A person leaves her home and walks 6 miles due east and then 4 miles southeast?How far away from home is she?What is her net displacement(determine the magnitude and direction of the displacement vector)?
east ... 6 + 2√2
south ... 2√2
use Pythagoras to find the distance
the direction angle is in Quad IV ... tan(Θ) = south / east
D^2 = 6^2 + 4^2 = 52
D = 2√13 or appr 7.2 units from home
angle:
as Scott said, tanØ = 4/6
Ø = appr 33.7°
depending on your notation, direction angle = -33.7°
or bearing of 123.7°
or E 33.7° S
To find out how far away the person is from home, we can use the Pythagorean theorem.
Let's break down the person's journey into two parts:
1. Walking 6 miles due east.
2. Walking 4 miles southeast.
1. Walking 6 miles due east:
Since the person is walking due east, there is no change in the north-south direction. Therefore, we only consider the displacement in the east-west direction, which is +6 miles.
2. Walking 4 miles southeast:
When moving southeast, we can break down the displacement into two components, east-west and north-south. Since the person is moving southeast, we can use a right-angled triangle to find the components.
The distance covered in the east-west direction is given by the adjacent side of the right triangle, which is 4 miles * cos(45°) = 2.83 miles in the positive east direction.
The distance covered in the north-south direction is given by the opposite side of the right triangle, which is 4 miles * sin(45°) = 2.83 miles in the south direction.
Now let's find the total displacement:
In the east-west direction, the person covered a distance of +6 miles and +2.83 miles, totaling +8.83 miles.
In the north-south direction, the person covered a distance of -2.83 miles.
Using the Pythagorean theorem, we can calculate the magnitude of the displacement:
Magnitude = √(east-west)^2 + (north-south)^2
Magnitude = √(8.83^2) + (-2.83^2)
Magnitude = √(77.9689 + 8.0089)
Magnitude = √(86.9778)
Magnitude ≈ 9.34 miles
The magnitude of the displacement is approximately 9.34 miles.
To determine the direction of the displacement vector, we can use trigonometry. We take the inverse tangent of the east-west displacement over the north-south displacement:
Direction = arctan(north-south/east-west)
Direction = arctan(-2.83/8.83)
Direction ≈ -18.12°
The magnitude of the displacement is approximately 9.34 miles, and the direction is approximately -18.12° (south of east).
Therefore, the person is approximately 9.34 miles away from home, with a net displacement of 9.34 miles at -18.12° relative to east.
To find the distance from home, we can use the Pythagorean theorem. The person walks 6 miles due east, and then 4 miles southeast. This forms a right triangle, where the 6-mile distance is the horizontal side, and the 4-mile distance is the vertical side.
Using the Pythagorean theorem, we can calculate the hypotenuse (distance from home):
Distance^2 = (6 miles)^2 + (4 miles)^2
Distance^2 = 36 + 16
Distance^2 = 52
Distance = √52
Distance ≈ 7.21 miles (rounded to two decimal places)
To calculate the net displacement (magnitude and direction), we need to consider both the distance and direction the person traveled. The horizontal displacement is 6 miles east, and the vertical displacement is 4 miles south.
Using vector addition, we can calculate the net displacement:
Net Displacement = √(6^2 + (-4)^2)
Net Displacement = √(36 + 16)
Net Displacement = √52
Net Displacement ≈ 7.21 miles (rounded to two decimal places)
To determine the direction, we use trigonometry. We can calculate the angle using the inverse tangent (arctan) function:
Angle = arctan(4 miles / 6 miles)
Angle = arctan(2/3)
Angle ≈ 33.69 degrees
Therefore, the net displacement is approximately 7.21 miles in a direction of 33.69 degrees south of east.