Question

A person starts at A (-6 m, 5 m) and walks to B (-6 m, -5 m). Then the person turns

and walks to C (5 m, -5 m). The person turns again and walks to D (5 m, 0 m.
Finally, the person turns and walks to E (-2 m, 0 m). Determine the distance and
displacement (both magnitude and CCW direction) for this 3-stage motion.
-10
6
4
2
0
-2
-4
-6
-10
-8
-8
-6
A
-4
-2
E
0
2
4
D
6
8
B
-6 -4 -2 0 2 4 6 8
10
10
6
4
2
0
-2
वृं
-6

Answer 1

The distance traveled is 33 m and the displacement is sqrt(41) m in a certain direction.

Step-by-step explanation:

The distance between two points can be found using the distance formula:

d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Let's calculate the distance for each stage:

1. AB: dAB = sqrt((-6 - (-6))^2 + (-5 - 5)^2) = sqrt(0 + 100) = 10 m
2. BC: dBC = sqrt((5 - (-6))^2 + (-5 - (-5))^2) = sqrt(121 + 0) = 11 m
3. CD: dCD = sqrt((5 - 5)^2 + (0 - (-5))^2) = sqrt(0 + 25) = 5 m
4. DE: dDE = sqrt((-2 - 5)^2 + (0 - 0)^2) = sqrt((-7)^2 + 0) = 7 m

The total distance traveled is the sum of the distances for each stage:

Total distance = dAB + dBC + dCD + dDE = 10 m + 11 m + 5 m + 7 m = 33 m

The displacement is the straight-line distance from the starting point to the ending point, along with the direction.

The starting point is (-6 m, 5 m) and the ending point is (-2 m, 0 m).

The displacement can be found using the distance formula:

displacement = sqrt((x2 - x1)^2 + (y2 - y1)^2)

displacement = sqrt((-2 - (-6))^2 + (0 - 5)^2) = sqrt(16 + 25) = sqrt(41)

The magnitude of the displacement is sqrt(41) m, and the counter-clockwise (CCW) direction can be derived from the angle formed by the line connecting the starting and ending points with the positive x-axis.

Learn more about Distance and displacement in a 3-stage motion