/*! 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 31 Find the direction and magnitude... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the direction and magnitude of the vectors. (a) \(\overrightarrow{\mathrm{A}}=(5.0 \mathrm{m}) \hat{\mathrm{x}}+(-2.0 \mathrm{m}) \hat{\mathrm{y}}\) (b) \(\overrightarrow{\mathrm{B}}=(-2.0 \mathrm{m}) \hat{\mathrm{x}}+(5.0 \mathrm{m}) \hat{\mathrm{y}},\) and (c) \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\)

Short Answer

Expert verified
(a) A: 5.39 m, -21.8°; (b) B: 5.39 m, 111.8°; (c) C: 4.24 m, 45°.

Step by step solution

01

Find Magnitude of Vector A

The magnitude of a vector \( \overrightarrow{A} = a_x \hat{x} + a_y \hat{y} \) is calculated using the formula \( \| \overrightarrow{A} \| = \sqrt{a_x^2 + a_y^2} \). For vector A, \( a_x = 5.0 \text{ m} \) and \( a_y = -2.0 \text{ m} \), so the magnitude is \( \| \overrightarrow{A} \| = \sqrt{(5.0)^2 + (-2.0)^2} = \sqrt{25 + 4} = \sqrt{29} \approx 5.39 \text{ m} \).
02

Find Direction of Vector A

The direction \( \theta \) of a vector is found using \( \theta = \tan^{-1} \left( \frac{a_y}{a_x} \right) \). For vector A, \( \tan^{-1} \left( \frac{-2.0}{5.0} \right) \approx -21.8^\circ \). Ensure to determine the correct quadrant: since \( a_x > 0 \) and \( a_y < 0 \), it's in the fourth quadrant.
03

Find Magnitude of Vector B

Similarly, for vector B with \( b_x = -2.0 \text{ m} \) and \( b_y = 5.0 \text{ m} \), the magnitude is \( \| \overrightarrow{B} \| = \sqrt{(-2.0)^2 + (5.0)^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.39 \text{ m} \).
04

Find Direction of Vector B

The direction \( \theta \) for vector B is \( \theta = \tan^{-1} \left( \frac{b_y}{b_x} \right) = \tan^{-1} \left( \frac{5.0}{-2.0} \right) \approx 111.8^\circ \). Since \( b_x < 0 \) and \( b_y > 0 \), vector B is in the second quadrant.
05

Calculate Vector C = A + B

To find \( \overrightarrow{A} + \overrightarrow{B} \), add the components: \( c_x = a_x + b_x = 5.0 - 2.0 = 3.0 \) m and \( c_y = a_y + b_y = -2.0 + 5.0 = 3.0 \) m. Thus, \( \overrightarrow{C} = 3.0 \hat{x} + 3.0 \hat{y} \).
06

Find Magnitude of Vector C

The magnitude of \( \overrightarrow{C} \) is \( \| \overrightarrow{C} \| = \sqrt{3.0^2 + 3.0^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \approx 4.24 \text{ m} \).
07

