Displacement vector A with arrow points due east and has a magnitude of 2.62 km. Displacement vector B with arrow points due north and has a magnitude of 2.26 km. Displacement vector C with arrow points due west and has a magnitude of 1.4 km. Displacement vector D points due south and has a magnitude of 2.8 km. Find the magnitude and direction (relative to due east) of the resultant vector A with arrow + B with arrow + C with arrow + D.
In x-y coordinates,
A = (2.62,0)
B = (0,2.26)
C = (-1.4,0)
D = (0,-2.8)
A+B+C+D = (1.22,-0.54)
That is 1.33 at E23.88°S
To find the magnitude and direction of the resultant vector, we need to add all the individual vectors together.
Let's first draw a diagram to visualize the vectors. We can represent the vectors as arrows, with their directions and magnitudes labeled.
Vector A (2.62 km) points due east (+x direction)
Vector B (2.26 km) points due north (+y direction)
Vector C (1.4 km) points due west (-x direction)
Vector D (2.8 km) points due south (-y direction)
Now, let's add the vectors together. We can add the x-components (east and west vectors) and the y-components (north and south vectors) separately.
The x-component:
Vector A + Vector C = 2.62 km east - 1.4 km west = 2.62 km - 1.4 km = 1.22 km east
The y-component:
Vector B + Vector D = 2.26 km north - 2.8 km south = 2.26 km - 2.8 km = -0.54 km south
Now, we have the resulting vector in terms of its x and y components: (1.22 km east, -0.54 km south)
To find the magnitude of the resultant vector, we can use the Pythagorean theorem:
Magnitude = sqrt((x-component)^2 + (y-component)^2)
Magnitude = sqrt((1.22 km)^2 + (-0.54 km)^2)
Magnitude = sqrt(1.4884 km^2 + 0.2916 km^2)
Magnitude = sqrt(1.78 km^2)
Magnitude ≈ 1.335 km
To find the direction of the resultant vector relative to due east, we can use the inverse tangent function:
Direction = arctan((y-component)/(x-component))
Direction = arctan((-0.54 km)/(1.22 km))
Direction ≈ -0.41 rad
To convert the direction from radians to degrees, we can use the conversion factor: 1 radian ≈ 57.3 degrees.
Direction ≈ -0.41 rad * 57.3 degrees/rad
Direction ≈ -23.5 degrees
Therefore, the magnitude of the resultant vector is approximately 1.335 km and its direction (relative to due east) is approximately -23.5 degrees.