Question

A girl walks 2 km East, turns left and walks 5 km, then turns left again and walks 2 km. How far is she from the starting point?

• A. 0 km
• B. 2 km
• C. 3 km
• D. 5 km

Hint:

In direction questions, use a coordinate grid or sketch to simplify displacement calculations and apply Pythagoras when needed.

Answer 1

The Correct Option is C

Phase 1: Trajectory Mapping
Movement 1: 2 km East, arriving at Point A.
Movement 2: Turn left (North), 5 km, arriving at Point B.
Movement 3: Turn left again (West), 2 km, ending at Point C.
Phase 2: Positional Analysis Relative to Origin
Horizontal displacement: 2 km East followed by 2 km West results in a net horizontal displacement of 0 km.
Vertical displacement: 5 km North with no Southward movement results in a net vertical displacement of 5 km.
Conclusion: Her final position is 5 km North of her starting point.
Clarification: Return to the East-West line of origin
\(\Rightarrow\) \text{She realigns vertically with her starting point}
→ Net East-West movement = 0
→ Net North-South movement = 5 km
Final Adjustment: The last 2 km movement was West from Point B. This places her directly above the starting point. The calculation for Northward distance is 5 km (total North) - 2 km (Westward offset from Point B's Northward position which cancels out the initial Eastward movement, effectively bringing her back to the vertical line of the start). Reconsidering the final move: 2 km West from Point B (which is 5km North of the start) means she ends up 5 km North of the starting point, directly on the vertical line passing through the start. The previous calculation was incorrect. The Net East-West displacement is 0. The Net North-South displacement is 5 km. Therefore, the final position is 5 km North of the starting point.
Hence, distance from starting point: \[\text{Distance} = 5 \text{ km}\]