Find Direction of Vector C

The angle \( \theta \) is found by \( \theta = \tan^{-1} \left( \frac{c_y}{c_x} \right) = \tan^{-1} (1) = 45^\circ \). Since \( c_x > 0 \) and \( c_y > 0 \), vector C is in the first quadrant.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Vector Magnitude
When dealing with vectors, one of the first tasks is finding the magnitude. Magnitude provides a scalar value that indicates the length or size of a vector, irrespective of its direction. It is crucial for understanding how much of something you have, like how far a car is to travel or the amount of force applied in a certain direction. To calculate the magnitude of a vector, say \( \overrightarrow{A} = a_x \hat{x} + a_y \hat{y} \), use the formula \( \| \overrightarrow{A} \| = \sqrt{a_x^2 + a_y^2} \). This formula derives from the Pythagorean Theorem, treating the components as sides of a right triangle.
For example, in our exercise, vector \( \overrightarrow{A} \) has components 5.0 m along the x-axis and -2.0 m along the y-axis. Calculating, we find the magnitude to be approximately 5.39 m. This same formula applies to vector \( \overrightarrow{B} \), resulting in a similar magnitude due to its symmetric component values of -2.0 m and 5.0 m. These calculations are foundational for engaging in vector problems, setting the stage for further analysis regarding direction and combination.
Vector Direction
Understanding the direction of a vector is crucial as it tells us which way the vector is pointing in a coordinate plane. The direction is often expressed as an angle relative to a reference axis, typically the positive x-axis. To find this angle \( \theta \), you use the tangent inverse function: \( \theta = \tan^{-1} \left( \frac{a_y}{a_x} \right) \).
In the case of vector \( \overrightarrow{A} \) from the exercise, the direction angle worked out to be about -21.8 degrees. The negative sign indicates it points below the x-axis, placing it in the fourth quadrant. Meanwhile, for vector \( \overrightarrow{B} \), the direction is around 111.8 degrees, corresponding to the second quadrant as it is angled above the negative x-axis. Knowing the direction enables us to predict how vectors interact with each other and their contributions to resultant effects.
Vector Addition
Vector addition is a fundamental operation in vector analysis, allowing us to combine effects or movements indicated by two or more vectors. To add vectors, combine their respective components. For vectors \( \overrightarrow{A} = a_x \hat{x} + a_y \hat{y} \) and \( \overrightarrow{B} = b_x \hat{x} + b_y \hat{y} \), the resultant vector \( \overrightarrow{C} = (a_x + b_x) \hat{x} + (a_y + b_y) \hat{y} \).
Take the example from our exercise: Adding vectors \( \overrightarrow{A} \) and \( \overrightarrow{B} \) gives \( \overrightarrow{C} = 3.0 \hat{x} + 3.0 \hat{y} \). This resultant vector sums up the effects of the two and integrates their magnitudes and directions into a single representation. Calculating the magnitude and direction afterward provides complete information about this new vector. Here, the magnitude is approximately 4.24 m, and the direction is 45 degrees, placing it neatly in the first quadrant.
By mastering vector addition, complex tasks such as flight navigation or force balancing become more straightforward, offering a clear result of accumulated influences.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You are driving up a long, inclined road. After 1.2 miles you notice that signs along the roadside indicate that your elevation has increased by \(530 \mathrm{ft}\) (a) What is the angle of the road above the horizontal? (b) How far do you have to drive to gain an additional 150 ft of elevation?

You slide a box up a loading ramp that is \(10.0 \mathrm{ft}\) long. At the top of the ramp the box has risen a height of \(3.00 \mathrm{ft}\). What is the angle of the ramp above the horizontal?

Two airplanes taxi as they approach the terminal. Plane 1 taxies with a speed of \(12 \mathrm{m} / \mathrm{s}\) due north. Plane 2 taxies with a speed of \(7.5 \mathrm{m} / \mathrm{s}\) in a direction \(20^{\circ}\) north of west. (a) What are the direction and magnitude of the velocity of plane 1 relative to plane \(2 ?\) (b) What are the direction and magnitude of the velocity of plane 2 relative to plane \(1 ?\)

Vector \(\overrightarrow{\mathbf{A}}\) points in the negative \(x\) direction and has a magnitude of 22 units. The vector \(\overrightarrow{\mathbf{B}}\) points in the positive \(y\) direction. (a) Find the magnitude of \(\overrightarrow{\mathbf{B}}\) if \(\overline{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) has a magnitude of 37 units. (b) Sketch \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathrm{B}}\).

The components of a vector \(\overrightarrow{\mathbf{A}}\) satisfy \(A_{x}<0\) and \(A_{y}<0 .\) Is the direction angle of \(\overrightarrow{\mathbf{A}}\) between \(0^{\circ}\) and \(90^{\circ},\) between \(90^{\circ}\) and \(180^{\circ},\) between \(180^{\circ}\) and \(270^{\circ},\) or between \(270^{\circ}\) and \(360^{\circ} ?\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.