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Difficulty: HardDirection and Distance Test

Shortly after sunrise, a wildlife photographer begins tracking an animal. She starts walking in a straight line, noticing that her shadow falls exactly to her left. After walking 40 m40\text{ m}, she turns 135∘135^\circ to her right and walks 302 m30\sqrt{2}\text{ m}. Finally, she turns 90∘90^\circ to her right and walks another 102 m10\sqrt{2}\text{ m} to reach the animal. What is her shortest distance and direction from her starting point?

  1. 20 m20\text{ m} towards EastAnswer
  2. B
    20 m20\text{ m} towards West
  3. C
    205 m20\sqrt{5}\text{ m} towards North-East
  4. D
    20 m20\text{ m} towards South

Answer

The photographer is 20 m20\text{ m} towards East from her starting point.
Based on sunrise shadow rules, the initial direction is North. Walking 40 m40\text{ m} North to (0,40)(0, 40), turning 135∘135^\circ right to travel 302 m30\sqrt{2}\text{ m} South-East to (30,10)(30, 10), and finally turning 90∘90^\circ right to travel 102 m10\sqrt{2}\text{ m} South-West results in a final coordinate of (20,0)(20, 0). This point is exactly 20 m20\text{ m} East of the starting point.

Step-by-Step Solution

1
Determine the initial facing direction using the position of the sun and the shadow.
The photographer is initially facing North.
At sunrise, the sun is in the East, so light travels West and shadows are cast to the West. Since her shadow falls to her left, her left side points West. Therefore, she must be facing North.
2
Plot the first movement on a coordinate plane.
Her position becomes (0,40)(0, 40).
Starting at the origin (0,0)(0,0), walking 40 m40\text{ m} North translates to an upward movement of +40+40 on the y-axis.
3
Calculate the second movement after a 135∘135^\circ right turn.
Her new position is (30,10)(30, 10).
Turning 135∘135^\circ to the right (clockwise) from North puts her facing South-East. Moving 302 m30\sqrt{2}\text{ m} South-East yields a displacement of Δx=302×cos⁡(−45∘)=30\Delta x = 30\sqrt{2} \times \cos(-45^\circ) = 30 and Δy=302×sin⁡(−45∘)=−30\Delta y = 30\sqrt{2} \times \sin(-45^\circ) = -30. Adding this to (0,40)(0, 40) gives (30,10)(30, 10).
4
Calculate the final movement after a 90∘90^\circ right turn.
Her final position is (20,0)(20, 0).
Turning 90∘90^\circ to the right from South-East puts her facing South-West. Moving 102 m10\sqrt{2}\text{ m} South-West yields a displacement of Δx=−10\Delta x = -10 and Δy=−10\Delta y = -10. Adding this to (30,10)(30, 10) gives (20,0)(20, 0).
5
Determine the final distance and direction from the origin.
The shortest distance is 20 m20\text{ m}, and the direction is East.
The final coordinate is (20,0)(20, 0). The distance from (0,0)(0,0) is 202+02=20 m\sqrt{20^2 + 0^2} = 20\text{ m}. Since it lies on the positive x-axis, the direction is East.

Key Concept

Vector displacement and cardinal alignment using sun-shadow rules.
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Topics

Direction and Distance Test

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