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Chapter 1: Problem 45

As preparation for this problem, consult Concept Simulation 1.1 at \(.\) On a safari, a team of naturalists sets out toward a research station located \(4.8 \mathrm{~km}\) away in a direction \(42^{\circ}\) north of east. After traveling in a straight line for \(2.4 \mathrm{~km}\), they stop and discover that they have been traveling \(22^{\circ}\) north of east, because their guide misread his compass. What are (a) the magnitude and (b) the direction (relative to due east) of the displacement vector now required to bring the team to the research station?

Short Answer

Expert verified
(a) The magnitude of the displacement is approximately 2.67 km, and (b) the direction is approximately 60.67掳 north of east.

Step by step solution

01

Determine the Initial Displacement Components

The team intended to travel 4.8 km, 42掳 north of east, but instead traveled 2.4 km, 22掳 north of east. First, let's find the eastward (x) and northward (y) components of this unintended travel. Using trigonometry: \( x_1 = 2.4 \times \cos(22掳) \) \( y_1 = 2.4 \times \sin(22掳) \).Substitute the values:\( x_1 \approx 2.23 \, \text{km} \) \( y_1 \approx 0.90 \, \text{km} \).
02

Determine the Target Displacement Components

Next, calculate the components of the vector from the starting point directly toward the ultimate destination:\( x_2 = 4.8 \times \cos(42掳) \) \( y_2 = 4.8 \times \sin(42掳) \).Substitute the values:\( x_2 \approx 3.57 \, \text{km} \) \( y_2 \approx 3.21 \, \text{km} \).
03

Calculate the Required Displacement Components

Determine the difference between the target displacement components and the traveled displacement components to find the required displacement:\( x_3 = x_2 - x_1 = 3.57 - 2.23 = 1.34 \, \text{km} \)\( y_3 = y_2 - y_1 = 3.21 - 0.90 = 2.31 \, \text{km} \).
04

Find Magnitude of the Required Displacement

The magnitude of the remaining displacement vector is calculated using the Pythagorean theorem:\( d = \sqrt{x_3^2 + y_3^2} = \sqrt{1.34^2 + 2.31^2} \).After calculation:\( d \approx 2.67 \, \text{km} \).
05

Find Direction of the Required Displacement

Determine the angle of the required displacement vector relative to the east using the tangent function:\( \theta = \tan^{-1} \left( \frac{y_3}{x_3} \right) = \tan^{-1} \left( \frac{2.31}{1.34} \right) \).After calculation:\( \theta \approx 60.67^{\circ} \text{ north of east} \).

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Key Concepts

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

Trigonometry in Physics
In physics, trigonometry is a vital tool for analyzing motions and forces. When dealing with problems like displacement vectors, understanding angles and sides of triangles becomes essential. This application helps break down complex motions into more manageable components. By using basic trigonometric functions like sine, cosine, and tangent, we can compute the directions and magnitudes of vectors.
Conceptually, remember:
  • Sine relates the opposite side to the hypotenuse.
  • Cosine relates the adjacent side to the hypotenuse.
  • Tangent relates the opposite side to the adjacent side.
These functions prove instrumental when converting between angles and side lengths in vector problems.
Vector Components
Vectors are fundamental in representing quantities with both magnitude and direction. In our problem, the team鈥檚 intended and actual paths are shown as vectors. To analyze their journey, we break these vectors into components. This means translating the vector into horizontal (\(x\)) and vertical (\(y\)) parts.
Here's how it's done:
  • The eastward (or horizontal) component is calculated using cosine.
  • The northward (or vertical) component is calculated using sine.
By understanding vector components, you can effectively manage and solve problems by simplifying how vectors interact or change.
Pythagorean Theorem
The Pythagorean theorem is a cornerstone in physics for calculating the magnitude of a resultant vector. When you have right triangle sides, it helps you find the hypotenuse, representing the total displacement or any other resultant magnitude.
This is expressed as:
  • \[d = \sqrt{x^2 + y^2}\]
Here, \(d\) is the magnitude, while \(x\) and \(y\) are the vector components. By using this theorem, you can seamlessly transition from component form back to a single vector magnitude, which is crucial for determining exact distances or forces.
Angle Measurement
Angles play a crucial role in determining the accurate direction of vectors. In the context of our exercise, we measured angles as deviations from the east, reflecting compass directions.
To find the direction of a vector, the tangent function is often used:
  • \[\theta = \tan^{-1} \left( \frac{y}{x} \right)\]
This equation calculates \(\theta\), the angle of the resultant vector, offering insight into its precise direction. This understanding is key in navigation and many real-world applications, ensuring you're on the correct path.
Physics Problem Solving
Effective problem-solving in physics requires a systematic approach. Start by breaking down the problem into manageable steps, like resolving vectors into components or applying trigonometry. Visual aids, like diagrams or sketches, can also help.
  • Identify what you know and what's needed.
  • Use consistent units and validate calculations.
  • Double-check results for consistency with the physical context.
Keep iterating on the problem-solving process. With practice, tackling questions like these will improve both your understanding and efficiency.

