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Chapter 3: Problem 25

Vector A points in the negative \(y\) direction and has a magnitude of 5 units. Vector \(\overrightarrow{\mathrm{B}}\) has twice the magnitude and points in the positive \(\underline{x}\) direction. Find the direction and magnitude of (a) \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) (b) \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}},\) and (c) \(\overrightarrow{\mathrm{B}}-\overrightarrow{\mathbf{A}}\)

Short Answer

Expert verified
a) Magnitude: \(5\sqrt{5}\), Direction: \(-26.6^\circ\). b) Magnitude: \(5\sqrt{5}\), Direction: \(153.4^\circ\). c) Magnitude: \(5\sqrt{5}\), Direction: \(26.6^\circ\).

Step by step solution

01

Understand Vector A

Vector A is in the negative \(y\) direction with a magnitude of 5 units. This can be expressed as the vector \(\vec{A} = 0\hat{i} - 5\hat{j}\). Since there is no component in the \(x\) direction, the \(\hat{i}\) term is zero.
02

Understand Vector B

Vector B has double the magnitude of Vector A, 10 units, and points in the positive \(x\) direction. Therefore, it can be expressed as: \(\vec{B} = 10\hat{i} + 0\hat{j}\).
03

Calculate \(\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}}\)

Add the corresponding components of \(\vec{A}\) and \(\vec{B}\): \(\vec{A} + \vec{B} = (0 + 10)\hat{i} + (-5 + 0)\hat{j} = 10\hat{i} - 5\hat{j}\). The magnitude is \(\sqrt{10^2 + (-5)^2} = \sqrt{125} = 5\sqrt{5}\) units. The direction is calculated as \(\theta = \tan^{-1}\left(\frac{-5}{10}\right) = -26.6^\circ\) from the positive x-axis.
04

Calculate \(\overrightarrow{\mathbf{A}} - \overrightarrow{\mathbf{B}}\)

Subtract the components of \(\vec{B}\) from \(\vec{A}\): \(\vec{A} - \vec{B} = (0 - 10)\hat{i} + (-5 - 0)\hat{j} = -10\hat{i} - 5\hat{j}\). The magnitude is \(\sqrt{(-10)^2 + (-5)^2} = \sqrt{125} = 5\sqrt{5}\) units. The direction is \(\theta = \tan^{-1}\left(\frac{-5}{-10}\right) = 153.4^\circ\) from the positive \(x\) axis, in the third quadrant.
05

Calculate \(\overrightarrow{\mathrm{B}} - \overrightarrow{\mathbf{A}}\)

Subtract the components of \(\vec{A}\) from \(\vec{B}\): \(\vec{B} - \vec{A} = (10 - 0)\hat{i} + (0 + 5)\hat{j} = 10\hat{i} + 5\hat{j}\). The magnitude is \(\sqrt{10^2 + 5^2} = \sqrt{125} = 5\sqrt{5}\) units. The direction is \(\theta = \tan^{-1}\left(\frac{5}{10}\right) = 26.6^\circ\) from the positive \(x\) axis.

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Key Concepts

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

Vector Subtraction
Vector subtraction involves determining the difference between two vectors. It's similar to vector addition but with subtraction of the corresponding components.
  • To subtract vector \( \vec{B} \) from vector \( \vec{A} \), you compute \( \vec{A} - \vec{B} \).
  • This involves subtracting each component of vector \( \vec{B} \) from the respective component of vector \( \vec{A} \).
  • For example, if \( \vec{A} = 0\hat{i} - 5\hat{j} \) and \( \vec{B} = 10\hat{i} + 0\hat{j} \), then \( \vec{A} - \vec{B} = (0-10)\hat{i} + (-5-0)\hat{j} = -10\hat{i} - 5\hat{j} \).
  • The reverse subtraction, \( \vec{B} - \vec{A} \), will yield different components: \( (10-0)\hat{i} + (0+5)\hat{j} = 10\hat{i} + 5\hat{j} \).

This method not only changes the direction of the resultant vector but also its magnitude, depending on initial vectors' orientations.
Vector Components
Vectors in a plane have both an x-component and a y-component, expressed in terms of unit vectors \(\hat{i}\) and \(\hat{j}\), respectively.
  • The x-component is associated with horizontal movement, represented by \(\hat{i}\).
  • The y-component covers vertical movement, indicated by \(\hat{j}\).

Each vector is the sum of these components, formulated as \( \vec{V} = V_x \hat{i} + V_y \hat{j} \).
In our problem:
  • Vector \( \vec{A} \) has components \(0\hat{i} - 5\hat{j}\).
  • Vector \( \vec{B} \) shows \(10\hat{i} + 0\hat{j}\).

Understanding these components is crucial when performing vector addition or subtraction, as each direction is independently addressed. This detailed breakdown aids in visualizing vector operations, making them easier to solve and understand.
Magnitude and Direction
For any vector, magnitude refers to its length. It can be calculated using the Pythagorean theorem, while direction indicates the angle it forms with a reference axis, commonly the positive x-axis.
  • The magnitude of a vector \( \vec{V} = V_x \hat{i} + V_y \hat{j} \) is calculated as \( \sqrt{V_x^2 + V_y^2} \).
  • The direction is determined using trigonometric functions, like tangent: \( \theta = \tan^{-1}\left(\frac{V_y}{V_x}\right) \).

In the problem, after vector operations:
  • The magnitude for each resultant vector is \( 5\sqrt{5} \).
  • The direction angles differ based on vector subtraction order, being \( -26.6^\circ \), \( 153.4^\circ \), and \( 26.6^\circ \) respectively.

This calculated angle can suggest the direction relative to the positive x-axis, often giving insights into vector rotations and other spatial adjustments.

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Most popular questions from this chapter

Vector A points in the negative \(x\) direction. Vector \(\overrightarrow{\mathrm{B}}\) points at an angle of \(30.0^{\circ}\) above the positive \(x\) axis. Vector \(C\) has a magnitude of \(15 \mathrm{m}\) and points in a direction \(40.0^{\circ}\) below the positive \(x\) axis. Given that \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}=0,\) find the magnitudes of \(\overline{\mathbf{A}}\) and \(\overline{\mathbf{B}}\).

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 drive a car 1500 ft to the east, then \(2500 \mathrm{ft}\) to the north. If the trip took 3.0 minutes, what were the direction and magnitude of your average velocity?

A jogger runs with a speed of \(3.25 \mathrm{m} / \mathrm{s}\) in a direction \(30.0^{\circ}\) above the \(x\) axis. (a) Find the \(x\) and \(y\) components of the jogger's velocity. (b) How will the velocity components found in part (a) change if the jogger 's speed is halved?

The \(x\) and \(y\) components of a vector \(\overrightarrow{\mathrm{r}}\) are \(r_{x}=14 \mathrm{m}\) and \(r_{y}=-9.5 \mathrm{m},\) respectively. Find (a) the direction and (b) the magnitude of the vector \(\overrightarrow{\mathbf{r}}\). ( \(\mathrm{c}\) ) If both \(r_{x}\) and \(r_{y}\) a re doubled, how do your answers to parts (a) and (b) change?
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