/*! 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 7 The \(x\) -component of vector \... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 7

The \(x\) -component of vector \(A\) is \(-25.0 \mathrm{~m}\) and the \(y\) -component is \(+40.0 \mathrm{~m}\). (a) What is the magnitude of \(\vec{A}\) ? (b) What is the angle between the direction of \(\vec{A}\) and the positive direction of \(x\) ?

Short Answer

Expert verified
The magnitude of \( \vec{A} \ \) is about 47.2 meters, and the angle with the positive x-direction is approximately 122掳.

Step by step solution

01

Identify Components

The given components of vector \( \vec{A} \ \) are \( A_x = -25.0 \mathrm{~m} \ \) and \( A_y = +40.0 \mathrm{~m} \ \).
02

Calculate the Magnitude of Vector \( \vec{A} \ \)

The magnitude of a vector \( \vec{A} \ \) can be found using the Pythagorean theorem: \ \[ |\vec{A}| = \sqrt{A_x^2 + A_y^2} \ \] Substituting the given values, we get \ \[ |\vec{A}| = \sqrt{(-25.0 \mathrm{~m})^2 + (40.0 \mathrm{~m})^2} = \sqrt{625 + 1600} = \sqrt{2225} \ \] Simplifying this, \ \[ |\vec{A}| \approx 47.2 \mathrm{~m} \ \]
03

Calculate the Angle

To find the angle \ \( \theta \ \) between the vector \( \vec{A} \ \) and the positive x-direction, use the tangent function: \ \[ \tan(\theta) = \frac{A_y}{A_x} \ \] Substituting the given values, we get \ \[ \tan(\theta) = \frac{40.0 \mathrm{~m}}{-25.0 \mathrm{~m}} = -1.6 \ \] To find \ \( \theta \ \), take the arctangent: \ \[ \theta = \tan^{-1}(-1.6) \approx -58^\text{掳} \ \] Since the vector is in the second quadrant (negative x, positive y), add 180掳 to get the correct angle: \ \[ \theta \approx 180^\text{掳} - 58^\text{掳} = 122^\text{掳} \ \]

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Key Concepts

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

Vector Components
Vectors have two main parts: magnitude and direction. When we break down a vector into its components, we essentially split it into its horizontal (x) and vertical (y) parts.
The given exercise tells us that the x-component of vector \(\overrightarrow{A}\) is -25.0 m and the y-component is +40.0 m. This means that if you were to visualize this vector, it would point 25 meters to the left and 40 meters up.
These components help us to understand how the vector behaves in a coordinate system and makes calculations easier.
Pythagorean Theorem
The Pythagorean Theorem is a mathematical formula used to find the magnitude of a vector.
The theorem states that for a right-angled triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides.
In the case of vectors, this means:
\(|\overrightarrow{A}| = \sqrt{A_x^2 + A_y^2}\)
For our problem, the components are -25.0 m and +40.0 m. Using the Pythagorean Theorem, we calculate the magnitude of the vector by substituting these values:
\(|\overrightarrow{A}| = \sqrt{(-25.0 \ \mathrm{m})^2 + (40.0 \ \mathrm{m})^2} = \sqrt{625 + 1600} = \sqrt{2225} \)
Upon simplifying, we get approximately:
\(|\overrightarrow{A}| \approx 47.2 \ \mathrm{m}\)
This magnitude tells us the length of the vector.
Arctangent Function
To find the direction of a vector, we often use the tangent function, which relates the angle of the vector to its components.
The formula is:
\(\tan(\theta) = \frac{A_y}{A_x}\)
We can solve for the angle \(\theta\) by finding the arctangent (inverse tangent) of the ratio of the y-component to the x-component:
\(\theta = \tan^{-1}\biggl(\frac{40.0 \ \mathrm{m}}{-25.0 \ \mathrm{m}}\biggr) \approx -58^\text{掳}\)
This can give us an angle measured counterclockwise from the positive x-axis.
However, because our vector falls in the second quadrant (negative x and positive y), we need to adjust our angle by adding 180掳:
\(\theta = 180^\text{掳} - 58^\text{掳} = 122^\text{掳}\)
This adjustment gives us the true direction of the vector.
Quadrants in Coordinate System
Understanding which quadrant a vector lies in is crucial for correctly determining its direction.
A coordinate system is divided into four quadrants:
  • Quadrant I: +x, +y
  • Quadrant II: -x, +y
  • Quadrant III: -x, -y
  • Quadrant IV: +x, -y

In our example, since the x-component is -25.0 m and the y-component is +40.0 m, the vector lies in the second quadrant.
This impacts the calculation of the angle, as shown when we added 180掳 to get \(122^\text{掳}\). It's important to know which quadrant to use the correct angle.

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Most popular questions from this chapter

For the vectors $$\vec{a}=(3.0 \mathrm{~m}) \hat{\mathrm{i}}+(4.0 \mathrm{~m}) \hat{\mathrm{j}} \text { and } \vec{b}=(5.0 \mathrm{~m}) \hat{\mathrm{i}}+(-2.0 \mathrm{~m}) \hat{\mathrm{j}},$$ give \(\vec{a}+\vec{b}\) in (a) unit-vector notation, and as (b) a magnitude and (c) an angle (relative to \(\hat{\mathrm{i}}\) ). Now give \(\vec{b}-\vec{a}\) in (d) unit-vector notation, and as (e) a magnitude and (f) an angle.

\(\vec{A}\) has the magnitude \(12.0 \mathrm{~m}\) and is angled \(60.0^{\circ}\) counterclockwise from the positive direction of the \(x\) axis of an \(x y\) coordinate system. Also, \(\vec{B}=(12.0 \mathrm{~m}) \hat{\mathrm{i}}+(8.00 \mathrm{~m}) \hat{\mathrm{j}}\) on that same coordinate system. We now rotate the system, counterclockwise about the origin by \(20.0^{\circ}\), to form an \(x^{\prime} y^{\prime}\) system. On this new system, what are (a) \(\vec{A}\) and (b) \(\vec{B}\), both in unit-vector notation?

A vector \(\vec{B}\), with a magnitude of \(8.0 \mathrm{~m}\), is added to a vector \(\vec{A}\), which lies along an \(x\) axis. The sum of these two vectors is a third vector that lies along the \(y\) axis and has a magnitude that is twice the magnitude of \(\vec{A}\). What is the magnitude of \(\vec{A}\) ?

Oasis \(B\) is \(25 \mathrm{~km}\) due east of oasis \(A\). Starting from oasis \(A\), a camel walks \(24 \mathrm{~km}\) in a direction \(15^{\circ}\) south of east and then walks \(8.0 \mathrm{~km}\) due north. How far is the camel then from oasis \(B\) ?

Find the (a) \(x\) - (b) \(y\) - and (c) \(z\) -components of the sum \(\Delta \vec{r}\) of the displacements \(\Delta \vec{c}\) and \(\Delta \vec{d}\) whose components in meters along the three axes are \(\Delta c_{x}=7.4, \Delta c_{y}=-3.8\) \(\Delta c_{z}=-6.1 ; \Delta d_{x}=4.4, \Delta d_{y}=-2.0, \Delta d_{z}=3.3\)
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