A man covered following displacements. Find the net displacement of the person
(i) 25m west
(ii)` 20 m 30^(@)` east of south
(iii)`5sqrt(3)m` south
(iv) 10 m east
(v) `10 m 60^(@)` north of west

Answer

To find the net displacement of the person based on the given displacements, we can break down each displacement into its components and then sum them up. Here’s a step-by-step solution:

### Step 1: Identify Each Displacement
1. **25 m West**: This displacement can be represented as:
- \( x_1 = -25 \, \text{m} \) (West is negative x-direction)
- \( y_1 = 0 \, \text{m} \)

2. **20 m at 30° East of South**: This displacement can be broken down into components:
- \( x_2 = 20 \sin(30^\circ) = 20 \times \frac{1}{2} = 10 \, \text{m} \) (East is positive x-direction)
- \( y_2 = -20 \cos(30^\circ) = -20 \times \frac{\sqrt{3}}{2} = -10\sqrt{3} \, \text{m} \) (South is negative y-direction)

3. **5√3 m South**: This displacement is:
- \( x_3 = 0 \, \text{m} \)
- \( y_3 = -5\sqrt{3} \, \text{m} \)

4. **10 m East**: This displacement is:
- \( x_4 = 10 \, \text{m} \)
- \( y_4 = 0 \, \text{m} \)

5. **10 m at 60° North of West**: This displacement can be broken down into components:
- \( x_5 = -10 \cos(60^\circ) = -10 \times \frac{1}{2} = -5 \, \text{m} \) (West is negative x-direction)
- \( y_5 = 10 \sin(60^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m} \) (North is positive y-direction)

### Step 2: Sum the Components
Now, we can sum all the x and y components:

**Total x-component**:
\[ x_{\text{total}} = x_1 + x_2 + x_3 + x_4 + x_5 \]
\[ = -25 + 10 + 0 + 10 - 5 = -10 \, \text{m} \]

**Total y-component**:
\[ y_{\text{total}} = y_1 + y_2 + y_3 + y_4 + y_5 \]
\[ = 0 - 10\sqrt{3} - 5\sqrt{3} + 0 + 5\sqrt{3} = -10\sqrt{3} \, \text{m} \]

### Step 3: Calculate the Net Displacement
The net displacement can be calculated using the Pythagorean theorem:
\[ d = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2} \]
\[ = \sqrt{(-10)^2 + (-10\sqrt{3})^2} \]
\[ = \sqrt{100 + 300} = \sqrt{400} = 20 \, \text{m} \]

### Step 4: Determine the Direction
To find the direction of the net displacement, we can use the tangent function:
\[ \tan(\theta) = \frac{y_{\text{total}}}{x_{\text{total}}} \]
\[ \tan(\theta) = \frac{-10\sqrt{3}}{-10} = \sqrt{3} \]
\[ \theta = 60^\circ \]

Since both components are negative, the direction is in the third quadrant, which corresponds to:
\[ \text{Direction} = 60^\circ \text{ South of West} \]

### Final Answer
The net displacement of the person is:
\[ \text{Net Displacement} = 20 \, \text{m} \text{ at } 60^\circ \text{ South of West} \]

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