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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 鈥渇orward鈥) of the knight鈥檚 overall displacement for the series of three moves?

Short Answer

Expert verified

a) Magnitude of the knight鈥檚 overall displacement is 5.39m

b) Angle of the knight鈥檚 overall displacement is21.8

Step by step solution

01

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.

B=Bxi+ByjB=B=Bx2+By2=tan-1BxBxTherearefiveforwardmoves,onerightwardmoveandthreeleftwardmoves.

02

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 ibe directed forward and jbe directed leftward. So, the resultant is

5.00mi+2.00mj

03

To find magnitude of B→

The magnitude of the vector5.00mi+2.00mj is

5.00mi+2.00mj=5.385m=5.39m

04

To find angle between B→ and x axis

The angle between the components of vector5.00mi+2.00mj is given by

tan-12.0m5.0m=21.8

Therefore, the angle of the knight鈥檚 overall displacement for the series of three moves is21.8 . As per our assumption, this angle would be the angle with the x axis, and it would be left of forward direction.

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