/*! 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 17 The cye of a hurricane passes ov... [FREE SOLUTION] | 91Ó°ÊÓ

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The cye of a hurricane passes over Grand Bahama Island in a direction \(60.0^{\circ}\) north of west with a speed of \(41.0 \mathrm{~km} / \mathrm{h}\). Three hours later the course of the hur. ricane suddenly shifts due north, and its speed slows to \(25.0 \mathrm{~km} / \mathrm{h}\). How far from Grand Bahama is the hurricane \(4.50 \mathrm{~h}\) after it passes over the island?

Short Answer

Expert verified
The hurricane is \(153.1 \mathrm{~km}\) away from Grand Bahama Island 4.5 hours after it passed over the island.

Step by step solution

01

Calculate the first part of the journey

The hurricane travels for 3 hours in a direction \(60^{\circ}\) north of west, with a speed of \(41.0 \mathrm{~km/h}\). We can resolve this into the westward and northward components. \nThe westward component of this journey is \(41.0 \mathrm{~km/h} \times 3 \mathrm{~h} \times \cos(60^{\circ}) = 61.5 \mathrm{~km}\). The northward component of this journey is \(41.0 \mathrm{~km/h} \times 3 \mathrm{~h} \times \sin(60^{\circ}) = 106.3 \mathrm{~km}\).
02

Calculate the second part of the journey

The hurricane then changes direction and travels due north for \(4.5 - 3 = 1.5\) hours with a speed of \(25.0 \mathrm{~km/h}\). The distance traveled in this leg of the journey is \(25.0 \mathrm{~km/h} \times 1.5 \mathrm{~h} = 37.5 \mathrm{~km}\). There is no westward component in this part.
03

Calculate total distance

The total distance of the hurricane from Grand Bahama can be calculated by adding the westward and northward components of both parts of the journey. \nThe westward component is \(61.5 \mathrm{~km}\) and the northward component is \(106.3 \mathrm{~km} + 37.5 \mathrm{~km} = 143.8 \mathrm{~km}\). \nNow we can use the Pythagorean theorem to calculate the total distance: \(\sqrt{(61.5 \mathrm{~km})^2 + (143.8 \mathrm{~km})^2 } = 153.1 \mathrm{~km}\).

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Key Concepts

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

Component Vectors
When solving problems involving directions and speeds, it's crucial to understand **component vectors**. Each movement or direction can be broken down into parts or components. These are typically resolved into horizontal (west-east) and vertical (north-south) components.
  • The westward component indicates movement along the east-west axis.
  • The northward component indicates movement along the north-south axis.
To resolve a vector into components, we use trigonometric functions:
  • *Cosine* for the horizontal (or westward) component: \(x = v \cdot \cos(\theta)\)
  • *Sine* for the vertical (or northward) component: \(y = v \cdot \sin(\theta)\)
Where \(v\) is the velocity or speed, and \(\theta\) is the angle of direction. Understanding component vectors allows us to break down complex movements into more manageable calculations, such as calculating the individual northward and westward movements of the hurricane as shown in the solution.
Pythagorean Theorem
The **Pythagorean theorem** is a fundamental concept in geometry used to determine the relationship between the sides of a right triangle. This theorem states: \[c^2 = a^2 + b^2\], where \(c\) is the hypotenuse, and \(a\) and \(b\) are the other two sides.
In the problem given, once we determine the northward and westward distances traveled, we can use the Pythagorean theorem to calculate the straight-line distance the hurricane traveled from its starting point.
  • The 'horizontal' leg consists of the total westward distance.
  • The 'vertical' leg consists of the total northward distance.
After calculating both component distances, the hypotenuse (or direct distance from the start point) represents the total distance traveled by the hurricane. It simplifies complex path calculations into a straightforward equation.
Problem-Solving Techniques
When tackling problems like these, certain **problem-solving techniques** become invaluable. Breaking down the problem into smaller steps makes it simpler. Follow a systematic approach:
- **Understand the Problem:** Recognize what is asked — in this case, the distance of the hurricane from a point. - **Break Down Movement into Components:** Separate the journey into parts and understand each component thoroughly. - **Use Appropriate Methods and Formulas:** Utilize trigonometric functions and formulas suitable for the task. - **Perform Calculations Carefully:** Solve for components and check accuracy at each stage. - **Combine Results:** Add or use further formulas like the Pythagorean theorem, to find the required distance. By organizing the process into these steps, problem-solving regarding vector mechanics becomes less daunting. Practice and familiarity with these techniques will enhance skills efficiently.

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