Final answer:
The plane's total displacement vector is approximately 65.0 km at an angle of 40.77° north of east, which is rounded off to near 45° north of east. Therefore, the correct option is d) 65.0 km, 45° north of east.
Step-by-step explanation:
To determine the total displacement vector of the plane after the two legs of its journey, we need to separate each leg into its northward and eastward components, and then sum these components.
For the first leg of the journey, the plane flies 40.0 km at 60° north of east:
- Eastward component = 40.0 km * cos(60°) = 20.0 km
- Northward component = 40.0 km * sin(60°) = 34.64 km
For the second leg of the journey, the plane flies 30.0 km at 15° north of east:
- Eastward component = 30.0 km * cos(15°) = 29.0 km
- Northward component = 30.0 km * sin(15°) = 7.79 km
Now, sum the components:
- Total eastward = 20.0 km + 29.0 km = 49.0 km
- Total northward = 34.64 km + 7.79 km = 42.43 km
The total displacement vector can be found using Pythagoras' theorem:
R = √(eastward² + northward²)
R = √(49.0² + 42.43²) km
R ≈ 64.7 km
To find the direction of the displacement vector, we take the arctan of the northward component divided by the eastward component:
Direction = arctan(northward/eastward)
Direction = arctan(42.43/49.0)
Direction ≈ 40.77° north of east
Therefore, the plane's displacement vector is approximately 65.0 km at an angle of 40.77° north of east, which is approximately 45° north of east. So the closest match to the options given is d) 65.0 km, 45° north of east.