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Difficulty: MediumDirection and Distance Test

At sunrise, a city planner standing at a central intersection observes that her shadow falls directly to her right side. From this point, she walks 15 m15\text{ m} straight ahead, then turns 135∘135^\circ clockwise and walks 102 m10\sqrt{2}\text{ m}. Next, she turns 45∘45^\circ anti-clockwise and walks 10 m10\text{ m}. Finally, she turns 90∘90^\circ clockwise and walks 5 m5\text{ m} to reach her destination. What is her shortest distance and direction from the starting point?

  1. 20 m20\text{ m}, WestAnswer
  2. B
    20 m20\text{ m}, East
  3. C
    25 m25\text{ m}, South-West
  4. D
    105 m10\sqrt{5}\text{ m}, South-West

Answer

The shortest distance is 20 m20\text{ m} and the direction relative to the starting point is West.
At sunrise, the sun is in the East, casting shadows toward the West. For a person's shadow to fall to their right side, West must be to their right, implying they are facing South. Setting the starting intersection as (0,0)(0,0), moving 15 m15\text{ m} South reaches (0,−15)(0, -15). Turning 135∘135^\circ clockwise changes facing to North-West; walking 102 m10\sqrt{2}\text{ m} contributes −10 m-10\text{ m} horizontally and +10 m+10\text{ m} vertically to reach (−10,−5)(-10, -5). Turning 45∘45^\circ anti-clockwise reorients facing to West; walking 10 m10\text{ m} West reaches (−20,−5)(-20, -5). Finally, turning 90∘90^\circ clockwise reorients facing to North; walking 5 m5\text{ m} North reaches (−20,0)(-20, 0). The net displacement from the start is 20 m20\text{ m} in the West direction.

Step-by-Step Solution

1
Determine initial facing direction using sunrise shadow orientation
Facing South
At sunrise, the sun is in the East and shadows fall toward the West. If the shadow is to her right side, her right must point West, which means she is initially facing South.
2
Calculate displacement after walking 15 m15\text{ m} straight ahead (South)
Position (0,−15)(0, -15)
Moving 15 m15\text{ m} South from origin (0,0)(0,0) gives a vertical change of −15 m-15\text{ m} along the y-axis.
3
Execute 135∘135^\circ clockwise turn from South and walk 102 m10\sqrt{2}\text{ m}
Position (−10,−5)(-10, -5)
Turning 135∘135^\circ clockwise from South (180°) reorients her to North-West (315°). Walking 102 m10\sqrt{2}\text{ m} gives Δx=−102cos⁡(45∘)=−10\Delta x = -10\sqrt{2}\cos(45^\circ) = -10 and Δy=+102sin⁡(45∘)=+10\Delta y = +10\sqrt{2}\sin(45^\circ) = +10. Updated coordinates: (0−10,−15+10)=(−10,−5)(0-10, -15+10) = (-10, -5).
4
Execute 45∘45^\circ anti-clockwise turn from North-West and walk 10 m10\text{ m}
Position (−20,−5)(-20, -5)
Turning 45∘45^\circ anti-clockwise from North-West reorients her due West. Walking 10 m10\text{ m} West updates coordinates to (−10−10,−5)=(−20,−5)(-10 - 10, -5) = (-20, -5).
5
Execute 90∘90^\circ clockwise turn from West and walk 5 m5\text{ m}
Position (−20,0)(-20, 0)
Turning 90∘90^\circ clockwise from West reorients her due North. Walking 5 m5\text{ m} North updates coordinates to (−20,−5+5)=(−20,0)(-20, -5 + 5) = (-20, 0).
6
Calculate final net displacement and direction from origin
Shortest distance = 20 m20\text{ m}, Direction = West
The straight-line distance from (0,0)(0,0) to (−20,0)(-20,0) is (−20)2+02=20 m\sqrt{(-20)^2 + 0^2} = 20\text{ m}, directed due West.

Key Concept

Vector displacement combining cardinal, sub-cardinal, and angular turns with shadow-based orientation
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Topics

Direction and Distance Test

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