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Chapter 4: Problem 39

In a game of lawn chess, where pieces are moved between the centers of squares that are each \(1.00 \mathrm{~m}\) on edge, a knight is moved in the following way: (1) two squares forward, one square rightward; (2) two squares leftward, one square forward; (3) two squares forward, one square leftward. What are (a) the magnitude and (b) the angle (relative to "forward") of the knight's overall displacement for the series of three moves?

Short Answer

Expert verified
The magnitude is approximately 2.24 meters, and the angle is 26.57 degrees relative to the forward direction.

Step by step solution

01

- Understand the Knight's Moves

The knight moves in three steps. Identify the changes in position for each move. Move 1: (2 squares forward, 1 square rightward), Move 2: (2 squares leftward, 1 square forward), Move 3: (2 squares forward, 1 square leftward).
02

- Convert Moves to Coordinates

Assume starting point is at the origin \((0,0)\). Move 1: (0,0) to (2,1), Move 2: (2,1) to (0,2), Move 3: (0,2) to (2,1).
03

- Calculate Overall Displacement

The overall displacement is the vector from the start (0,0) to the final position (2,1). Hence, the resulting displacement vector is \( (2,1) \).
04

- Calculate Magnitude of Displacement

Use the Pythagorean theorem to calculate the magnitude \(\text{Magnitude} = \sqrt{(2)^2 + (1)^2} = \sqrt{5} \approx 2.24 \text{ meters}\).
05

- Calculate Angle of Displacement

Use the inverse tangent function to calculate the angle: \(\theta = \tan^{-1} \left( \frac{1}{2} \right) = 26.57^\circ \). The angle is measured relative to the forward direction.

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Key Concepts

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

Displacement
In physics, displacement refers to the change in position of an object from its initial point to its final point. It is a vector quantity, meaning it has both magnitude and direction. For example, if a knight moves on a chessboard from point A to point B, the straight line connecting these points represents the displacement. In our exercise, the knight's final position results in a displacement vector, which we calculated using specific steps. Understanding displacement is crucial because it helps determine the shortest path between two points.
Vector Calculation
Vector calculation involves determining both magnitude and direction. Vectors are represented as arrows; the length signifies the magnitude, and the arrow points in the direction. When solving problems like the knight's moves, we summarize multiple vector steps into one resulting vector. For instance:
  • First Move: Up 2 squares, right 1 square
  • Second Move: Left 2 squares, up 1 square
  • Third Move: Up 2 squares, left 1 square
By adding these vectors together, we can find the knight's overall displacement vector. This simplified approach helps us understand the final position more clearly.
Coordinate System
A coordinate system provides a framework for locating points in a space. In our problem, we use a simple two-dimensional system with x (horizontal) and y (vertical) axes. Starting from the origin (0,0), each move of the knight can be translated into coordinates. For example:
  • Starting point: (0,0)
  • First Move: (2,1)
  • Second Move: (0,2)
  • Third Move: (2,1)
These positions help us visualize and calculate the knight's path effectively. Understanding how to use a coordinate system is crucial in physics for simplifying and solving spatial problems.
Pythagorean Theorem
The Pythagorean Theorem is a mathematical tool used to calculate the magnitude of a displacement vector. It states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. For our problem, the displacement vector \( (2,1) \) can be solved as follows:
  • \

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Most popular questions from this chapter

In the sum \(\vec{A}+\vec{B}=\) \(\vec{C}\), vector \(\vec{A}\) has a magnitude of \(12.0\) \(\mathrm{m}\) and is angled \(40.0^{\circ}\) counterclockwise from the \(+x\) direction, and vector \(\vec{C}\) has a magnitude of \(15.0 \mathrm{~m}\) and is angled \(20.0^{\circ}\) counterclockwise from the \(-x\) direction. What are (a) the magnitude and (b) the angle (relative to \(+x)\) of \(\vec{B}\) ?

In each case below, sketch the velocity vector. Find the magnitude and direction of motion with respect to the \(x\) axis of the coordinate system: (a) \(\vec{v}=(2.45 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(3.67 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) (b) \(\vec{v}=(-2.45 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(5.20 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\)

What are (a) the \(x\) -component and (b) the \(y\) -component of a vector \(\vec{a}\) in the \(x y\) plane if its direction is \(250^{\circ}\) counterclockwise from the positive direction of the \(x\) axis and its magnitude is \(7.3 \mathrm{~m}\) ?

Consider how the components of a vector in the plane change if I change the reference point. Suppose I start with a coordinate system with an origin at \(O\). An arbitrary vector \(\vec{r}=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}\) with coordinates \((x, y)\) specifies a point in this system. Suppose also that I have another point \(O^{\prime}\) specified in this coordinate system by a vector \(\vec{A}=A_{x} \hat{\mathrm{i}}+A_{y} \hat{\mathrm{j}}\). If I change my origin to \(O^{\prime}\) (without rotating the axes), what would the coordinates be for the point specified by \(\vec{r}\) ?

(a) In unit-vector notation, what is the sum of $$\vec{a}=(4.0 \mathrm{~m}) \hat{\mathrm{i}}+(3.0 \mathrm{~m}) \hat{\mathrm{j}} \text { and } \vec{b}=(-13.0 \mathrm{~m}) \hat{\mathrm{i}}+(7.0 \mathrm{~m}) \hat{\mathrm{j}} ?$$ What are (b) the magnitude and (c) the direction of \(\vec{a}+\vec{b}\) (relative to \(\hat{\mathrm{i}}\) ?
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