A man leaves from his office for his home. He walks towards East. After moving…
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A man leaves from his office for his home. He walks towards East. After moving a distance of 20 m, he turns South and walks 10 m. Then he walks 35 m towards the west and further 5 m towards the north. He then turns towards east and walks 15 m. What is the straight distance (in metres) between his initial and final position?
- A.
0 m
- B.
5 m
- C.
10 m
- D.
11 m
Attempted by 6 students.
Show answer & explanation
Correct answer: B
When a path consists of movements along only two perpendicular directions (here East-West and North-South), the straight-line distance between the start and end points is found by adding the signed displacements along each axis separately, then combining the two net displacements with the Pythagorean theorem. If one axis nets to zero, the straight distance is simply the magnitude of the net displacement along the other axis.
East 20 m: +20 m along the East-West axis.
South 10 m: −10 m along the North-South axis (South taken as negative).
West 35 m: −35 m along the East-West axis.
North 5 m: +5 m along the North-South axis.
East 15 m: +15 m along the East-West axis.
Net East-West displacement = 20 − 35 + 15 = 0 m. Net North-South displacement = −10 + 5 = −5 m, i.e. 5 m South of the start.
Since the East-West component is zero, the final point lies directly along the North-South line from the start. So the straight-line distance between the initial and final positions is 5 m.
Cross-check with coordinates: taking the start as (0, 0), the path passes through (20, 0) → (20, −10) → (−15, −10) → (−15, −5) → (0, −5). The distance from (0, 0) to (0, −5) is √(02 + 52) = 5 m, confirming the result.