/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Interactive LearningWare 3.1 at ... [FREE SOLUTION] | 91Ó°ÊÓ

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Interactive LearningWare 3.1 at reviews the approach taken in problems such as this one. A bird watcher meanders through the woods, walking \(0.50 \mathrm{~km}\) due east, \(0.75 \mathrm{~km}\) due south, and \(2.15 \mathrm{~km}\) in a direction \(35.0^{\circ}\) north of west. The time required for this trip is \(2.50 \mathrm{~h}\). Determine the magnitude and direction (relative to due west) of the bird watcher's (a) displacement and (b) average velocity. Use kilometers and hours for distance and time, respectively.

Short Answer

Expert verified
Displacement: 1.35 km at 20.84° north of west; Average velocity: 0.54 km/h at 20.84° north of west.

Step by step solution

01

Breaking Down the Displacement Vectors

The bird watcher walks in three segments. Let's represent each as a vector: 1. Walking 0.50 km due east is represented by vector \( \vec{A} = (0.50 \text{ km}, 0) \).2. Walking 0.75 km due south is represented by vector \( \vec{B} = (0, -0.75 \text{ km}) \).3. Walking 2.15 km at 35.0° north of west is split into components: - West component: \( \cos(35.0°) \times 2.15 \text{ km} = 1.76 \text{ km (west)} \) - North component: \( \sin(35.0°) \times 2.15 \text{ km} = 1.23 \text{ km (north)} \) Thus, \( \vec{C} = (-1.76 \text{ km}, 1.23 \text{ km}) \).
02

Calculating Total Displacement

Add the vectors \( \vec{A}, \vec{B}, \text{and} \ \vec{C} \) to find the total displacement vector \( \vec{D} \): \[ \vec{D} = \vec{A} + \vec{B} + \vec{C} = (0.50 \text{ km} - 1.76 \text{ km}, -0.75 \text{ km} + 1.23 \text{ km}) \] This simplifies to: \( \vec{D} = (-1.26 \text{ km}, 0.48 \text{ km}) \).
03

Calculating Magnitude of Displacement

The magnitude of the displacement \(|\vec{D}|\) is calculated using the Pythagorean theorem:\[ |\vec{D}| = \sqrt{(-1.26)^2 + (0.48)^2} \]\[ |\vec{D}| \approx \sqrt{1.5876 + 0.2304} = \sqrt{1.818} \approx 1.35 \text{ km} \].
04

Determining the Direction of Displacement

To find the direction of \( \vec{D} \), calculate the angle \( \theta \) relative to due west using:\[ \theta = \arctan\left(\frac{0.48}{1.26}\right) \approx \arctan(0.38095) \approx 20.84° \] Thus, the direction is 20.84° north of west.
05

Calculating Average Velocity

The average velocity \( \vec{V}_{avg} \) is the total displacement \( \vec{D} \) divided by the total time (2.50 hours):\[ \vec{V}_{avg} = \frac{1.35 \text{ km}}{2.50 \text{ h}} \approx 0.54 \text{ km/h} \]The direction is the same as the displacement, 20.84° north of west.

