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A dog searching for a bone walks \(3.50 \mathrm{m}\) south, then runs \(8.20 \mathrm{m}\) at an angle \(30.0^{\circ}\) north of east, and finally walks \(15.0 \mathrm{m}\) west. Find the dog's resultant displacement vector using graphical techniques.

Short Answer

Expert verified
To solve this exercise, you have to represent each movement of the dog as a vector and then add those vectors together by using graphical techniques. You should consider the direction and magnitude of each vector while addition. After adding the vectors, you will get the resultant vector.

Step by step solution

01

Understand the Scenario

The dog's movement has been divided into three actions. Each action can be represented with a vector. The first vector \( A \) represents the movement 3.50 m South which can be represented as a line drawn to the bottom in our graph. The 2nd vector \( B \) represents the movement of 8.20 m at a 30 degree angle North of East which means it's moving upward to the right. The last vector \( C \), 15 m to the West is a straight leftward movement.
02

Find Components of Vectors

Each of these vectors will have two components, horizontal and vertical. We find vector \( B \)'s components by: \( Bx = 8.20 \cos(30.0^{\circ}) \) and \( By = 8.20 \sin(30.0^{\circ}) \). Vector \( B \) is the only vector which has both horizontal and vertical components. Vectors \( A \) and \( C \) only have one component each, vertical and horizontal respectively.
03

Add Horizontal and Vertical Components

We get the resultant displacement vector by adding these components together. The resultant horizontal component \( Rx \) can be calculated as \( Rx = Bx + C = 8.20 \cos(30.0^{\circ}) - 15.0 \). Similarly, the resultant vertical component \( Ry \) would be \( Ry = A + By = -3.50 + 8.20 \sin(30.0^{\circ}) \) where southward is taken as the negative direction.
04

Find Magnitude and Direction of Resultant Vector

The magnitude of the resultant displacement vector can be calculated as \( R = \sqrt{Rx^{2} + Ry^{2}} \) while direction can be found using the tangent function \( \theta = \tan^{-1}(Ry/Rx) \). Keep in mind to add or subtract 180 degrees depending on the quadrant of the resultant vector.

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Key Concepts

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

Vector Decomposition
In physics, vector decomposition is the process of breaking a vector into its constituent components, typically along the x (horizontal) and y (vertical) axes. This is crucial for solving problems involving multiple vectors, particularly when they do not align with the standard axes or when they are at an angle.

Take the scenario of our dog's journey, for instance. When the dog moves at an angle, this movement can be decomposed into two parts: one part in the horizontal direction and another in the vertical direction. Let's focus on the vector representing the dog's movement north of east at a 30-degree angle. By using trigonometric functions, we can decompose this into horizontal (\( B_x \) ) and vertical (\( B_y \) ) components using the formulas: \begin{align*}B_x &= 8.20 \times \text{cos}(30^{\times}) \text{, and} \B_y &= 8.20 \times \text{sin}(30^{\times}).onumber\br\brThe advantage of vector decomposition lies in its ability to simplify complex vector analysis into more manageable calculations, especially when combined with graphical vector addition or other mathematical techniques.
Graphical Vector Addition
Graphical vector addition allows us to visually determine the resultant vector when multiple vectors are involved. It involves placing vectors tip to tail and drawing a new vector (the resultant) from the start of the first vector to the tip of the last vector.

In the context of the dog's movement, each leg of its journey is a vector—southward, northeastward at an angle, and westward. By sketching these vectors in order and connecting the starting point to the end point, we can find the dog's overall displacement. The graphical method is a powerful tool because it gives an intuitive and immediate sense of the magnitude and direction of the resultant vector.

However, for greater precision, especially in complex problems or when exact values are needed, it is often followed up with analytical methods using trigonometry and the Pythagorean theorem for exact results.
Trigonometry in Physics
Trigonometry plays a significant role in physics, especially when dealing with vectors. Angles and distances in vector problems are often related through trigonometric functions like sine, cosine, and tangent. These functions are essential for translating between the direction and magnitude of vectors and their horizontal and vertical components.

For instance, in calculating the dog's displacement vector, trigonometric functions help to determine the components of the northeastward vector. Additionally, once the horizontal (\( R_x \) ) and vertical (\( R_y \) ) components of the resultant displacement are found, trigonometry once again aids in finding the final magnitude and direction through the equations: \begin{align*}R &= \text{sqrt}({R_x^2} + {R_y^2}) \text{, and} \theta &= \text{tan}^{-1}(R_y/R_x).onumber\br\brTrigonometry thus bridges the gap between graphical representation and the computation required for precise answers in physics problems, ensuring accurate and comprehensible solutions.

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Most popular questions from this chapter

Instructions for finding a buried treasure include the fol. lowing: Go 75.0 paces at \(240^{\circ},\) turn to \(135^{\circ}\) and walk 125 paces, then travel 100 paces at \(160^{\circ} .\) The angles are measured counterclockwise from an axis pointing to the east, the \(+x\) direction. Determine the resultant displacement from the starting point.

A surveyor measures the distance across a straight river by the following method: starting directly across from a tree on the opposite bank, she walks \(100 \mathrm{m}\) along the riverbank to cstablish a basclinc. Then she sights across to the tree. The angle from her bascline to the tree is \(35.0^{\circ} .\) How wide is the river?

The vector \(A\) has \(x, y,\) and \(z\) components of \(8.00,12.0,\) and -4.00 units, respectively. (a) Write a vector expression for A in unit-vector notation. (b) Obtain a unit-vector expression for a vector \(\mathbf{B}\) one fourth the length of A pointing in the same direction as A. (c) Obtain a unit-vector expres. sion for a vector \(\mathbf{C}\) three times the length of A pointing in the dircction opposite the dircction of A.

Find the sum of these four vector forces: \(12.0 \mathrm{N}\) to the right at \(35.0^{\circ}\) above the horizontal, \(31.0 \mathrm{N}\) to the left at \(55.0^{\circ}\) above the horizontal, \(8.40 \mathrm{N}\) to the left at \(35.0^{\circ} \mathrm{be}\) low the horizontal, and \(24.0 \mathrm{N}\) to the right at \(55.0^{\circ}\) below the horizontal. Follow these steps: Make a drawing of this situation and sclect the best axcs for \(x\) and \(y\) so you have the least number of components. Then add the vectors by the component method.

Vector A has a negative \(x\) component 3.00 units in length and a positive \(y\) component 2.00 units in length. (a) Deter. mine an expression for \(\mathbf{A}\) in unit-vector notation. (b) Determine the magnitude and direction of \(\mathbf{A}\) (c) What vector \(\overline{\mathbf{B}}\) when added to \(\mathbf{A}\) gives a resultant vector with no \(x\) component and a negative \(y\) component 4.00 units in length?

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