Understanding the Problem: Distance from Starting Point

The question asks for the final distance of Abhijeet from his house after a series of movements. He starts at his house and makes four distinct moves, each covering a distance of 30 m in a specific diagonal direction.

To solve this, we need to understand how movements in different directions combine. We can think of each diagonal movement as having components in the North-South and East-West directions.

Analyzing Each Movement and its Components

Let's assume the house is at the origin (0,0). We can break down each movement into its East (positive x) and North (positive y) components. Movements West are negative x, and movements South are negative y.

• North-West (NW): This direction is exactly between North and West. A 30 m movement in this direction means moving 30 m multiplied by $\cos(45^\circ)$ towards West and 30 m multiplied by $\sin(45^\circ)$ towards North. Since $\cos(45^\circ) = \sin(45^\circ) = \frac{1}{\sqrt{2}}$, the components are $-30 \times \frac{1}{\sqrt{2}}$ in the East direction and $+30 \times \frac{1}{\sqrt{2}}$ in the North direction.
• South-West (SW): This direction is exactly between South and West. A 30 m movement means moving $30 \times \frac{1}{\sqrt{2}}$ towards West and $30 \times \frac{1}{\sqrt{2}}$ towards South. The components are $-30 \times \frac{1}{\sqrt{2}}$ in the East direction and $-30 \times \frac{1}{\sqrt{2}}$ in the North direction.
• South-East (SE): This direction is exactly between South and East. A 30 m movement means moving $30 \times \frac{1}{\sqrt{2}}$ towards East and $30 \times \frac{1}{\sqrt{2}}$ towards South. The components are $+30 \times \frac{1}{\sqrt{2}}$ in the East direction and $-30 \times \frac{1}{\sqrt{2}}$ in the North direction.
• North-East (NE): This direction is exactly between North and East. A 30 m movement means moving $30 \times \frac{1}{\sqrt{2}}$ towards East and $30 \times \frac{1}{\sqrt{2}}$ towards North. The components are $+30 \times \frac{1}{\sqrt{2}}$ in the East direction and $+30 \times \frac{1}{\sqrt{2}}$ in the North direction.

Step-by-Step Displacement Calculation

Let's add up the components of each movement to find the total displacement from the starting point (house). We can represent the components as $(\Delta x, \Delta y)$. Let $d = \frac{30}{\sqrt{2}}$.

Now, let's find the total displacement by summing the components:

• Total East-West Displacement ($\Delta X_{total}$): $(-d) + (-d) + (+d) + (+d) = -2d + 2d = 0$
• Total North-South Displacement ($\Delta Y_{total}$): $(+d) + (-d) + (-d) + (+d) = 0d + 0d = 0$

The final position relative to the starting point (house) is $(0,0)$.

Calculating the Final Distance from the House

The final position is at the coordinates $(0,0)$. The starting point (house) was also at $(0,0)$. The distance between the final position and the starting position is the magnitude of the total displacement vector $(\Delta X_{total}, \Delta Y_{total})$.

Distance $= \sqrt{(\Delta X_{total})^2 + (\Delta Y_{total})^2}$

Distance $= \sqrt{(0)^2 + (0)^2} = \sqrt{0 + 0} = \sqrt{0} = 0$ m.

Therefore, Abhijeet is 0 m away from his house. He has returned to his starting point.

Revision Table: Direction and Displacement Concepts

Additional Information: Path vs. Displacement

It is important to distinguish between the total distance traveled and the final displacement. In this problem, Abhijeet traveled a total distance of $30 m + 30 m + 30 m + 30 m = 120 m$. However, because he ended up back at his starting point, his total displacement from the house is 0 m. The question specifically asks how far away he is from his house, which refers to the magnitude of the displacement from the starting point.

The sequence of movements forms a closed loop, bringing Abhijeet back to where he began. Visualizing this path on a coordinate system or as a series of vectors starting from the origin confirms that the tail of the first vector is at (0,0) and the head of the last vector also ends at (0,0).