In two successive chess moves, a player first moves his queen four squares forward, then moves the queen three steps to the left (from the player’s view). Assume each square is 4.0 on a side.
To determine the overall displacement of the queen, we can break down the two moves individually and then combine them.
First move: The player moves the queen four squares forward. Since each square is 4.0 on a side, the displacement in the forward direction is 4.0 x 4 = 16.0 units.
Second move: The player moves the queen three steps to the left. Since each square is 4.0 on a side, the displacement in the left direction is 4.0 x 3 = 12.0 units.
To find the overall displacement, we need to combine the displacements in both directions. Since the queen moves forward first and then to the left, we can treat these displacements as two-dimensional vectors.
To combine the displacements, we use vector addition. We can represent the forward displacement as a vector along the positive y-axis (y-direction) with a magnitude of 16.0 units. The left displacement can be represented as a vector along the negative x-axis (x-direction) with a magnitude of 12.0 units.
Now, we can add these two vectors to find the resultant displacement.
Using Pythagoras' theorem, we can determine the magnitude of the resultant displacement:
Magnitude of the resultant displacement = √(16.0^2 + 12.0^2)
= √(256 + 144)
= √400
= 20.0 units
Therefore, the overall displacement of the queen after the two successive moves is 20.0 units.