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Difficulty: MediumDirection and Distance Test

A wildlife biologist starts tracking a tagged animal from a base station while facing North. She first walks 14 m14\text{ m} due North, then turns 90∘90^\circ clockwise and walks 18 m18\text{ m}. Next, she turns 135∘135^\circ clockwise and walks 102 m10\sqrt{2}\text{ m}. Finally, she turns 45∘45^\circ counter-clockwise and walks 10 m10\text{ m} straight ahead to reach her observation point. What is the shortest distance between her observation point and the base station, and in which direction is she relative to the base station?

  1. 10 m10\text{ m}, South-EastAnswer
  2. B
    10 m10\text{ m}, South-West
  3. C
    14 m14\text{ m}, South-East
  4. D
    20 m20\text{ m}, North-East

Answer

10 m10\text{ m}, South-East
The correct answer is 10 m10\text{ m}, South-East. Taking the base station as origin (0,0)(0,0), the initial moves place the biologist at (18,14)(18, 14). Turning 135∘135^\circ clockwise from East points her South-West, shifting her coordinates by (−10,−10)(-10, -10) to (8,4)(8, 4). Turning 45∘45^\circ counter-clockwise from South-West points her due South, shifting her coordinates by (0,−10)(0, -10) to (8,−6)(8, -6). Using the Pythagorean theorem, 82+(−6)2=10 m\sqrt{8^2 + (-6)^2} = 10\text{ m}, located in the South-East quadrant relative to the origin.

Step-by-Step Solution

1
Calculate displacement after Step 1 and Step 2
Position is (18,14)(18, 14) relative to origin (0,0)(0,0)
Walking 14 m14\text{ m} North places her at (0,14)(0, 14) facing North. Turning 90∘90^\circ clockwise faces East; walking 18 m18\text{ m} places her at (18,14)(18, 14) facing East.
2
Calculate displacement after Step 3 angular turn
Position is (8,4)(8, 4) facing South-West
From East, a 135∘135^\circ clockwise turn reorients facing to South-West. Moving 102 m10\sqrt{2}\text{ m} SW gives displacement components Δx=−102cos⁡(45∘)=−10 m\Delta x = -10\sqrt{2}\cos(45^\circ) = -10\text{ m} and Δy=−102sin⁡(45∘)=−10 m\Delta y = -10\sqrt{2}\sin(45^\circ) = -10\text{ m}. New position is (18−10,14−10)=(8,4)(18 - 10, 14 - 10) = (8, 4).
3
Calculate displacement after Step 4 angular turn
Final Position is (8,−6)(8, -6) facing South
From South-West (225∘225^\circ), a 45∘45^\circ counter-clockwise turn changes facing to South (270∘270^\circ). Moving 10 m10\text{ m} South gives Δx=0\Delta x = 0, Δy=−10 m\Delta y = -10\text{ m}. Final coordinates are (8,4−10)=(8,−6)(8, 4 - 10) = (8, -6).
4
Determine shortest distance and directional vector relative to base station
Distance =10 m= 10\text{ m}, Direction = South-East
Shortest distance d=82+(−6)2=64+36=10 md = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = 10\text{ m}. Since x=+8x = +8 (East) and y=−6y = -6 (South), the final direction relative to origin is South-East.

Key Concept

Vector displacement decomposition involving cardinal directions and angular rotations.
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Topics

Direction and Distance Test

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