 Suppose you walk 11.5 m in a direction exactly 15° south of west then you walk 24 m in a direction exactly 44° west of north.


To solve this problem, we can break down the two displacements into their respective components and then find the resultant displacement.

### Step 1: Breaking the displacements into components
We will use trigonometry to split each displacement into **x (horizontal)** and **y (vertical)** components.

#### Displacement 1: 11.5 m, 15° south of west
- Since this direction is south of west, the **x-component** will be negative (west is negative x-direction) and the **y-component** will also be negative (south is negative y-direction).
  
  \[
  x_1 = 11.5 \times \cos(15^\circ)
  \]
  \[
  y_1 = 11.5 \times \sin(15^\circ)
  \]
  
  After calculation, the components are:
  \[
  x_1 = -11.5 \times \cos(15^\circ)
  \]
  \[
  y_1 = -11.5 \times \sin(15^\circ)
  \]

#### Displacement 2: 24 m, 44° west of north
- For this displacement, the **x-component** will be negative (west) and the **y-component** will be positive (north).

  \[
  x_2 = 24 \times \sin(44^\circ)
  \]
  \[
  y_2 = 24 \times \cos(44^\circ)
  \]

### Step 2: Summing the components
Now, sum the **x** and **y** components to get the total displacement.

\[
x_{\text{total}} = x_1 + x_2
\]
\[
y_{\text{total}} = y_1 + y_2
\]

### Step 3: Resultant displacement
To find the magnitude of the resultant displacement, use the Pythagorean theorem:

\[
d = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2}
\]

And to find the direction of the resultant displacement (relative to the positive x-axis):

\[
\theta = \tan^{-1}\left(\frac{y_{\text{total}}}{x_{\text{total}}}\right)
\]

Let me calculate that.The magnitude of the resultant displacement is approximately **31.24 meters**, and the direction is **152.78°**, measured counterclockwise from the positive x-axis (or west of north).

Do you have any further questions or would you like more details?

Here are five related questions to expand the topic:
1. How would the resultant displacement change if the second walk were longer?
2. How can we use vector addition to solve similar problems?
3. What is the significance of direction in determining vector components?
4. How would the displacement differ if we started by walking north of west instead?
5. How could you find the total time taken if the speed for each walk was given?

**Tip:** Always remember to consider both magnitude and direction when working with vectors.

Suppose you walk 11.5 m in a direction exactly 15° south of west then you walk 24 m in a direction exactly 44° west of north.

Learn how to calculate the resultant displacement when walking 11.5 meters at 15° south of west and 24 meters at 44° west of north. This solution uses vector addition, trigonometric ratios, and the Pythagorean theorem. Ideal for students in Grades 9-12.

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