/*! 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 2 A sailboat runs before the wind ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 2

A sailboat runs before the wind with a constant speed of \(4.2 \mathrm{m} / \mathrm{s}\) in a direction \(32^{\circ}\) north of west. How far (a) west and (b) north has the sailboat traveled in 25 min?

Short Answer

Expert verified
The sailboat has traveled approximately 5330.63 meters west and 3339.67 meters north.

Step by step solution

01

Convert Time to Seconds

To find the total distance, we need to first convert the time from minutes to seconds. We know there are 60 seconds in one minute. Thus, \[ t = 25 \text{ min} \times 60 = 1500 \text{ seconds}. \]
02

Calculate Total Distance Traveled

Using the boat's speed and the total time, we can find the total distance traveled. The formula for distance when speed and time are known is \( d = v \times t \). Here, \( v = 4.2 \text{ m/s} \) and \( t = 1500 \text{ s} \). \[ d = 4.2 \times 1500 = 6300 \text{ meters}. \]
03

Find the Westward Component

The boat travels at an angle of \( 32^{\circ} \) north of west. To find the component of this distance traveled to the west, we use the cosine function: \( d_{\text{west}} = d \cdot \cos(32^{\circ}) \). \[ d_{\text{west}} = 6300 \cdot \cos(32^{\circ}) \approx 5330.63 \text{ meters}. \]
04

Find the Northward Component

Similarly, to find the distance traveled north, we use the sine function: \( d_{\text{north}} = d \cdot \sin(32^{\circ}) \). \[ d_{\text{north}} = 6300 \cdot \sin(32^{\circ}) \approx 3339.67 \text{ meters}. \]

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Key Concepts

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

Trigonometry in Physics
Trigonometry plays a vital role in physics to solve problems involving angles and distances. When dealing with motion at an angle, such as a sailboat navigating north of west, breaking the vectors into horizontal and vertical components becomes essential. This is achieved using trigonometric functions like sine and cosine.
  • The cosine function helps find the horizontal or x-component of a vector. It鈥檚 used when calculating how far something travels west in this problem.
  • The sine function helps find the vertical or y-component of a vector. It鈥檚 used here to determine the distance the sailboat travels north.
Understanding these concepts allows us to decompose any vector into its basic parts: one along the axis of the angle's direction and one perpendicular to it. This method is crucial in physics to predict an object's position based on its speed and direction.
Speed and Distance Calculation
To calculate how far a moving object travels, knowing its speed and time of travel is essential. Speed represents how fast an object's position changes; distance is the total pathway covered.
When given speed in meters per second as in the sailboat scenario, converting time from minutes to seconds ensures the units are coherent for calculating distance:
  • Use the formula: \( d = v \times t \), where \( v \) is speed and \( t \) is time.
  • In this problem, speed was \( 4.2 \, \text{m/s} \) and the time was \( 1500 \) seconds, leading to a distance of \( 6300 \) meters.
Accurate conversion of units is crucial for correct calculations. It ensures the final distance represents a true measure of the path traveled over time, regardless of the calculation or problem domain.
Directional Motion Analysis
Directional motion analysis involves breaking down movement into components to understand the path better. While the sailboat moves at an angle north of west, analyzing motion involves assessing how far it travels in each direction:
  • Determine the westward distance using the cosine function: \( d_{\text{west}} = d \cdot \cos(32^{\circ}) \)
  • Similarly, calculate the northward distance with the sine function: \( d_{\text{north}} = d \cdot \sin(32^{\circ}) \)
  • The calculations result in approximately \( 5330.63 \) meters west and \( 3339.67 \) meters north.
These calculations help us understand not just how far something moves but also in what direction. This breakdown is essential for applications in navigation and understanding how objects move through space.

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Most popular questions from this chapter

An archer shoots an arrow horizontally at a target \(15 \mathrm{m}\) away. The arrow is aimed directly at the center of the target, but it hits \(52 \mathrm{cm}\) lower. What was the initial speed of the arrow?

A hockey puck just clears the 2.00-m-high boards on its way out of the rink. The base of the boards is \(20.2 \mathrm{m}\) from the point where the puck is launched. (a) Given the launch angle of the puck, \(\theta\), outline a strategy that you can use to find its initial speed, \(v_{0}\). (b) Use your strategy to find \(v_{0}\) for \(\theta=15.0^{\circ}\).

The "hang time" of a punt is measured to be \(4.50 \mathrm{s}\). If the ball was kicked at an angle of \(63.0^{\circ}\) above the horizontal and was caught at the same level from which it was kicked, what was its initial speed?

A soccer ball is kicked with an initial speed of \(10.2 \mathrm{m} / \mathrm{s}\) in a direction \(25.0^{\circ}\) above the horizontal. Find the magnitude and direction of its velocity (a) \(0.250 \mathrm{s}\) and (b) 0.500 s after being kicked. (c) Is the ball at its greatest height before or after 0.500 s? Explain.

A train moving with constant velocity travels \(170 \mathrm{m}\) north in \(12 \mathrm{s}\) and an undetermined distance to the west. The speed of the train is \(32 \mathrm{m} / \mathrm{s}\). (a) Find the direction of the train's motion relative to north. (b) How far west has the train traveled in this time?
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