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Difficulty: MediumDirection and Distance Test

An autonomous security robot performs test runs along four distinct patrol routes starting from a central command post at coordinate (0,0)(0,0). Match each multi-step movement route described in List-I with its corresponding net displacement (shortest straight-line distance and direction relative to the command post) in List-II.

  • Route I: Moves 10 m10\text{ m} North, turns 90∘90^\circ right and moves 4 m4\text{ m}, then turns 135∘135^\circ right and moves 42 m4\sqrt{2}\text{ m}.6 m6\text{ m}, due North
  • Route II: Moves 16 m16\text{ m} East, turns 90∘90^\circ left and moves 12 m12\text{ m}, then turns 90∘90^\circ left and moves 11 m11\text{ m}.13 m13\text{ m}, North-East
  • Route III: Moves 7 m7\text{ m} South, turns 90∘90^\circ right and moves 24 m24\text{ m}, then turns 90∘90^\circ right and moves 14 m14\text{ m}.25 m25\text{ m}, North-West
  • Route IV: Moves 12 m12\text{ m} West, turns 135∘135^\circ left and moves 82 m8\sqrt{2}\text{ m}, then turns 45∘45^\circ left and moves 4 m4\text{ m}.8 m8\text{ m}, due South

Answer

Route I matches 6 m6\text{ m}, due North; Route II matches 13 m13\text{ m}, North-East; Route III matches 25 m25\text{ m}, North-West; Route IV matches 8 m8\text{ m}, due South.
Each patrol route resolves to a distinct final coordinate when tracked on a Cartesian plane where East/North are positive and West/South are negative. Route I reaches (0,6)(0,6), which is 6 m6\text{ m} due North. Route II reaches (5,12)(5,12), giving a distance of 52+122=13 m\sqrt{5^2+12^2}=13\text{ m} North-East. Route III reaches (−24,7)(-24,7), giving a distance of (−24)2+72=25 m\sqrt{(-24)^2+7^2}=25\text{ m} North-West. Route IV reaches (0,−8)(0,-8), which is 8 m8\text{ m} due South.

Step-by-Step Solution

1
Calculate net coordinates for Route I
Path vector: North 10 m→(0,10)10\text{ m} \rightarrow (0,10). Right turn (90∘90^\circ East) 4 m→(4,10)4\text{ m} \rightarrow (4,10). Right turn (135∘135^\circ South-West) for 42 m→Δx=−42cos⁡45∘=−44\sqrt{2}\text{ m} \rightarrow \Delta x = -4\sqrt{2}\cos 45^\circ = -4, Δy=−42sin⁡45∘=−4\Delta y = -4\sqrt{2}\sin 45^\circ = -4. Final coordinate: (4−4,10−4)=(0,6)(4-4, 10-4) = (0,6). Displacement: 6 m6\text{ m} North.
Decompose angular motion into horizontal and vertical components.
2
Calculate net coordinates for Route II
Path vector: East 16 m→(16,0)16\text{ m} \rightarrow (16,0). Left turn (90∘90^\circ North) 12 m→(16,12)12\text{ m} \rightarrow (16,12). Left turn (90∘90^\circ West) 11 m→(5,12)11\text{ m} \rightarrow (5,12). Distance =52+122=13 m= \sqrt{5^2 + 12^2} = 13\text{ m}. Direction: positive x and positive y indicate North-East.
Apply the Pythagorean theorem x2+y2\sqrt{x^2 + y^2} to find straight-line distance.
3
Calculate net coordinates for Route III
Path vector: South 7 m→(0,−7)7\text{ m} \rightarrow (0,-7). Right turn (90∘90^\circ West) 24 m→(−24,−7)24\text{ m} \rightarrow (-24,-7). Right turn (90∘90^\circ North) 14 m→(−24,7)14\text{ m} \rightarrow (-24, 7). Distance =(−24)2+72=576+49=25 m= \sqrt{(-24)^2 + 7^2} = \sqrt{576 + 49} = 25\text{ m}. Direction: negative x and positive y indicate North-West.
Sum vertical displacements (−7+14=+7)(-7 + 14 = +7) and horizontal displacement (−24)(-24) to resolve net position.
4
Calculate net coordinates for Route IV
Path vector: West 12 m→(−12,0)12\text{ m} \rightarrow (-12,0). Left turn (135∘135^\circ South-East) for 82 m→Δx=+82cos⁡45∘=+88\sqrt{2}\text{ m} \rightarrow \Delta x = +8\sqrt{2}\cos 45^\circ = +8, Δy=−82sin⁡45∘=−8→(−4,−8)\Delta y = -8\sqrt{2}\sin 45^\circ = -8 \rightarrow (-4,-8). Left turn (45∘45^\circ East) 4 m→(−4+4,−8)=(0,−8)4\text{ m} \rightarrow (-4+4, -8) = (0,-8). Displacement: 8 m8\text{ m} South.
Calculate exact component breakdown for sub-cardinal movement vector.

Key Concept

Direction and Distance Test - Coordinate Vector Resolution
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Topics

Direction and Distance Test

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