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Key Concepts

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

Displacement
Displacement refers to the overall change in position of an object. It is a vector quantity, meaning it has both magnitude and direction. In the scenario of the bird watcher, each path through the woods contributes to the total displacement. While distance measures the total ground covered, displacement focuses on the shortest path between the starting and ending points.
The bird watcher's journey is broken into three vectors: walking east, south, and in a direction north of west. By adding these vectors together using vector addition, we can find the resultant vector that represents the bird watcher's total displacement.
Mathematically, the displacement is calculated by summing up the east-west and north-south components separately. The components are added as follows:
  • East-west direction: 0.50 km (east) - 1.76 km (west)
  • North-south direction: -0.75 km (south) + 1.23 km (north)
Converting these into a single vector yields the displacement vector (\(-1.26 \text{ km}, 0.48 \text{ km}\)). This vector gives both the magnitude and direction of the bird watcher's displacement from the starting to the final position.
Average Velocity
Average velocity is a measure of how quickly an object changes its position, taking into account the direction of movement. It is distinct from speed, which only considers magnitude. To find the average velocity of the bird watcher, we use the total displacement vector (\(-1.26 \text{ km}, 0.48 \text{ km}\)) calculated earlier and the total time of the journey, which is 2.50 hours.
Thus, the average velocity vector (\( \text{\vec{V}}_{avg} \)) is obtained by dividing each component of the displacement by the total time. For magnitude:
  • \( \text{Magnitude of } \vec{V}_{avg} = \frac{1.35 \text{ km}}{2.50 \text{ h}} \approx 0.54 \text{ km/h} \)
The direction remains the same as the direction determined for the displacement vector, which is 20.84° north of west. Understanding average velocity helps one identify not just how fast, but also in which direction the bird watcher is moving on average during the trip.
Pythagorean Theorem
The Pythagorean theorem is a mathematical principle employed to determine the length of the sides of a right triangle. It states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
In this exercise, the theorem helps us calculate the magnitude of the displacement vector (\(|\vec{D}|\)). This is done by treating the east-west and north-south components as the two shorter sides of a right triangle:
  • \( |\vec{D}| = \sqrt{(-1.26)^2 + (0.48)^2} \)
  • \( |\vec{D}| \approx \sqrt{1.5876 + 0.2304} \approx \sqrt{1.818} \approx 1.35 \text{ km} \)
By using the Pythagorean theorem, we derive the shortest possible straight-line distance between the bird watcher's starting and ending points, capturing the essence of displacement.
Vector Components
Vector components break down a vector into smaller, more manageable parts. These parts usually align with specific coordinate axes, such as east-west and north-south in this scenario. It simplifies complex vector problems by transforming a vector into scalar components that can be easily addedVector components are particularly helpful when the direction of a vector is not aligned with standard axes, as shown by the 2.15 km walk at 35° north of west. Here, the vector is broken down as:
  • Westward component: \( \cos(35°) \times 2.15 \text{ km} \approx 1.76 \text{ km} \)
  • Northward component: \( \sin(35°) \times 2.15 \text{ km} \approx 1.23 \text{ km} \)
The decomposition facilitates vector addition by outlining clear steps for combining various directions of movement into one coherent displacement vector. Understanding and using vector components is essential in physics for analyzing and solving vector-related problems efficiently.

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Most popular questions from this chapter

(a) Does Barbara see him moving toward the east or toward the west? (b) Does Barbara see him moving toward the north or toward the south? (c) Considering your answers to parts (a) and (b), how does Barbara see Neil moving relative to herself, toward the east and north, toward the east and south, toward the west and north, or toward the west and south? Justify your answers in each case. With respect to the ground, Barbara is skating due south at a speed of \(4.0 \mathrm{~m} / \mathrm{s}\). With respect to the ground, Neil is skating due west at a speed of \(3.2 \mathrm{~m} / \mathrm{s} .\) Find Neil's velocity (magnitude and direction relative to due west) as seen by Barbara. Make sure that your answer agrees with your answer to part (c) of the Concept Questions.

In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for \(0.050 \mathrm{~s}\), during which time it experiences an acceleration of \(340 \mathrm{~m} / \mathrm{s}^{2}\). The ball is launched at an angle of \(51^{\circ}\) above the ground. Determine the horizontal and vertical components of the launch velocity.

Before starting this problem consult Interactive Solution \(\underline{3.67}\) at . A golfer, standing on a fairway, hits a shot to a green that is elevated \(5.50 \mathrm{~m}\) above the point where she is standing. If the ball leaves her club with a velocity of \(46.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(35.0^{\circ}\) above the ground, find the time that the ball is in the air before it hits the green.

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