a person moves 30 m north , then 20 m east and finally 30 root 2 south west what is his displacement from the original position
To find the displacement from the original position, we need to determine the net distance and direction of the person's movement.
Let's break down the person's movements:
1. Moving 30 m north: This means the person is moving in the positive y-axis direction. Let's call this vector A.
2. Moving 20 m east: This means the person is moving in the positive x-axis direction. Let's call this vector B.
3. Moving 30√2 south-west: This means the person is moving in the negative y-axis and negative x-axis direction at an angle of 45 degrees. Let's call this vector C.
Now, we can calculate the net vector by adding all vectors together.
Vector A = 30 m north = (0, 30)
Vector B = 20 m east = (20, 0)
Vector C = 30√2 south-west = (-30√2*cos(45°), -30√2*sin(45°)) = (-30, -30)
Net vector = Vector A + Vector B + Vector C
= (0, 30) + (20, 0) + (-30, -30)
= (0 + 20 - 30, 30 + 0 - 30)
= (-10, 0)
Therefore, the person's displacement from the original position is 10 m to the west (negative x-axis direction).
Disp. = 30i + 20 + 30*sqrt2[225o]
Diep. = 30i + 20 + -30-30i = -10m = 10m West.
For a different approach, see Related Questions: Sun, 7-29-12, 11:05 AM.