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Vector \(\overrightarrow{\mathbf{A}}\) has a magnitude of 12.3 units and points due west. Vector \(\overrightarrow{\mathbf{B}}\) points due north. (a) What is the magnitude of \(\overrightarrow{\mathbf{B}}\) if \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) has a magnitude of 15.0 units? (b) What is the direction of \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) relative to due west? (c) What is the magnitude of \(\overrightarrow{\mathbf{B}}\) if \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\) has a magnitude of 15.0 units? (d) What is the direction of \(\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}\) relative to due west?

Short Answer

Expert verified
(a) 8.59 units, (b) 34.8° north of west, (c) 8.59 units, (d) 34.8° south of west.

Step by step solution

01

Analyze Vector Addition Formula

To find the magnitude of vector \( \overrightarrow{\mathbf{B}} \) when \( \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}} \) has a magnitude of 15.0 units, we use the Pythagorean theorem given vectors at right angles. This gives \( |\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}| = \sqrt{A^2 + B^2} \).
02

Applying the Magnitude Formula for Part (a)

Given \( \overrightarrow{\mathbf{A}} = 12.3 \) units and \( |\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}| = 15.0 \) units, substitute these values into the equation:\[15.0 = \sqrt{12.3^2 + B^2}\]Square both sides to eliminate the square root:\[225 = 12.3^2 + B^2\]
03

Solve for \( B \)

Calculate \( 12.3^2 = 151.29 \) and solve for \( B^2 \):\[225 = 151.29 + B^2\]Subtract 151.29 from both sides:\[B^2 = 73.71\]Take the square root to find \( B \):\[B = \sqrt{73.71} \approx 8.59\text{ units}\]
04

Determine Direction of \( \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}} \)

Using the tangent function, \( \tan \theta = \frac{B}{A} \), calculate the angle \( \theta \) relative to due west.\[\tan \theta = \frac{8.59}{12.3}\]\[\theta = \arctan(0.69837) \approx 34.8^\circ \text{ north of west.}\]
05

Analyze Vector Subtraction Formula

For the magnitude of \( \overrightarrow{\mathbf{B}} \) when \( \overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}} \) has a magnitude of 15.0 units, use the formula:\[|\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}| = \sqrt{A^2 + B^2}\] which is similar to the addition formula.
06

Applying the Magnitude Formula for Part (c)

Given \( \overrightarrow{\mathbf{A}} = 12.3 \) units and \( |\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}| = 15.0 \) units, substitute these values into the equation and solve for \( B \). \[15.0 = \sqrt{12.3^2 + B^2}\]Essentially the same calculations as Step 3, thus, \( B \approx 8.59 \text{ units}\).
07

Determine Direction of \( \overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}} \)

Just as in addition, use \( \tan \theta = \frac{B}{A} \) where \( B \approx 8.59 \) units. However, subtract vector direction indicates \( 34.8^\circ \) but south of west:\[ \theta = \arctan(0.69837) \approx 34.8^\circ \text{ south of west.} \]

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Key Concepts

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

Magnitude Calculation
When working with vectors, understanding how to calculate their magnitude is crucial. The magnitude of a vector represents its length or size, regardless of its direction. To find the magnitude of a vector, especially when vectors are at right angles, we use the Pythagorean theorem. If you have vector \( \overrightarrow{\mathbf{A}} \) with a magnitude of 12.3 units and its total combined magnitude with vector \( \overrightarrow{\mathbf{B}} \) is given as 15.0 units, you can apply the Pythagorean theorem:
\[ |\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}| = \sqrt{A^2 + B^2} \]
By plugging in the known values, you isolate \( B \) and solve for it. This is done by first squaring the total magnitude and subtracting \( A^2 \) from it. Then, taking the square root will give you the magnitude of vector \( B \), which is essential in vector addition problems.
  • Magnitude shows how much of something exists.
  • It is always a positive number.
  • Calculated using the Pythagorean theorem for right angle vectors.
Vector Direction
Vectors have both magnitude and direction, forming their full identity. The direction refers to where the vector is pointing, which can be described using angles relative to a reference direction. In our problem, to find the direction of the resultant vector \( \overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} \), we use trigonometric functions.
The direction of a vector is often measured as an angle from a specified line, such as due west or due north. Using the tangent function:
\[ \tan \theta = \frac{B}{A} \]
This allows us to calculate the angle \( \theta \) relative to the given direction. By taking the arctan, the inverse of tan, we determine the angle, giving the vector's precise direction such as "north of west" or "south of west."
  • Direction specifies the way the vector is pointing.
  • Measured using angles from a reference direction.
  • Important for interpreting vector orientation in space.
Pythagorean Theorem
The Pythagorean theorem is a fundamental concept in mathematics, allowing us to find the relationship between the sides of a right triangle. When applied to vectors, it provides an efficient way to calculate the resultant magnitude. Consider vectors at right angles; the formula becomes applicable as it relates to their components.
  • The theorem states: \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse.
  • In vector terms, \( c \) is the magnitude of the resultant vector.
  • It helps determine the length of the vector if two sides are known.
For example, in our given problem, it allows us to calculate the unknown vector component by manipulating the equation:\[ \sqrt{A^2 + B^2} = 15.0 \]With this tool, solving vector addition and subtraction becomes systematic and straightforward, offering solutions to complex geometric and physical problems.

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Most popular questions from this chapter

You are driving into St. Louis, Missouri, and in the distance you see the famous Gateway to the West arch. This monument rises to a height of \(192 \mathrm{m}\). You estimate your line of sight with the top of the arch to be \(2.0^{\circ}\) above the horizontal. Approximately how far (in kilometers) are you from the base of the arch?

Azelastine hydrochloride is an antihistamine nasal spray. A standard size container holds one fluid ounce (oz) of the liquid. You are searching for this medication in a European drugstore and are asked how many milliliters (mL) there are in one fluid ounce. Using the following conversion factors, determine the number of milliliters in a volume of one fluid ounce: 1 gallon \((\) gal \()=128\) oz, \(3.785 \times 10^{-3}\) cubic meters \(\left(\mathrm{m}^{3}\right)=1\) gal, and \(1 \mathrm{mL}=10^{-6} \mathrm{m}^{3}.\)

Vector \(\overrightarrow{\mathbf{A}}\) has a magnitude of 6.00 units and points due east. Vector \(\overrightarrow{\mathbf{B}}\) points due north. (a) What is the magnitude of \(\overrightarrow{\mathbf{B}},\) if the vector \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) points \(60.0^{\circ}\) north of east? (b) Find the magnitude of \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\).

The magnitude of the force vector \(\overrightarrow{\mathbf{F}}\) is 82.3 newtons. The \(x\) component of this vector is directed along the \(+x\) axis and has a magnitude of 74.6 newtons. The \(y\) component points along the \(+y\) axis. (a) Find the direction of \(\overrightarrow{\mathbf{F}}\) relative to the \(+x\) axis. (b) Find the component of \(\overrightarrow{\mathbf{F}}\) along the \(+y\) axis.

An ocean liner leaves New York City and travels \(18.0^{\circ}\) north of east for \(155 \mathrm{km} .\) How far east and how far north has it gone? In other words, what are the magnitudes of the components of the ship's displacement vector in the directions (a) due east and (b) due north?

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