- a. Ax = 21.02 km (East)
- b. Ay = 34.91 km (North)
- c. Bx = -38.91 km (West)
- d. By = 83.62 km (North)
- e. Rx = -17.89 km (West)
- f. Ry = 118.53 km (North)
- g. Resultant's magnitude ≈ 120.5 km
- h. Resultant's direction ≈ -80.5 degrees (measured clockwise from East)
To solve this problem, we first need to resolve each displacement vector into its x and y components using trigonometry.
For Path A:
- Displacement magnitude = 40.7 km
- Direction = 59.1 degrees N of East
Resolve into components:
- Ax = 40.7 km * cos(59.1 degrees) = 40.7 km * 0.515 = 21.02 km (East)
- Ay = 40.7 km * sin(59.1 degrees) = 40.7 km * 0.857 = 34.91 km (North)
For Path B:
- Displacement magnitude = 92.2 km
- Direction = 24.5 degrees N of W (which is equivalent to 65.5 degrees W of N)
Resolve into components:
- Bx = 92.2 km * cos(65.5 degrees) = 92.2 km * 0.422 = 38.91 km (West)
- By = 92.2 km * sin(65.5 degrees) = 92.2 km * 0.906 = 83.62 km (North)
Now, we sum the x and y components to find the resultant components:
- Rx = Ax + Bx = 21.02 km - 38.91 km = -17.89 km
- Ry = Ay + By = 34.91 km + 83.62 km = 118.53 km
The negative sign for Rx indicates a direction towards the West.
Finally, we calculate the magnitude and direction of the resultant:
- Resultant's magnitude = [tex]\sqrt{(Rx^2 + Ry^2)[/tex] = [tex]\sqrt{(-17.89 km)^2 + (118.53 km)^2} \approx 120.5 km[/tex]
- Resultant's direction = arctan(Ry/Rx) = arctan(118.53 km / -17.89 km) ≈ -80.5 degrees (measured clockwise from East)
So, the results are:
- a. Ax = 21.02 km (East)
- b. Ay = 34.91 km (North)
- c. Bx = -38.91 km (West)
- d. By = 83.62 km (North)
- e. Rx = -17.89 km (West)
- f. Ry = 118.53 km (North)
- g. Resultant's magnitude ≈ 120.5 km
- h. Resultant's direction ≈ -80.5 degrees (measured clockwise from East)
