Question

A man walks 2 km towards East and then he turns to South and walks 6 km. Again he turns to East and walks 4 km, after this he turns to North and walk 14 km. How far is he from his starting point?

• A. \( 10 \) km
• B. \( 15 \) km
• C. \( 20 \) km
• D. \( 25 \) km

Hint:

\textbf{Direction and Distance Problems.} To solve these problems, it is often helpful to visualize the movements using a coordinate system or by drawing a diagram. Break down the journey into segments and calculate the net displacement in the North-South and East-West directions. The final distance from the starting point can then be found using the Pythagorean theorem.

Answer 1

The Correct Option is A

Let the starting point of the man be O. We can visualize his movements on a coordinate plane. (A) He walks 2 km towards East. Let this point be A. The displacement is (2, 0). (B) He turns to South and walks 6 km. Let this point be B. The change in displacement is (0, -6). His position is (2, -6). (C) He turns to East and walks 4 km. Let this point be C. The change in displacement is (4, 0). His position is (2 + 4, -6) = (6, -6). (D) He turns to North and walks 14 km. Let this point be D. The change in displacement is (0, 14). His position is (6, -6 + 14) = (6, 8). The final position of the man is at coordinates (6, 8) with respect to his starting point (0, 0). The distance from his starting point can be calculated using the distance formula: $$ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$ Here, \( (x_1, y_1) = (0, 0) \) and \( (x_2, y_2) = (6, 8) \). $$ \text{Distance} = \sqrt{(6 - 0)^2 + (8 - 0)^2} $$ $$ \text{Distance} = \sqrt{6^2 + 8^2} $$ $$ \text{Distance} = \sqrt{36 + 64} $$ $$ \text{Distance} = \sqrt{100} $$ $$ \text{Distance} = 10 \text{ km} $$ The man is 10 km away from his starting point.