5. A man walks 50 m [30° N of E], then 70 m [60° S of E], and finally 30 m [45° N of W]. What is the displacement of the man?

To find the total displacement, we need to break down each leg of the man's journey into its components along the east (x) and north (y) directions. We will then sum these components to find the resultant displacement vector.

### Step 1: Break down each displacement into x and y components.

1. **First displacement: 50 m [30° N of E]**
   - The angle is measured from the east, going 30° toward the north.
   - \( x_1 = 50 \cos(30^\circ) = 50 \times \frac{\sqrt{3}}{2} = 43.3 \, \text{m} \) (eastward)
   - \( y_1 = 50 \sin(30^\circ) = 50 \times \frac{1}{2} = 25 \, \text{m} \) (northward)

2. **Second displacement: 70 m [60° S of E]**
   - The angle is measured from the east, going 60° toward the south.
   - \( x_2 = 70 \cos(60^\circ) = 70 \times \frac{1}{2} = 35 \, \text{m} \) (eastward)
   - \( y_2 = 70 \sin(60^\circ) = 70 \times \frac{\sqrt{3}}{2} = 60.6 \, \text{m} \) (southward, negative direction)

3. **Third displacement: 30 m [45° N of W]**
   - The angle is measured from the west, going 45° toward the north.
   - \( x_3 = 30 \cos(45^\circ) = 30 \times \frac{1}{\sqrt{2}} = 21.2 \, \text{m} \) (westward, negative direction)
   - \( y_3 = 30 \sin(45^\circ) = 30 \times \frac{1}{\sqrt{2}} = 21.2 \, \text{m} \) (northward)

### Step 2: Sum the components in each direction.
- **Total x-component**:
  \[
  x_{\text{total}} = x_1 + x_2 + x_3 = 43.3 + 35 - 21.2 = 57.1 \, \text{m}
  \]
- **Total y-component**:
  \[
  y_{\text{total}} = y_1 + y_2 + y_3 = 25 - 60.6 + 21.2 = -14.4 \, \text{m}
  \]

### Step 3: Calculate the resultant displacement.
The magnitude of the resultant displacement is given by:
\[
d = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2} = \sqrt{(57.1)^2 + (-14.4)^2} = \sqrt{3260.41 + 207.36} = \sqrt{3467.77} \approx 58.9 \, \text{m}
\]

### Step 4: Calculate the direction of the displacement.
The direction is given by the angle \( \theta \) from the east, using the arctangent function:
\[
\theta = \tan^{-1}\left(\frac{y_{\text{total}}}{x_{\text{total}}}\right) = \tan^{-1}\left(\frac{-14.4}{57.1}\right) = \tan^{-1}(0.252) \approx -14.2^\circ
\]
Since the angle is negative, the direction is **14.2° South of East**.

### Final Answer:
The man's displacement is approximately **58.9 m [14.2° S of E]**.

Do you have any questions or would you like more details on the steps?

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Here are 5 related questions to deepen understanding:
1. How would the displacement change if the man walked 100 m due east first?
2. What is the difference between distance traveled and displacement in this case?
3. How can you apply vector addition to other fields such as physics or engineering?
4. What if one of the displacements were in 3D space, how would the calculation differ?
5. How would you calculate the displacement if the angles were measured in radians?

**Tip:** When working with vectors, always break them into components to simplify calculations.

A man walks 50 m [30° N of E], then 70 m [60° S of E], and finally 30 m [45° N of W]. What is the displacement of the man?

Explore how to calculate the displacement of a man walking in different directions using vector addition and trigonometry. This step-by-step solution covers vector components, the Pythagorean Theorem, and trigonometric identities. Suitable for high school students in Grades 10-12.

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