Answer :
The car's resultant displacement is approximately 45.6 km at an angle of about 56.7 degrees north of east, calculated using the Pythagorean theorem and trigonometry.
To find the magnitude and direction of the car's resultant displacement, we can treat the two segments of the journey as vector components. The car travels 26.2 km due north and then 36.9 km in another direction.
1. Magnitude of Displacement: Using the Pythagorean theorem, we can find the magnitude of the resultant displacement (D):
D = sqrt{(26.2 {km})^2 + (36.9 {km})^2}
D = sqrt{686.44 {km}^2 + 1360.61 {km}^2}
D = sqrt{2047.05 {km}^2}
D = 45.3 km
The magnitude of the car's resultant displacement is approximately 45.3 km.
2. Direction of Displacement: To find the direction, we can use trigonometry. The car's northward displacement is along the y-axis, and the eastward displacement is along the x-axis. The angle (\( \theta \)) between the resultant displacement and the east direction (positive x-axis) can be found using the inverse tangent (arctan) function:
theta = arctan {26.2 km} / {36.9 km}
theta = 35.4^o
The direction is measured counterclockwise from the east direction. To find the compass direction (north of east), we subtract this angle from 90 degrees:
Direction} = 90^o - 35.4^o
Direction = 54.6^o north of east
So, the car's resultant displacement is approximately 45.3 km at an angle of about 54.6 degrees north of east.
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