/*! 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 19 A vector has an \(x\) component ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A vector has an \(x\) component of \(-25.0\) units and a \(y\) component of 40.0 units. Find the magnitude and direction of this vector.

Short Answer

Expert verified
The magnitude of the vector is 47.2 units, and its direction is approximately 122.0 degrees from the positive x-axis.

Step by step solution

01

Calculate Magnitude

To find the magnitude of a vector with components in the x and y directions, use the Pythagorean theorem. The magnitude (or length) can be calculated by \(\sqrt{x^2 + y^2}\), where \(x\) and \(y\) are the vector's components. In this case, the magnitude is \(\sqrt{(-25.0)^2 + (40.0)^2}\).
02

Perform the Calculation for Magnitude

Square the x and y components and then add them together: \( (-25.0)^2 + (40.0)^2 = 625 + 1600 = 2225 \). Take the square root of the result to find the magnitude: \(\sqrt{2225} = 47.2 \)
03

Calculate the Direction

To find the direction of the vector, calculate the angle it makes with the positive x-axis using the tangent inverse function, often known as arctan or tan-1. The angle \(\theta\) is given by \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\), so \(\theta = \tan^{-1}\left(\frac{40.0}{-25.0}\right)\).
04

Perform the Calculation for Direction

Calculate the arctan of \(40.0/-25.0\) to find the angle. Be careful to note that the vector is in the second quadrant (since the x-component is negative and the y-component is positive), so you'll need to adjust the angle accordingly if your calculator provides a negative result.
05

Adjust the Angle for Proper Quadrant

If the calculator gives a negative angle, add \(180\) degrees to find the angle relative to the positive x-axis. For instance, if \(\tan^{-1}\left(\frac{40.0}{-25.0}\right)\) gives \(\theta = -58.0\) degrees, the actual direction is \(\theta + 180 = 122.0\) degrees.

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.

Pythagorean Theorem
The Pythagorean theorem is a fundamental concept in mathematics, particularly in geometry. It states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This can be expressed as the formula: \[ c^2 = a^2 + b^2 \] where \(c\) represents the length of the hypotenuse, and \(a\) and \(b\) represent the lengths of the triangle's other two sides.

When dealing with vectors, this theorem can be used to calculate the magnitude or length of a vector, which represents the hypotenuse of a triangle formed by its \(x\) and \(y\) components as the adjacent and opposite sides. The addition of vectors corresponds to the complementary sides of the triangle, and their resultant magnitude is the length of the hypotenuse. In the exercise provided, the vector's magnitude is calculated using the Pythagorean theorem with its \(x\) and \(y\) components.
Trigonometric Functions
Trigonometric functions are the foundation of trigonometry and are used to describe the relationships between the angles and sides of triangles. The three basic trigonometric functions are sine (sin), cosine (cos), and tangent (tan), and they each correspond to ratios of sides of a right-angled triangle. The tangent of an angle within a right triangle is the ratio of the opposite side to the adjacent side: \[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]

In the context of vector magnitude and direction, the tangent function is used to find the angle of the vector relative to the horizontal axis. This angle is found by taking the inverse tangent (arc-tan or tan-1) of the ratio of the vector's \(y\) component to its \(x\) component. The exercise requires you to calculate this angle to determine the direction of the vector, considering the proper quadrant the vector resides in.
Unit Vector Components
Unit vectors are vectors with a magnitude of one. They are often used to indicate direction and serve as a basis for any vector in a coordinate system, as any vector can be expressed as a combination of unit vectors scaled by the appropriate magnitude. In two-dimensional space, the standard unit vectors are \(\hat{i}\) and \(\hat{j}\) which represent the \(x\)-axis and \(y\)-axis directions, respectively.

To describe a vector in terms of unit vectors, you can deconstruct it into its components along the axes. For instance, a vector \(v\) with components of \(x\) and \(y\) can be written as \(v = x\hat{i} + y\hat{j}\). Understanding unit vector components is essential when computing vector operations, including addition, subtraction, and multiplication (dot and cross products).
Vector Calculus
Vector calculus is an essential branch of mathematics that extends calculus to vector fields. It involves various operations with vectors, such as differentiation and integration, and helps in understanding physical concepts in multiple dimensions. In the area of vector magnitude and direction, vector calculus provides a systematic approach to manipulating and combining vectors to find results such as magnitude, direction, and even work done by a force.

In the given exercise, we scratch the surface of vector calculus by finding the magnitude and direction, the initial steps in describing any vector field. The calculated magnitude tells us how strong the vector is, while the direction tells us where it is pointed in space, which is critical information when solving more complex problems involving vectors in physics and engineering.

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

A rectangular parallelepiped has dimensions \(a, b,\) and \(c,\) as in Figure \(\mathrm{P} 3.64\) . (a) Obtain a vector expression for the face diagonal vector \(\mathbf{R}_{1} .\) What is the magnitude of this vector? (b) Obtain a vector expression for the body diagonal vector \(\mathbf{R}_{2} .\) Note that \(\mathbf{R}_{1}, c \mathbf{k},\) and \(\mathbf{R}_{2}\) make a right triangle and prove that the magnitude of \(\mathbf{R}_{2}\) is \(\sqrt{a^{2}+b^{2}+c^{2}}\)

The instantaneous position of an object is specified by its position vector \(\mathbf{r}\) leading from a fixed origin to the location of the point object. Suppose that for a certain object the position vector is a function of time, given by \(\mathbf{r}=4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 t \hat{\mathbf{k}},\) where \(r\) is in meters and \(t\) is in seconds. Evaluate \(d \mathbf{r} / d t .\) What does it represent about the object?

Find the sum of these four vector forces: 12.0 \(\mathrm{N}\) to the right at \(35.0^{\circ}\) above the horizontal, 31.0 \(\mathrm{N}\) to the left at \(55.0^{\circ}\) above the horizontal, 8.40 \(\mathrm{N}\) to the left at \(35.0^{\circ} \mathrm{be}-\) low the horizontal, and 24.0 \(\mathrm{N}\) to the right at \(55.0^{\circ}\) below the horizontal. Follow these steps: Make a drawing of this situation and select the best axes for \(x\) and \(y\) so you have the least number of components. Then add the vectors by the component method.

As it passes over Grand Bahama Island, the eye of a hurricane is moving 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 hurricane suddenly shifts due north, and its speed slows to 25.0 \(\mathrm{km} / \mathrm{h}\) . How far from Grand Bahama is the eye 4.50 \(\mathrm{h}\) after it passes over the island?

While exploring a cave, a spelunker starts at the entrance and moves the following distances. She goes 75.0 \(\mathrm{m}\) north, 250 \(\mathrm{m}\) east, 125 \(\mathrm{m}\) at an angle \(30.0^{\circ}\) north of east, and 150 \(\mathrm{m}\) south. Find the resultant displacement from the cave entrance.

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.