Displacement vector points due east and has a magnitude of 4.58 km. Displacement vector points due north and has a magnitude of 6.25 km. Displacement vector points due west and has a magnitude of 8.65 km. Displacement vector points due south and has a magnitude of 1.84 km. Find (a) the magnitude of the resultant vector + + + , and (b) its direction as a positive angle relative to due west.
a. D = 4.58 + 6.25i - 8.65 - 1.84i =
-4.07 + 4.41i, Q2
D = sqrt(4.07^2+4.41^2) km.
b. Tan A = Y/X = 4.41/-4.07 = -1.08354
A = -47.3o = 47.3o N. of W.
To find the magnitude of the resultant vector, we can add the vectors component-wise.
Let's define the x-axis as pointing east and the y-axis as pointing north.
The x-component of the resultant vector is the sum of the x-components of the individual vectors.
x-component = 4.58 km (due east) + 0 km (due north) + (-8.65 km) (due west) + 0 km (due south)
= 4.58 km - 8.65 km
= -4.07 km
The y-component of the resultant vector is the sum of the y-components of the individual vectors.
y-component = 0 km (due east) + 6.25 km (due north) + 0 km (due west) + (-1.84 km) (due south)
= 6.25 km - 1.84 km
= 4.41 km
Using the Pythagorean theorem, we can calculate the magnitude of the resultant vector:
Magnitude of the resultant vector = sqrt((-4.07 km)^2 + (4.41 km)^2)
= sqrt(16.5649 km^2 + 19.4881 km^2)
= sqrt(36.053 km^2)
= 6.00 km (rounded to two decimal places)
Therefore, the magnitude of the resultant vector is approximately 6.00 km.
To find the direction of the resultant vector as a positive angle relative to due west, we can use trigonometry.
tan(theta) = (y-component / x-component)
theta = atan(y-component / x-component)
= atan(4.41 km / -4.07 km)
≈ -47.90°
Since we want the angle as a positive value relative to due west, we can add 180° to it.
theta = -47.90° + 180°
≈ 132.10°
Therefore, the direction of the resultant vector as a positive angle relative to due west is approximately 132.10°.
To find the magnitude of the resultant vector, we can use the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In this case, we can treat each displacement vector as a side of a right triangle, with the magnitude of each vector being the length of the corresponding side.
Let's label the vectors as follows:
- Displacement vector due east: E = 4.58 km
- Displacement vector due north: N = 6.25 km
- Displacement vector due west: W = 8.65 km
- Displacement vector due south: S = 1.84 km
Now, let's calculate the magnitude of the resultant vector using the Pythagorean theorem:
Resultant magnitude = √(E^2 + N^2 + W^2 + S^2)
Substituting the values we have:
Resultant magnitude = √(4.58^2 + 6.25^2 + 8.65^2 + 1.84^2)
Calculating this expression, we find that the magnitude of the resultant vector is approximately 13.78 km.
Now, to find the direction of the resultant vector relative to due west, we can use trigonometry. Specifically, we can use the tangent function.
Let's define the angle between the resultant vector and due west as θ.
θ = arctan(N - S / E + W)
Substituting the values we have:
θ = arctan(6.25 - 1.84 / 4.58 + 8.65)
Calculating this expression, we find that the angle θ is approximately 18.2 degrees.
Therefore, the answers to the given problem are:
(a) The magnitude of the resultant vector is approximately 13.78 km.
(b) The direction of the resultant vector relative to due west is approximately 18.2 degrees.