Problem 19 - Fundamentals of Physics

Question

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?

Step-by-Step Solution

Answer

The displacement magnitude is approximately 2.24 meters, at an angle of 26.6° relative to "forward."

Step 1: Identify Initial Position and Moves

The initial position of the knight is at the origin point (0, 0). The moves can be described in terms of vectors. The first move is (2 forward, 1 rightward), the second move is (2 leftward, 1 forward), and the third move is (2 forward, 1 leftward). This gives us vectors for each move.

Step 2: Translate Moves into Vectors

Convert the moves into vector form. Move 1 is represented as \( (2, 1) \), Move 2 as \( (-2, 1) \), and Move 3 as \( (2, -1) \) in Cartesian coordinates, where forward is along the x-axis and rightward is along the y-axis.

Step 3: Calculate Total Displacement Vector

Add the vectors from each move to find the total displacement. Total displacement vector = Move 1 + Move 2 + Move 3 = \( (2, 1) + (-2, 1) + (2, -1) = (2, 1) \).

Step 4: Find the Magnitude of Displacement

The magnitude of the displacement vector \( (2, 1) \) can be calculated using the formula \( \sqrt{x^2 + y^2} \). Thus, the magnitude is \( \sqrt{2^2 + 1^2} = \sqrt{5} \approx 2.24 \text{ meters} \).

Step 5: Determine the Angle of Displacement

Calculate the angle of displacement relative to the forward direction (x-axis) using the tangent inverse. \( \theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{1}{2}\right) \). This gives an angle of approximately \( 26.6^\circ \).

Key Concepts

Vector Addition

In kinematics, vectors are fundamental. They represent quantities with both magnitude and direction. Here, we deal with the knight’s movement in a chess game using vector addition. When a knight moves in a series of directions, each move can be described by a vector. To find the overall displacement of the knight, we need to add these vectors.

Adding vectors involves summing their components. For example, if the knight's movements are given as vectors \((2, 1)\) for the first move, \((-2, 1)\) for the second, and \((2, -1)\) for the third move, you simply add the corresponding components together:

• First move: from the origin to \((2, 1)\)
• Second move: adds \((-2, 1)\) resulting in \((0, 2)\)
• Third move: adds \((2, -1)\) resulting in \((2, 1)\)

This technique of vector addition provides the total displacement of the knight on the board.

Displacement

Displacement is a key concept in physics that describes a change in position. It’s not the same as distance. Instead, displacement is the shortest path between the starting point and the ending point, taking direction into account.

In our chess problem, we track the knight’s movement from its starting position to its final spot on the board. The initial position is at the origin point \((0, 0)\). The final position after calculating the total vector addition is \((2, 1)\). Therefore, the displacement of the knight is represented by the vector \((2, 1)\).

This vector tells us how much the knight has moved and in which direction relative to its starting position.

Magnitude and Direction Calculation

Once we determine the displacement vector \((2, 1)\), the next step is to find its magnitude and direction. The magnitude is calculated using the Pythagorean theorem, which applies to vectors in a Cartesian plane.

The formula for the magnitude \(R\) is:
\[R = \sqrt{x^2 + y^2}\]
Plugging in our values, we have \(x = 2\) and \(y = 1\):

• \(R = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}\)
• \(R \approx 2.24 \text{ meters}\)

To find the angle \(\theta\) of the displacement relative to the forward direction (the x-axis), use the tangent inverse function:
\[\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{1}{2}\right)\]
This gives an angle of approximately \(26.6^\circ\).

Thus, the knight's move results in a displacement with a magnitude of approximately 2.24 meters and a direction of 26.6 degrees from the forward axis.

Cartesian Coordinates

Cartesian coordinates are used to precisely specify the position of a point in a plane. They consist of a pair of numerical values \((x, y)\), where \(x\) represents the horizontal position, and \(y\) represents the vertical position.

In the context of the chess problem, the knight's moves are mapped into a Cartesian coordinate system. Initially, the knight starts at the origin \((0, 0)\). Each move can be translated into a vector with specific \(x\) and \(y\) coordinates. For instance:

• Move 1: \((2, 1)\)
• Move 2: \((-2, 1)\)
• Move 3: \((2, -1)\)

By plotting these vectors, we can visualize how the knight navigates the board. The Cartesian system helps simplify and calculate the resulting displacement, facilitating our understanding of the knight's journey across the chessboard.