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Question

Difficulty: MediumDirection and Distance Test

Match each multi-step movement sequence described in List-I with its corresponding net displacement (distance and direction) relative to the initial starting point in List-II.

  • Walks 12 m12\text{ m} North, turns right and walks 9 m9\text{ m}, then turns right again and walks 12 m12\text{ m}.9 m9\text{ m} East
  • Walks 5 m5\text{ m} East, turns left and walks 12 m12\text{ m}, then turns left again and walks 5 m5\text{ m}.12 m12\text{ m} North
  • Walks 15 m15\text{ m} South, turns left and walks 8 m8\text{ m}, then turns left again and walks 15 m15\text{ m}.8 m8\text{ m} East
  • Walks 10 m10\text{ m} West, turns left and walks 24 m24\text{ m}, then turns left again and walks 10 m10\text{ m}.24 m24\text{ m} South

Answer

The item describing a walk of 12 m12\text{ m} North, 9 m9\text{ m} right (East), and 12 m12\text{ m} right (South) matches 9 m9\text{ m} East; the item describing 5 m5\text{ m} East, 12 m12\text{ m} left (North), and 5 m5\text{ m} left (West) matches 12 m12\text{ m} North; the item describing 15 m15\text{ m} South, 8 m8\text{ m} left (East), and 15 m15\text{ m} left (North) matches 8 m8\text{ m} East; and the item describing 10 m10\text{ m} West, 24 m24\text{ m} left (South), and 10 m10\text{ m} left (East) matches 24 m24\text{ m} South.
Each scenario represents a three-leg trajectory where the first leg and the third leg are of equal magnitude but opposite directions, completely canceling each other out. Thus, the net displacement vector is equal in magnitude and direction to the second leg of the journey.

Step-by-Step Solution

1
Analyze Path 1: Initial position (0,0)(0,0).
Move North by 12 m→(0,12)12\text{ m} \rightarrow (0,12). Turn right (facing East) and move 9 m→(9,12)9\text{ m} \rightarrow (9,12). Turn right (facing South) and move 12 m→(9,0)12\text{ m} \rightarrow (9,0). Net vector is (9,0)(9,0), which is 9 m9\text{ m} East.
Sequential turns of 90∘90^\circ clockwise from North lead to East and then South.
2
Analyze Path 2: Initial position (0,0)(0,0).
Move East by 5 m→(5,0)5\text{ m} \rightarrow (5,0). Turn left (facing North) and move 12 m→(5,12)12\text{ m} \rightarrow (5,12). Turn left (facing West) and move 5 m→(0,12)5\text{ m} \rightarrow (0,12). Net vector is (0,12)(0,12), which is 12 m12\text{ m} North.
Sequential turns of 90∘90^\circ counter-clockwise from East lead to North and then West.
3
Analyze Path 3: Initial position (0,0)(0,0).
Move South by 15 m→(0,−15)15\text{ m} \rightarrow (0,-15). Turn left (facing East) and move 8 m→(8,−15)8\text{ m} \rightarrow (8,-15). Turn left (facing North) and move 15 m→(8,0)15\text{ m} \rightarrow (8,0). Net vector is (8,0)(8,0), which is 8 m8\text{ m} East.
Sequential left turns while facing South lead to East, then North.
4
Analyze Path 4: Initial position (0,0)(0,0).
Move West by 10 m→(−10,0)10\text{ m} \rightarrow (-10,0). Turn left (facing South) and move 24 m→(−10,−24)24\text{ m} \rightarrow (-10,-24). Turn left (facing East) and move 10 m→(0,−24)10\text{ m} \rightarrow (0,-24). Net vector is (0,−24)(0,-24), which is 24 m24\text{ m} South.
Sequential left turns while facing West lead to South, then East.

Key Concept

Orthogonal Vector Addition and Cancelation in Two Dimensions
Estimated Time:1m 30s
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Topics

Direction and Distance Test

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