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Most popular questions from this chapter

Before starting this problem, review Interactive Solution \(\underline{1.29}\) at. Vector \(\overrightarrow{\mathrm{A}}\) has a magnitude of 12.3 units and points due west. Vector \(\vec{B}\) points due north. (a) What is the magnitude of \(\overrightarrow{\mathbf{B}}\) if \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\) has a magnitude of 15.0 units? (b) What is the direction of \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\) relative to due west? (c) What is the magnitude of \(\overrightarrow{\mathrm{B}}\) if \(\overrightarrow{\mathrm{A}} \overrightarrow{\mathrm{B}}\) has a magnitude of 15.0 units? (d) What is the direction of \(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}\) relative to due west?

Multiple-Concept Example 9 reviews the concepts that play a role in this problem. As an aid in working this problem, consult Concept Simulation 1.1 at. Two forces are applied to a tree stump to pull it out of the ground. Force \(\overrightarrow{\mathbf{F}}_{\mathbf{A}}\) has a magnitude of 2240 newtons and points \(34.0^{\circ}\) south of east, while force \(\overrightarrow{\mathbf{F}}_{\mathbf{B}}\) has a magnitude of 3160 newtons and points due south. Using the component method, find the magnitude and direction of the resultant force \(\overrightarrow{\mathbf{F}}_{\mathbf{A}}+\overrightarrow{\mathbf{F}}_{\mathbf{B}}\) that is applied to the stump. Specify the direction with respect to due east.

The \(x\) vector component of a displacement vector \(\overrightarrow{\mathbf{r}}\) has a magnitude of \(125 \mathrm{~m}\) and points along the negative \(x\) axis. The \(y\) vector component has a magnitude of \(184 \mathrm{~m}\) and points along the negative \(y\) axis. Find the magnitude and direction of \(\overrightarrow{\mathbf{r}}\). Specify the direction with respect to the negative \(x\) axis.

Multiple-Concept Example 9 deals with the concepts that are important in this problem. A grasshopper makes four jumps. The displacement vectors are (1) \(27.0 \mathrm{~cm}\), due west; (2) \(23.0 \mathrm{~cm}, 35.0^{\circ}\) south of west; (3) \(28.0 \mathrm{~cm}, 55.0^{\circ}\) south of east; and (4) \(35.0 \mathrm{~cm}, 63.0^{\circ}\) north of east. Find the magnitude and direction of the resultant displacement. Express the direction with respect to due west.

Three deer, \(A, B,\) and \(C,\) are grazing in a field. Deer \(B\) is located \(62 \mathrm{~m}\) from deer \(A\) at an angle of \(51^{\circ}\) north of west. Deer \(\mathrm{C}\) is located \(77^{\circ}\) north of east relative to deer \(\mathrm{A}\). The distance between deer \(\mathrm{B}\) and \(\mathrm{C}\) is \(95 \mathrm{~m}\). What is the distance between deer \(\mathrm{A}\) and \(\mathrm{C} ?\)
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