A man walks 7 km in the North direction. He then walks 4 km Eastward, and then 4 km Southward. How far and in which direction is he now from his original position?
5 km in South-East direction
5 km in North-East direction
15 km towards East direction
7 km towards North direction
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CISF-AC-EXE – 2024
1. He walks 7 km in the North direction. His position is now (0, 7).
2. He then walks 4 km Eastward. His position changes by +4 in the x-direction. His new position is (0 + 4, 7) = (4, 7).
3. He then walks 4 km Southward. His position changes by -4 in the y-direction. His new position is (4, 7 – 4) = (4, 3).
His final position is (4, 3). His original position was (0, 0).
The displacement from the original position is a vector from (0,0) to (4,3).
The distance from the original position is the magnitude of this displacement vector, which can be calculated using the distance formula (or Pythagorean theorem):
Distance = $\sqrt{(\text{final x} – \text{initial x})^2 + (\text{final y} – \text{initial y})^2}$
Distance = $\sqrt{(4 – 0)^2 + (3 – 0)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ km.
The final position (4, 3) is in the positive x and positive y quadrant relative to the origin. In terms of directions, positive x is East and positive y is North. Therefore, the direction is North-East.