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Difficulty: Very hardDirection and Distance Test

Four police officers on patrol follow different multi-step tactical movement routes involving angular rotations and shadow orientations. Match each officer's movement sequence (List-I) with their net displacement vector relative to their starting point (List-II).

  • Officer A starts facing East and walks 12 m12\text{ m}. He turns 135∘135^\circ clockwise and walks 102 m10\sqrt{2}\text{ m}. Next, he turns 90∘90^\circ anti-clockwise and walks 52 m5\sqrt{2}\text{ m}. Finally, he turns 135∘135^\circ anti-clockwise and walks 15 m15\text{ m}.7 m7\text{ m} directly East
  • At sunrise, Officer B stands facing his shadow. He walks 20 m20\text{ m} forward, turns 45∘45^\circ anti-clockwise and walks 102 m10\sqrt{2}\text{ m}. He then turns 135∘135^\circ clockwise and walks 26 m26\text{ m}, before making a final 90∘90^\circ clockwise turn and walking 18 m18\text{ m}.20 m20\text{ m} North-West
  • Officer C walks 12 m12\text{ m} North, turns 90∘90^\circ to his right and walks 24 m24\text{ m}. He then turns 135∘135^\circ clockwise and walks 102 m10\sqrt{2}\text{ m}. Finally, he turns 45∘45^\circ anti-clockwise and walks 16 m16\text{ m}.142 m14\sqrt{2}\text{ m} South-East
  • At sunset, Officer D stands facing his shadow. He walks 9 m9\text{ m} forward, turns 45∘45^\circ clockwise and walks 82 m8\sqrt{2}\text{ m}. He then turns 90∘90^\circ clockwise and walks 52 m5\sqrt{2}\text{ m}, followed by a 135∘135^\circ clockwise turn and a walk of 4 m4\text{ m}.15 m15\text{ m} South-East

Answer

Officer A matches 7 m7\text{ m} directly East; Officer B matches 20 m20\text{ m} North-West; Officer C matches 142 m14\sqrt{2}\text{ m} South-East; Officer D matches 15 m15\text{ m} South-East.
Each officer's sequence correctly resolves into orthogonal components when accounting for shadow direction (sunrise shadow = West, sunset shadow = East) and angular turn directions (clockwise vs anti-clockwise). Officer A reaches (7,0)(7, 0), Officer B reaches (−12,16)(-12, 16) with distance 20 m20\text{ m}, Officer C reaches (14,−14)(14, -14) with distance 142 m14\sqrt{2}\text{ m}, and Officer D reaches (12,−9)(12, -9) with distance 15 m15\text{ m}.

Step-by-Step Solution

1
Set up a standard Cartesian coordinate system where East is +x+x, West is −x-x, North is +y+y, and South is −y-y. Establish shadow directions: sunrise shadow points West; sunset shadow points East.
Sunrise shadow facing = West; Sunset shadow facing = East.
Sun rises in the East (shadows cast West) and sets in the West (shadows cast East).
2
Resolve Officer A's displacement steps into orthogonal components: Step 1 (12 m East) = (+12,0)(+12, 0); Step 2 (135∘135^\circ CW from East = SW, 102 m10\sqrt{2}\text{ m}) = (−10,−10)(-10, -10); Step 3 (90∘90^\circ Anti-CW from SW = SE, 52 m5\sqrt{2}\text{ m}) = (+5,−5)(+5, -5); Step 4 (135∘135^\circ Anti-CW from SE = North, 15 m15\text{ m}) = (0,+15)(0, +15). Sum components.
∑x=12−10+5+0=7 m\sum x = 12 - 10 + 5 + 0 = 7\text{ m}, ∑y=0−10−5+15=0 m\sum y = 0 - 10 - 5 + 15 = 0\text{ m}. Displacement = 7 m7\text{ m} East.
Direct vector addition yields net position (7,0)(7, 0).
3
Resolve Officer B's displacement steps starting facing West: Step 1 (20 m West) = (−20,0)(-20, 0); Step 2 (45∘45^\circ Anti-CW from West = SW, 102 m10\sqrt{2}\text{ m}) = (−10,−10)(-10, -10); Step 3 (135∘135^\circ CW from SW = North, 26 m26\text{ m}) = (0,+26)(0, +26); Step 4 (90∘90^\circ CW from North = East, 18 m18\text{ m}) = (+18,0)(+18, 0). Sum components.
∑x=−20−10+18=−12 m\sum x = -20 - 10 + 18 = -12\text{ m}, ∑y=−10+26=+16 m\sum y = -10 + 26 = +16\text{ m}. Net magnitude (−12)2+162=20 m\sqrt{(-12)^2 + 16^2} = 20\text{ m} North-West.
Pythagorean magnitude 144+256=20 m\sqrt{144 + 256} = 20\text{ m} in NW quadrant.
4
Resolve Officer C's displacement steps: Step 1 (12 m North) = (0,+12)(0, +12); Step 2 (24 m East) = (+24,0)(+24, 0); Step 3 (135∘135^\circ CW from East = SW, 102 m10\sqrt{2}\text{ m}) = (−10,−10)(-10, -10); Step 4 (45∘45^\circ Anti-CW from SW = South, 16 m16\text{ m}) = (0,−16)(0, -16). Sum components.
∑x=24−10=14 m\sum x = 24 - 10 = 14\text{ m}, ∑y=12−10−16=−14 m\sum y = 12 - 10 - 16 = -14\text{ m}. Magnitude 142+(−14)2=142 m\sqrt{14^2 + (-14)^2} = 14\sqrt{2}\text{ m} South-East.
Equal magnitude orthogonal components in +x+x and −y-y yield a 45∘45^\circ SE vector.
5
Resolve Officer D's displacement steps starting facing East: Step 1 (9 m East) = (+9,0)(+9, 0); Step 2 (45∘45^\circ CW from East = SE, 82 m8\sqrt{2}\text{ m}) = (+8,−8)(+8, -8); Step 3 (90∘90^\circ CW from SE = SW, 52 m5\sqrt{2}\text{ m}) = (−5,−5)(-5, -5); Step 4 (135∘135^\circ CW from SW = North, 4 m4\text{ m}) = (0,+4)(0, +4). Sum components.
∑x=9+8−5=12 m\sum x = 9 + 8 - 5 = 12\text{ m}, ∑y=−8−5+4=−9 m\sum y = -8 - 5 + 4 = -9\text{ m}. Magnitude 122+(−9)2=15 m\sqrt{12^2 + (-9)^2} = 15\text{ m} South-East.
Pythagorean triplet (9,12,15)(9, 12, 15) in SE quadrant.

Key Concept

Vector resolution of multi-step angular displacements and shadow-based directional orientation.
Estimated Time:3m 0s
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Topics

Direction and Distance Test

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