In a school compound a girl walks 40km due east from the laboratory to the staff room to submit her report…

The question has some typos: "40km" should be 40 m, and "the10m" is a stray fragment. I'll read it as 40 m east, then 30 m north, and find the displacement from the laboratory.

Step 1: Analyze the question Displacement is the straight-line distance from the starting point to the final position, together with direction. It is a vector, so it is not the total distance walked.

Step 2: The concept When two displacements are at right angles (east and north), we use Pythagoras' theorem:

R² = (East)² + (North)²

Step 3: Solution

• East: 40 m
• North: 30 m

R² = 40² + 30² R² = 1600 + 900 = 2500 R = √2500 = 50 m

A note on the wording: the question says she "returns" before walking north. If she truly returned to the laboratory, the 30 m north would give a displacement of just 30 m, which is not an option. The intended meaning is clearly that she walks 30 m north from the staff room, which gives the 3-4-5 triangle (30-40-50).

The direction is given by tan θ = 30/40 = 0.75, so θ ≈ 37° north of east.

Answer: C. 50 m

Exam tip: Whenever you see 30 and 40 (or 3 and 4) at right angles, think of the 3-4-5 triple, so the resultant is 50. The other common triples are 5-12-13 and 8-15-17. Also remember that distance walked (70 m, option D) is not the same as displacement (50 m). Examiners like to use the total distance as a distractor.