Final answer:
The algebraically determined resultant vector of Janet's walk is about 504.8 meters at an angle of 7.93 degrees north of east.
Step-by-step explanation:
To find the resultant vector of Janet's walk, we need to consider her movements as three separate vectors and then find their sum. Janet's first displacement is 450 m due south, the second is 200 m due east, and the third is 600 m at 60 degrees northeast.
Vector A (southward):
A = 450 m South
Ax = 0 (no movement in the east-west direction)
Ay = -450 m (movement is southward, thus negative in the 'up-down' coordinate system)
Vector B (eastward):
B = 200 m East
Bx = +200 m (movement is eastward, thus positive in the 'left-right' coordinate system)
By = 0 (no movement in the north-south direction)
Vector C (northeastward at 60 degrees):
C = 600 m at 60° northeast
To find the components, we use cos for the horizontal component and sin for the vertical component:
Cx = C * cos(60°) = 600 * cos(60°) = 600 * 0.5 = 300 m
Cy = C * sin(60°) = 600 * sin(60°) = 600 * (√3/2) ≈ 600 * 0.866 = 519.6 m
Now, we sum the components of each vector:
- Resultant vector in the x-direction (east-west): Rx = Ax + Bx + Cx = 0 + 200 + 300 = 500 m
- Resultant vector in the y-direction (north-south): Ry = Ay + By + Cy = -450 + 0 + 519.6 ≈ 69.6 m
To find the magnitude and direction of the resultant vector R, we can use the Pythagorean theorem and trigonometry:
- Magnitude of R: |R| = √(Rx² + Ry²) = √(500² + 69.6²) ≈ √(250000 + 4841.76) ≈ √(254841.76) ≈ 504.8 m
- Direction of R: θ = arctan(Ry/Rx) ≈ arctan(69.6/500) ≈ 7.93° north of east
The resultant displacement of Janet is therefore approximately 504.8 m at 7.93° north of east.