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(a) Determine the magnitude and direction of the sum of the three vectors \(\overrightarrow{\mathbf{v}}_{1}=4.0 \hat{\mathbf{i}}-8.0 \hat{\mathbf{j}}, \overrightarrow{\mathbf{v}}_{2}=\hat{\mathbf{i}}+\hat{\mathbf{j}},\) and \(\overrightarrow{\mathbf{v}}_{3}=-2.0 \hat{\mathbf{i}}+4.0 \hat{\mathbf{j}} .(b)\) Determine \(\overrightarrow{\mathbf{v}}_{1}-\overrightarrow{\mathbf{v}}_{2}+\overrightarrow{\mathbf{v}}_{3}\)

Short Answer

Expert verified
(a) Magnitude: \(3\sqrt{2}\), Direction: \(315^\circ\); (b) \(1.0 \hat{\mathbf{i}} - 5.0 \hat{\mathbf{j}}\)

Step by step solution

01

Add the Three Vectors

To find the sum of the vectors \(\overrightarrow{\mathbf{v}}_{1}\), \(\overrightarrow{\mathbf{v}}_{2}\), and \(\overrightarrow{\mathbf{v}}_{3}\), add their respective components. The sum of the vectors is given by:\[\begin{align*}\overrightarrow{\mathbf{v}}_{\text{sum}} &= (4.0 + 1.0 - 2.0) \hat{\mathbf{i}} + (-8.0 + 1.0 + 4.0) \hat{\mathbf{j}} \&= 3.0 \hat{\mathbf{i}} - 3.0 \hat{\mathbf{j}}.\end{align*}\]
02

Calculate the Magnitude of the Sum

The magnitude of a vector \(\overrightarrow{\mathbf{v}} = a \hat{\mathbf{i}} + b \hat{\mathbf{j}}\) is calculated using the formula:\[||\overrightarrow{\mathbf{v}}|| = \sqrt{a^2 + b^2}\]For \(\overrightarrow{\mathbf{v}}_{\text{sum}} = 3.0 \hat{\mathbf{i}} - 3.0 \hat{\mathbf{j}}\):\[||\overrightarrow{\mathbf{v}}_{\text{sum}}|| = \sqrt{(3.0)^2 + (-3.0)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}.\]
03

Determine the Direction of the Sum

The direction of the vector is given by the angle \(\theta\) relative to the positive \(\hat{\mathbf{i}}\)-axis, calculated using:\[\theta = \tan^{-1}\left(\frac{b}{a}\right)\]Substitute \(a = 3.0\) and \(b = -3.0\):\[\theta = \tan^{-1}\left(\frac{-3.0}{3.0}\right) = \tan^{-1}(-1) = -45^\circ\]Since the vector lies in the fourth quadrant, we add 360° to get the positive angle: \[\theta = 360^\circ - 45^\circ = 315^\circ.\]
04

Calculate the Resultant of \(\overrightarrow{\mathbf{v}}_{1} - \overrightarrow{\mathbf{v}}_{2} + \overrightarrow{\mathbf{v}}_{3}\)

Subtract \(\overrightarrow{\mathbf{v}}_{2}\) from \(\overrightarrow{\mathbf{v}}_{1}\) and then add \(\overrightarrow{\mathbf{v}}_{3}\):\[\begin{align*}\overrightarrow{\mathbf{v}}_{\text{result}} &= (4.0 \hat{\mathbf{i}} - 8.0 \hat{\mathbf{j}}) - (1.0 \hat{\mathbf{i}} + 1.0 \hat{\mathbf{j}}) + (-2.0 \hat{\mathbf{i}} + 4.0 \hat{\mathbf{j}}) \&= (4.0 - 1.0 - 2.0) \hat{\mathbf{i}} + (-8.0 - 1.0 + 4.0) \hat{\mathbf{j}} \&= 1.0 \hat{\mathbf{i}} - 5.0 \hat{\mathbf{j}}.\end{align*}\]

