91Ó°ÊÓ

When dealing with vector displacement, breaking each vector into components is crucial. Each vector has two components: horizontal (x-axis) and vertical (y-axis). Understanding these components helps simplify calculations and problems.
Displacement vectors have directions, and we assign positive or negative values based on these directions:

• West and south are negative directions.
• East and north are positive directions.

Let's look at an example for better clarity. Consider the displacement segment: 23.0 cm, 35.0° south of west. To find the vector components:

• The x-component is calculated using: - Formula: \[-23.0 \cos(35.0°)\] - Calculated: -18.83 \, \text{cm}
• The y-component is calculated using: - Formula: \[-23.0 \sin(35.0°)\] - Calculated: -13.19 \, \text{cm}

Remember, using cosine deals with horizontal, and sine with vertical components. Calculating each vector's components helps in finding the resultant displacement.

To find the magnitude of the resultant displacement, we use the total vector components obtained from each individual vector. The process involves a simple, yet critical application of the Pythagorean theorem. This helps us determine the overall displacement in a more straightforward manner.
Firstly, sum up all the x-components and y-components:

• Total x-component =
  (-27.0) + (-18.83) + 16.07 + 15.85 = -13.91 \, \text{cm}
• Total y-component =
  (0) + (-13.19) + (-22.91) + 31.13 = -5.03 \, \text{cm}

Once we have these totals, we calculate the magnitude, denoted as \(R\), using:\[R = \sqrt{(-13.91)^2 + (-5.03)^2}\]Plugging in the values and solving, we find:\[R \approx 14.91 \, \text{cm}\]This value represents the length of the resultant displacement vector, or essentially, how far the object has moved in total.

Determining the direction of the resultant displacement helps in understanding the path taken in movement. After calculating the total x and y components, we use trigonometric functions to identify the direction with respect to a reference point, often due west.
The arctangent function (\(\tan^{-1}\)) is particularly useful here, as it relates the angle of the resultant vector to the respective components. Here's how we do it:\[\theta = \tan^{-1}\left(\frac{-5.03}{-13.91}\right)\]Solving for \(\theta\), we find:\[\theta \approx 19.84^\circ\]This value represents an angle where both components are negative. In such a case, the direction is identified as being south of west. This means the resultant vector leans towards the southern hemisphere, measured from due west.By understanding both the magnitude and direction, we encapsulate the overall movement in a two-dimensional plane. Remember, direction is always reported with respect to a standard cardinal direction like due west or due east.