‘R’ walks 1 km. to east and then he turns to south and walks 5 km. Aga

‘R’ walks 1 km. to east and then he turns to south and walks 5 km. Again he turns to east and walks 2 km. After this he turns to north and walks 9 km. How far is he from his starting point ?

3 km.
4 km.
5 km.
7 km.
This question was previously asked in
UPSC CAPF – 2009
The correct option is C.
Let the starting point be the origin (0,0) on a 2D plane, with East along the positive x-axis and North along the positive y-axis.
1. R walks 1 km to the east: The position is (1, 0).
2. He turns to south and walks 5 km: South is in the negative y direction. The position becomes (1, 0 – 5) = (1, -5).
3. Again he turns to east and walks 2 km: East is in the positive x direction. The position becomes (1 + 2, -5) = (3, -5).
4. After this he turns to north and walks 9 km: North is in the positive y direction. The position becomes (3, -5 + 9) = (3, 4).

The final position is (3, 4) and the starting position is (0, 0).
The distance from the starting point is the straight-line distance between (0, 0) and (3, 4).
Using the distance formula: Distance = $\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$
Distance = $\sqrt{(3 – 0)^2 + (4 – 0)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ km.

This problem is a classic example of vector addition or displacement calculation. Each leg of the journey is a displacement vector. Adding the displacements (1,0), (0,-5), (2,0), (0,9) gives the total displacement vector (3,4). The distance from the start is the magnitude of the total displacement vector.
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