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Difficulty: MediumDirection and Distance Test

A maintenance engineer at a solar power facility begins an inspection from the central monitoring unit facing East. He first walks 15 m15\text{ m} straight ahead, then turns 90∘90^\circ clockwise and walks 20 m20\text{ m}. Next, he turns 135∘135^\circ anti-clockwise and walks 102 m10\sqrt{2}\text{ m}. He then turns 45∘45^\circ clockwise and walks 5 m5\text{ m}. Finally, he turns 90∘90^\circ anti-clockwise and walks 10 m10\text{ m} to reach his final inspection point. What is the shortest distance (in meters) between his final position and the central monitoring unit?

Answer: 30 m

Answer

The shortest distance between the final position and the central monitoring unit is 30 m30\text{ m}.
By resolving each directional move into Cartesian coordinates (x,y)(x, y), the initial segment gives (15,0)(15, 0), the second segment gives (15,−20)(15, -20), the third segment along North-East adds (10,10)(10, 10) to reach (25,−10)(25, -10), the fourth segment adds (5,0)(5, 0) to reach (30,−10)(30, -10), and the final segment adds (0,10)(0, 10) to land exactly at (30,0)(30, 0). The straight-line distance from (0,0)(0,0) to (30,0)(30,0) is 30 m30\text{ m}.

Step-by-Step Solution

1
Set up a 2D Cartesian coordinate system with the starting central monitoring unit at (0,0)(0,0) facing East along the positive x-axis.
Initial position after 15 m15\text{ m} East is (15,0)(15, 0).
Tracking displacement using coordinates avoids directional confusion.
2
Determine facing direction and displacement after a 90∘90^\circ clockwise turn.
Facing direction becomes South. Position after moving 20 m20\text{ m} South is (15,−20)(15, -20).
Turning 90∘90^\circ clockwise from East points directly South.
3
Calculate component displacements for a 135∘135^\circ anti-clockwise rotation from South.
Facing direction becomes North-East (45∘45^\circ). Moving 102 m10\sqrt{2}\text{ m} adds +10 m+10\text{ m} to x and +10 m+10\text{ m} to y, yielding position (25,−10)(25, -10).
Anti-clockwise rotation from South (270∘270^\circ) by 135∘135^\circ leads to 45∘45^\circ (North-East), where 102⋅12=10 m10\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 10\text{ m} along each axis.
4
Apply the next movement of 5 m5\text{ m} after a 45∘45^\circ clockwise turn.
Facing direction turns from North-East back to East. Position becomes (25+5,−10)=(30,−10)(25 + 5, -10) = (30, -10).
Turning 45∘45^\circ clockwise from North-East realigns the path to due East.
5
Apply final turn of 90∘90^\circ anti-clockwise and movement of 10 m10\text{ m}.
Facing direction becomes North. Position becomes (30,−10+10)=(30,0)(30, -10 + 10) = (30, 0).
Turning 90∘90^\circ anti-clockwise from East points North, adding +10 m+10\text{ m} to the y-coordinate.
6
Compute net displacement from starting point (0,0)(0,0) to final point (30,0)(30,0).
Shortest distance = (30−0)2+(0−0)2=30 m\sqrt{(30-0)^2 + (0-0)^2} = 30\text{ m}.
Applying the Euclidean distance formula gives the straight-line distance.

Key Concept

Multi-leg spatial vector addition using 2D Cartesian coordinates and trigonometric angular conversions.
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Topics

Direction and Distance Test

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