Question

In a game of lawn chess, where pieces are moved between the centers of squares that are each 1.00m 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

a) Magnitude of the knight’s overall displacement is 5.39m

b) Angle of the knight’s overall displacement is$21.8°$

Step-by-step solution

To understand the concept of the problem

consider each move of the knight as a vector and using the vector algebra, add the given vectors to find their resultant.Using the general formula for magnitude and angle, the magnitude of the vector and the angle between the resultant vector and the forward direction can be computed. To calculate magnitude and angle following equations to be used.

$$\begin{gathered} \overrightarrow{B}=\left(\overrightarrow{B_x}\right)\stackrel{⏜}{\mathrm{i}}+\left(\overrightarrow{B_y}\right)\stackrel{⏜}{\mathrm{j}} \\ B=\left|\overrightarrow{B}\right|=\sqrt{{\left(\overrightarrow{B_x}\right)}^2+{\left(\overrightarrow{B_y}\right)}^2} \\ \theta ={\tan }^{-1}\left(\frac{\overrightarrow{B_x}}{B_x}\right) \\ \mathrm{There}\mathrm{are}\mathrm{five}\mathrm{forward}\mathrm{moves},\mathrm{one}\mathrm{rightward}\mathrm{move}\mathrm{and}\mathrm{three}\mathrm{leftward}\mathrm{moves}. \end{gathered}$$

To find x and y components of B→

Since there are no backward moves, the total number of forward moves would be 5. There is one rightward move and three leftward moves, so one rightward move would be canceled out by one leftward move, since squares are equal and direction would be opposite. So effectively, we have 5 forward moves and 2 leftward moves.

Consider $\stackrel{⏜}{\mathrm{i}}$be directed forward and $\stackrel{⏜}{\mathrm{j}}$be directed leftward. So, the resultant is

$\left(5.00\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(2.00\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}}$

To find magnitude of B→

The magnitude of the vector$\left(5.00\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(2.00\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}}$ is

$\sqrt{\left(5.00\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(2.00\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}}}=5.385\mathrm{m}=5.39\mathrm{m}$

To find angle between B→ and x axis

The angle between the components of vector$\left(5.00\mathrm{m}\right)\stackrel{⏜}{\mathrm{i}}+\left(2.00\mathrm{m}\right)\stackrel{⏜}{\mathrm{j}}$ is given by

${\tan }^{-1}\left(\frac{2.0\mathrm{m}}{5.0\mathrm{m}}\right)=21.8°$

Therefore, the angle of the knight’s overall displacement for the series of three moves is$21.8°$ . As per our assumption, this angle would be the angle with the x axis, and it would be left of forward direction.