳 sin(62.0掳).
Understanding these components is crucial as they allow us to piece together the journey's overall direction and magnitude.
Trigonometry Applications
Trigonometry plays a vital role in solving vector problems. By using trigonometric functions such as sine (sin), cosine (cos), and tangent (tan), we can link angles with side lengths in right-angled triangles. Since vectors can be represented as triangles, trigonometry is our tool for solving these puzzles.

In this scenario, trigonometric functions help transform the ship's journey into manageable components. Once you have the angle and the hypotenuse鈥攖hat is, the magnitude or length of 285 km in this context鈥攖hese functions allow for straightforward calculations:
  • Cosine is used to determine the adjacent (or x-component) length.
  • Sine gives the length of the opposite side (or y-component).
This method is advantageous in evaluating separate parts of the journey independently, enabling us to focus on any net result needed.
Resultant Vector Calculation
After determining the components of the initial journey, combining them with the desired displacement helps find a resultant vector. In physics and engineering, the resultant vector sums all individual vector movements into one final vector, showing the total effect.

To solve the exercise, the ship's desired displacement vector needs to be achieved. This means combining its initial movement with a subsequent one that results in being 115 km east. By setting up an equation system, we ensure both components align to our goal:
  • The x-component needed is +115 km east, compensating for the -133 km west initially.
  • The y-component should counter any previous northward movement to end with 0 km net north-south.
The solution involves solving these equations using trigonometry and Pythagoras' theorem, allowing determination of the next distance and angle the ship must follow to achieve the desired position.

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