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Key Concepts

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

Magnitude of a Vector
Understanding the magnitude of a vector is like knowing its size or length in a coordinate space. Imagine you have a line representing a vector. The magnitude tells you exactly how long that line is. To find this, you can use the Pythagorean theorem. For a vector represented as \( \overrightarrow{\mathbf{v}} = a \hat{\mathbf{i}} + b \hat{\mathbf{j}} \), the magnitude \( ||\overrightarrow{\mathbf{v}}|| \) is calculated using:
  • \( ||\overrightarrow{\mathbf{v}}|| = \sqrt{a^2 + b^2} \)
In simple terms, you square each of the components, add them together, and then take the square root of that sum.
This formula helps you understand how to get from one point to another in the plane, with detailed knowledge of the distance covered.
Vector Direction
The direction of a vector is crucial as it shows where the vector points towards on a plane. Imagine it as the arrowhead of your vector line.
To determine the direction, you calculate the angle the vector makes with a standard direction, usually the positive x-axis (\( \hat{\mathbf{i}} \)-axis). Use:
  • \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \)
This gives you an angle \( \theta \).
If this angle is not within the 0° to 360° range, you adjust it depending on the vector's quadrant, according to:
  • For the fourth quadrant, add 360° to \( \theta \)
This ensures the angle is a positive measure heading clockwise from the positive x-axis.
Trigonometric Functions
Trigonometric functions are like tools that help us understand relationships between angles and sides of triangles. For vectors, the most common trig function is tangent (\( \tan \)) which relates to direction.
The tangent function helps in finding vector direction, using its formula:
  • \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \)
This formula lets you convert between the components of a vector and the angle it makes with the x-axis.
By understanding how tangent works, you can quickly figure out the exact tilt or orientation of your vector.
Vector Components
A vector can be thought of as a combination of pieces along the x-axis and y-axis, known as components. If you imagine breaking down a slope into its flat (horizontal) and steep (vertical) parts, that's essentially what vector components are.
For the vector \( \overrightarrow{\mathbf{v}} = a \hat{\mathbf{i}} + b \hat{\mathbf{j}} \), the components are \( a \) along the x-axis and \( b \) along the y-axis. They represent how much the vector moves in each direction.
Learning to add or subtract these components is like stacking blocks. You simply add each corresponding part, like:
  • \( \overrightarrow{\mathbf{v}}_{1} = 4 \hat{\mathbf{i}} - 8 \hat{\mathbf{j}} \)
  • \( \overrightarrow{\mathbf{v}}_{2} = 1 \hat{\mathbf{i}} + 1 \hat{\mathbf{j}} \)
For these, you'd add \( 4 + 1 \) in the x-direction, and \( -8 + 1 \) in the y-direction. The resulting vector shows your new direction and length. Understanding these helps to break down, reconstruct, and analyze any type of motion accurately.

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Most popular questions from this chapter

A person going for a morning jog on the deck of a cruise ship is running toward the bow (front) of the ship at \(2.0 \mathrm{~m} / \mathrm{s}\) while the ship is moving ahead at \(8.5 \mathrm{~m} / \mathrm{s}\). What is the velocity of the jogger relative to the water? Later, the jogger is moving toward the stern (rear) of the ship. What is the jogger's velocity relative to the water now?

(1I) An ant walks on a piece of graph paper straight along the \(x\) axis a distance of 10.0 \(\mathrm{cm}\) in 2.00 \(\mathrm{s}\) . It then turns left \(30.0^{\circ}\) and walks in a straight line another 10.0 \(\mathrm{cm}\) in 1.80 s Finally, it turns another \(70.0^{\circ}\) to the left and walks another 10.0 \(\mathrm{cm}\) in 1.55 \(\mathrm{s}\) . Determine \((a)\) the \(x\) and \(y\) components of the ant's average velocity, and \((b)\) its magnitude and direction.

\((a)\) A skier is accelerating down a \(30.0^{\circ}\) hill at \(1.80 \mathrm{~m} / \mathrm{s}^{2}\) (Fig. \(3-39\) ). What is the vertical component of her acceleration? (b) How long will it take her to reach the bottom of the hill, assuming she starts from rest and accelerates uniformly, if the elevation change is \(325 \mathrm{~m} ?\)

An airplane is heading due south at a speed of \(580 \mathrm{~km} / \mathrm{h}\). If a wind begins blowing from the southwest at a speed of \(90.0 \mathrm{~km} / \mathrm{h}\) (average), calculate (a) the velocity (magnitude and direction) of the plane, relative to the ground, and (b) how far from its intended position it will be after \(11.0 \mathrm{~min}\) if the pilot takes no corrective action. [Hint: First draw a diagram.

A swimmer is capable of swimming \(0.60 \mathrm{~m} / \mathrm{s}\) in still water. \((a)\) If she aims her body directly across a \(55-\mathrm{m}\) -wide river whose current is \(0.50 \mathrm{~m} / \mathrm{s}\), how far downstream (from a point opposite her starting point) will she land? (b) How long will it take her to reach the other side?

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