A person moves 30m north and then 20 m towards east and finally `30sqrt(2)`m in south-west direction. The displacement of the person from the origin will be

Options

A. 10m along north
B. 10m long south
C. 10m along wast
D. Zero

Answer

To solve the problem of finding the displacement of a person who moves 30 meters north, 20 meters east, and finally \(30\sqrt{2}\) meters in the south-west direction, we can use vector addition. Let's break down the problem step-by-step:

### Step 1: Represent the movements as vectors
1. **Movement 1:** 30 meters north.
- Vector: \( \vec{A} = 30 \hat{j} \)

2. **Movement 2:** 20 meters east.
- Vector: \( \vec{B} = 20 \hat{i} \)

3. **Movement 3:** \(30\sqrt{2}\) meters south-west.
- South-west direction means 45 degrees south of west.
- Vector: \( \vec{C} = 30\sqrt{2} (\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j}) \)
- Since it is south-west, both components will be negative:
\[ \vec{C} = 30\sqrt{2} \left(-\frac{1}{\sqrt{2}} \hat{i} - \frac{1}{\sqrt{2}} \hat{j}\right) = -30 \hat{i} - 30 \hat{j} \]

### Step 2: Sum the vectors to find the net displacement
\[ \vec{R} = \vec{A} + \vec{B} + \vec{C} \]
\[ \vec{R} = 30 \hat{j} + 20 \hat{i} + (-30 \hat{i} - 30 \hat{j}) \]
\[ \vec{R} = (20 \hat{i} - 30 \hat{i}) + (30 \hat{j} - 30 \hat{j}) \]
\[ \vec{R} = -10 \hat{i} + 0 \hat{j} \]

### Step 3: Calculate the magnitude of the displacement vector
Since the resultant vector is \(\vec{R} = -10 \hat{i}\), the magnitude is:
\[ |\vec{R}| = \sqrt{(-10)^2 + 0^2} = \sqrt{100} = 10 \text{ meters} \]

### Step 4: Determine the direction of the displacement
The displacement vector \(\vec{R} = -10 \hat{i}\) indicates that the displacement is 10 meters towards the west.

### Final Answer
The displacement of the person from the origin is 10 meters towards the west.

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