/*! 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 40 The nose of an ultralight plane ... [FREE SOLUTION] | 91Ó°ÊÓ

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The nose of an ultralight plane is pointed due south, and its airspeed indicator shows \(35 \mathrm{~m} / \mathrm{s}\). The plane is in a \(10 \mathrm{~m} / \mathrm{s}\) wind blowing toward the southwest relative to the earth. (a) In a vectoraddition diagram, show the relationship of \(\overrightarrow{\boldsymbol{v}}_{\mathrm{P} / \mathrm{E}}\) (the velocity of the plane relative to the earth) to the two given vectors. (b) Let \(x\) be east and \(y\) be north, and find the components of \(\overrightarrow{\boldsymbol{v}}_{\mathrm{P} / \mathrm{E}}\). (c) Find the magnitude and direction of \(\overrightarrow{\boldsymbol{v}}_{\mathrm{P} / \mathrm{E}}\)

Short Answer

Expert verified
The components of the velocity of the plane with respect to the Earth are -10/√2 m/s in the x-direction (west) and -(10/√2 + 35) m/s in the y-direction (south). The magnitude and direction of \(\overrightarrow{v}_{P/E}\) are calculated from these components.

Step by step solution

01

Illustrate Vector Addition Diagram

Draw a diagram illustrating the situation. The vector \(\overrightarrow{v}_{P/W}\) representing the velocity of the plane (P) with respect to the wind (W) is directed due south, and has a magnitude of \(35 m/s\). The vector \(\overrightarrow{v}_{W/E}\) representing the velocity of the wind with respect to the Earth (E) is directed southwest, and has a magnitude of \(10 m/s\). The resulting vector \(\overrightarrow{v}_{P/E}\) representing the velocity of the plane with respect to Earth is what we want to find.
02

Decompose Vector

Let's represent north as positive y-direction and east as the positive x-direction. As the wind is blowing toward the southwest, it has components in both south and west directions. Since southwest is exactly between south and west, we can represent the wind’s velocity as \(\overrightarrow{v}_{W/E} = -10/\sqrt{2} \hat{i} - 10/\sqrt{2} \hat{j}\). Plane's velocity with respect to wind \(\overrightarrow{v}_{P/W} = -35 \hat{j}\).
03

Find Components of \(\overrightarrow{v}_{P/E}\)

The velocity of the plane with respect to the Earth is the vector sum of the velocity of the plane with respect to the wind and the velocity of the wind with respect to the Earth. Therefore, \(\overrightarrow{v}_{P/E} = \overrightarrow{v}_{P/W} + \overrightarrow{v}_{W/E} = (-10/\sqrt{2} \hat{i} - 10/\sqrt{2} \hat{j}) + (-35 \hat{j})\). Hence, \(\overrightarrow{v}_{P/E} = -10/\sqrt{2} \hat{i} - (10/\sqrt{2} + 35) \hat{j}\).
04

Find Magnitude and Direction of \(\overrightarrow{v}_{P/E}\)

To find the magnitude of \(\overrightarrow{v}_{P/E}\), use the Pythagorean Theorem: \(|\overrightarrow{v}_{P/E}| = \sqrt{(-10/\sqrt{2})^2 + (10/\sqrt{2} + 35)^2}\). To find the direction, calculate the angle that \(\overrightarrow{v}_{P/E}\) makes with the positive y-axis. Use the formula: \(\tan(\Theta_p) = |v_{P/E}^x| / |v_{P/E}^y|\), where \(v_{P/E}^x\) and \(v_{P/E}^y\) are the x-and y-components of \(\overrightarrow{v}_{P/E}\) respectively.

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Key Concepts

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

Velocity Components
When dealing with vector addition in physics, specifically for problems involving velocity, it's crucial to break down each vector into its respective components. This helps us to easily analyze and understand the vector's effect in each direction.

In this exercise, we consider two vectors: the plane’s velocity relative to the wind and the wind's velocity relative to the Earth. Each vector can be split into components along the x-axis (east-west direction) and the y-axis (north-south direction). This process is called vector decomposition.
  • The plane's velocity, moving due south, is represented as a component only in the negative y-direction: \(-35 \hat{j}\) m/s.
  • The wind's velocity, heading southwest, has components both in the negative x and y directions. Given that it's heading exactly between south and west, these components are equal: \(-10/\sqrt{2} \hat{i} - 10/\sqrt{2} \hat{j}\) m/s.
By breaking down each velocity into its x and y components, we simplify the process of vector addition, which is crucial for finding the plane's resultant velocity with respect to the Earth.
Magnitude and Direction
Understanding both the magnitude and direction of a resultant vector is key to describing the overall motion of an object. Once we have the velocity components, we can find the resultant vector, which gives us this information.

The magnitude of the plane's velocity relative to the Earth is found using the Pythagorean Theorem. By squaring each component of the vector, then summing them and taking the square root, we get the overall magnitude:\[|\overrightarrow{v}_{P/E}| = \sqrt{(\frac{-10}{\sqrt{2}})^2 + (\frac{10}{\sqrt{2}} + 35)^2}\].

This calculation provides the speed of the plane relative to Earth.The direction of the resultant vector is equally important. It describes the angle at which the plane travels compared to a reference direction, typically the y-axis in these problems. We use the tangent function to find this angle:\[\tan(\Theta) = \frac{|v_{P/E}^{x}|}{|v_{P/E}^{y}|}\],where \(v_{P/E}^{x}\) and \(v_{P/E}^{y}\) are the x and y components of the plane's velocity relative to Earth. Solving for \(\Theta\) gives the angle of motion, offering a complete picture of the velocity vector.
Vector Decomposition
Vector decomposition is a fundamental technique in physics, particularly useful for understanding how different forces or velocities interact. By decomposing vectors into x and y components, we can better address problems involving complex movements like those in aviation or nautical contexts.

In this exercise, vector decomposition helps us distinguish and correctly interpret the plane's and wind's impact on overall direction and speed.
  • The wind's vector, being 10 m/s directed southwest, is split into two equal components because southwest is a 45-degree direction. Each of these components is calculated by multiplying the speed by \(\cos(45^{\circ})\) or \(\sin(45^{\circ})\), which is \(\frac{1}{\sqrt{2}}\).
  • The plane's vector in the south direction is straightforward, having only a y-component since it's moving directly south.
By setting the x and y axes as our reference directions, vector decomposition transforms complex vector operations into simple algebraic additions and subtractions, making it a reliable method in vector analysis.

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