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(a) What is the sum of \(\overrightarrow{\mathbf{A}}=3.0 \hat{\mathbf{x}}+5.0 \hat{\mathbf{y}}\) and \(\overrightarrow{\mathbf{B}}=1.0 \hat{\mathbf{x}}-3.0 \hat{\mathbf{y}}\) ? (b) What are the magnitude and direction of \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) ?

Short Answer

Expert verified
(a) The sum is \( 4.0 \hat{\mathbf{x}} + 2.0 \hat{\mathbf{y}} \). (b) Magnitude: \( 2\sqrt{5} \); Direction: \( 26.57^\circ \).

Step by step solution

01

Finding the Sum of Vectors

To find the sum of two vectors, we add the corresponding components. The vectors given are \( \overrightarrow{\mathbf{A}}=3.0 \hat{\mathbf{x}}+5.0 \hat{\mathbf{y}} \) and \( \overrightarrow{\mathbf{B}}=1.0 \hat{\mathbf{x}}-3.0 \hat{\mathbf{y}} \). Add the \( \hat{\mathbf{x}} \) components: \( 3.0 + 1.0 = 4.0 \). Add the \( \hat{\mathbf{y}} \) components: \( 5.0 - 3.0 = 2.0 \). Therefore, the sum is \( \overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} = 4.0 \hat{\mathbf{x}} + 2.0 \hat{\mathbf{y}} \).
02

Calculating the Magnitude of the Sum

The magnitude of a vector \( \overrightarrow{\mathbf{C}} = a \hat{\mathbf{x}} + b \hat{\mathbf{y}} \) is given by \( \sqrt{a^2 + b^2} \). For \( \overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}} \), the sum is \( 4.0 \hat{\mathbf{x}} + 2.0 \hat{\mathbf{y}} \). Therefore, the magnitude is \( \sqrt{(4.0)^2 + (2.0)^2} = \sqrt{16 + 4} = \sqrt{20} \). Therefore, the magnitude is \( 2\sqrt{5} \).
03

Determining the Direction of the Sum

The direction \( \theta \) of a vector \( \overrightarrow{\mathbf{C}} \) can be found using \( \tan \theta = \frac{b}{a} \), where \( b \) is the \( \hat{\mathbf{y}} \) component and \( a \) is the \( \hat{\mathbf{x}} \) component. For the sum \( 4.0 \hat{\mathbf{x}} + 2.0 \hat{\mathbf{y}} \), we calculate \( \tan \theta = \frac{2.0}{4.0} = \frac{1}{2} \). Therefore, \( \theta = \tan^{-1}\left(\frac{1}{2}\right) \approx 26.57^\circ \) above 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 Components
Vector components are a way to express a vector using its projections along the coordinate axes. Vectors can be broken down into two parts: one along the x-axis and one along the y-axis. These are called vector components. Each component of a vector is scaled by a unit vector (like \( \hat{\mathbf{x}} \) and \( \hat{\mathbf{y}} \)), which shows the direction along the respective axis.
  • The x-component represents how much the vector moves in the x-direction, while
  • The y-component signifies the motion in the y-direction.
In the exercise, we added the components of vectors \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\) to find their sum. By adding the x-components (3.0 + 1.0) and y-components (5.0 - 3.0), we effectively combined two vectors into one new vector: \( 4.0 \hat{\mathbf{x}} + 2.0 \hat{\mathbf{y}} \). This method simplifies vector addition and enables us to work in a more component-wise straightforward manner.
Magnitude of a Vector
The magnitude of a vector reveals its size or length. It is a measure of how far the vector extends, regardless of its direction, denoted often simply as \(|\overrightarrow{\mathbf{C}}|\). It's determined using the Pythagorean theorem for the components, reminiscent of finding the length of the hypotenuse of a right triangle. To calculate the magnitude of a vector \(\overrightarrow{\mathbf{C}} = a \hat{\mathbf{x}} + b \hat{\mathbf{y}}\), use the formula:\[\sqrt{a^2 + b^2}\]For our sum vector \( 4.0 \hat{\mathbf{x}} + 2.0 \hat{\mathbf{y}} \), its magnitude can be calculated as:\[\sqrt{(4.0)^2 + (2.0)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\]This result, \(2\sqrt{5}\), demonstrates the length of the resulting vector. Magnitude helps us understand the vector's influence or effect without considering its direction.
Direction of a Vector
The direction of a vector describes where the vector points in the coordinate system. Unlike magnitude, which is purely about size, the direction tells you in which direction the vector is acting. This is usually measured as the angle \(\theta\) with respect to the positive x-axis. To find the direction, trigonometry is often employed with the tangent function:\[\tan \theta = \frac{b}{a}\]where \(b\) is the y-component and \(a\) is the x-component of the vector.Applying this to our result, we have:\[\tan \theta = \frac{2.0}{4.0} = \frac{1}{2}\]Thus, the angle is calculated as:\[\theta = \tan^{-1}\left(\frac{1}{2}\right) \approx 26.57^\circ\]This angle tells us that the resultant vector \(4.0 \hat{\mathbf{x}} + 2.0 \hat{\mathbf{y}}\) is directed approximately 26.57 degrees above the positive x-axis. Knowing the direction is crucial for fully understanding vector behavior in physics and engineering contexts.

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Most popular questions from this chapter

A pitcher throws a fastball horizontally at a speed of \(140 \mathrm{~km} / \mathrm{h}\) toward home plate, \(18.4 \mathrm{~m}\) away. \((\mathrm{a})\) If the batter's combined reaction and swing times total \(0.350 \mathrm{~s}\), how long can the batter watch the ball after it has left the pitcher's hand before swinging? (b) In traveling to the plate, how far does the ball drop from its original horizontal line?

It is raining, and there is no wind. When you are sitting in a stationary car, the rain falls straight down relative to the car and the ground. But when you're driving, the rain appears to hit the windshield at an angle. (a) As the velocity of the car increases, this angle (1) also increases, (2) remains the same, (3) decreases. Why? (b) If the raindrops fall straight down at a speed of \(10 \mathrm{~m} / \mathrm{s},\) but appear to make an angle of \(25^{\circ}\) to the vertical, what is the speed of the car?

In two successive chess moves, a player first moves his queen two squares forward, then moves the queen three steps to the left (from the player's view). Assume each square is \(3.0 \mathrm{~cm}\) on a side. (a) Using forward (toward the player's opponent) as the positive \(y\) -axis and right as the positive \(x\) -axis, write the queen's net displacement in component form. (b) At what net angle was the queen moved relative to the leftward direction?

A boat that travels at a speed of \(6.75 \mathrm{~m} / \mathrm{s}\) in still water is to go directly across a river and back ( \(>\) Fig. 3.38 ). The current flows at \(0.50 \mathrm{~m} / \mathrm{s}\). (a) At what angle(s) must the boat be steered? (b) How long does it take to make the round trip? (Assume that the boat's speed is constant at all times, and neglect turnaround time.)

A stone thrown off a bridge \(20 \mathrm{~m}\) above a river has an initial velocity of \(12 \mathrm{~m} / \mathrm{s}\) at an angle of \(45^{\circ}\) above the horizontal (vFig. 3.33). (a) What is the range of the stone? (b) At what velocity does the stone strike the water?

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