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Difficulty: Very hardDirection and Distance Test

At sunrise, an environmental officer stands facing her own shadow. She then executes the following sequence of movements:
1. She turns 135∘135^\circ anti-clockwise and walks 102 m10\sqrt{2}\text{ m}.
2. She turns 90∘90^\circ to her right and walks 62 m6\sqrt{2}\text{ m}.
3. She turns 135∘135^\circ to her left and walks 15 m15\text{ m}.
4. She turns 135∘135^\circ to her left again and walks 72 m7\sqrt{2}\text{ m}.

What is her shortest direct distance from the starting point, and in which direction is she positioned relative to her starting point?

  1. 15 m15\text{ m}, North-EastAnswer
  2. B
    15 m15\text{ m}, South-East
  3. C
    25 m25\text{ m}, North-East
  4. D
    21 m21\text{ m}, North-West

Answer

15 m15\text{ m} in the North-East direction
At sunrise, a person facing her shadow faces West. Executing the four angular turns results in successive vectors: (10,10)(10, 10) North-East, (6,−6)(6, -6) South-East, (0,15)(0, 15) North, and (−7,−7)(-7, -7) South-West. Summing these vector components gives a net East-West displacement of X=9 mX = 9\text{ m} East and a net North-South displacement of Y=12 mY = 12\text{ m} North. Using the Pythagorean theorem, the direct distance is 92+122=15 m\sqrt{9^2 + 12^2} = 15\text{ m} in the North-East quadrant.

Step-by-Step Solution

1
Determine initial facing direction
Facing West (180∘180^\circ)
At sunrise, the Sun is in the East, so shadows fall toward the West. Facing her own shadow means she initially faces West.
2
Calculate vector displacement for Movement 1
Direction: North-East (45∘45^\circ), Vector: (+10,+10)(+10, +10)
From West (180∘180^\circ), turning 135∘135^\circ anti-clockwise gives 180∘−135∘=45∘180^\circ - 135^\circ = 45^\circ (North-East). Displacement =(102cos⁡45∘,102sin⁡45∘)=(10,10)= (10\sqrt{2}\cos 45^\circ, 10\sqrt{2}\sin 45^\circ) = (10, 10).
3
Calculate vector displacement for Movement 2
Direction: South-East (315∘315^\circ), Vector: (+6,−6)(+6, -6)
From North-East (45∘45^\circ), turning 90∘90^\circ right (clockwise) gives 45∘−90∘=−45∘=315∘45^\circ - 90^\circ = -45^\circ = 315^\circ (South-East). Displacement =(62cos⁡315∘,62sin⁡315∘)=(6,−6)= (6\sqrt{2}\cos 315^\circ, 6\sqrt{2}\sin 315^\circ) = (6, -6).
4
Calculate vector displacement for Movement 3
Direction: North (90∘90^\circ), Vector: (0,+15)(0, +15)
From South-East (−45∘-45^\circ), turning 135∘135^\circ left (anti-clockwise) gives −45∘+135∘=90∘-45^\circ + 135^\circ = 90^\circ (North). Displacement =(0,15)= (0, 15).
5
Calculate vector displacement for Movement 4
Direction: South-West (225∘225^\circ), Vector: (−7,−7)(-7, -7)
From North (90∘90^\circ), turning 135∘135^\circ left (anti-clockwise) gives 90∘+135∘=225∘90^\circ + 135^\circ = 225^\circ (South-West). Displacement =(72cos⁡225∘,72sin⁡225∘)=(−7,−7)= (7\sqrt{2}\cos 225^\circ, 7\sqrt{2}\sin 225^\circ) = (-7, -7).
6
Sum total net displacement (X,Y)(X, Y) and find direct distance and final cardinal direction
X=9 mX = 9\text{ m} East, Y=12 mY = 12\text{ m} North; Distance =15 m= 15\text{ m}, Direction = North-East
Total East displacement X=10+6+0−7=9 mX = 10 + 6 + 0 - 7 = 9\text{ m}. Total North displacement Y=10−6+15−7=12 mY = 10 - 6 + 15 - 7 = 12\text{ m}. Direct distance =92+122=81+144=225=15 m= \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15\text{ m}. Since both X>0X > 0 and Y>0Y > 0, the final direction relative to the start is North-East.

Key Concept

Direction and Distance Test - Vector Decomposition & Shadow Orientation
Estimated Time:3m 0s
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Topics

Direction and Distance Test

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