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Difficulty: HardDirection and Distance Test

An archaeological surveyor starting from a central benchmark walks 20 m20\text{ m} due East. He then turns 135∘135^\circ to his left and walks 142 m14\sqrt{2}\text{ m}. Next, he turns 45∘45^\circ to his left and walks 18 m18\text{ m}. Finally, he turns 90∘90^\circ to his right and walks 2 m2\text{ m} to reach an excavation site. What is the shortest direct distance (in meters) between the starting benchmark and the excavation site?

Answer: 20 m

Answer

The shortest direct distance between the starting benchmark and the excavation site is 20 meters.
By resolving each sequential movement into Cartesian coordinate components (x,y)(x, y), the final position relative to origin (0,0)(0,0) is found to be (−12 m,16 m)(-12\text{ m}, 16\text{ m}). The straight-line distance is computed using the distance formula (−12)2+(16)2=144+256=400=20 m\sqrt{(-12)^2 + (16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20\text{ m}.

Step-by-Step Solution

1
Plot the initial displacement on a Cartesian coordinate plane with origin (0,0)(0,0) at the starting benchmark.
Position after walking 20 m20\text{ m} East is (20,0)(20, 0) facing East (0∘0^\circ).
East represents the positive x-axis direction.
2
Calculate displacement after turning 135∘135^\circ left (facing North-West) and walking 142 m14\sqrt{2}\text{ m}.
Position becomes (20−14,0+14)=(6,14)(20 - 14, 0 + 14) = (6, 14).
In North-West direction, horizontal component is −142×12=−14 m-14\sqrt{2} \times \frac{1}{\sqrt{2}} = -14\text{ m} and vertical component is +142×12=+14 m+14\sqrt{2} \times \frac{1}{\sqrt{2}} = +14\text{ m}.
3
Calculate displacement after turning 45∘45^\circ left from North-West (facing West) and walking 18 m18\text{ m}.
Position becomes (6−18,14)=(−12,14)(6 - 18, 14) = (-12, 14).
Turning 45∘45^\circ left from North-West aligns facing direction with West (negative x-axis).
4
Calculate displacement after turning 90∘90^\circ right from West (facing North) and walking 2 m2\text{ m}.
Final position is (−12,14+2)=(−12,16)(-12, 14 + 2) = (-12, 16).
Turning 90∘90^\circ right from West aligns facing direction with North (positive y-axis).
5
Calculate the direct straight-line distance from starting point (0,0)(0,0) to final position (−12,16)(-12, 16).
Distance D=(−12)2+162=144+256=400=20 mD = \sqrt{(-12)^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\text{ m}.
Applying the Pythagorean theorem D=x2+y2D = \sqrt{x^2 + y^2} yields the shortest distance.

Key Concept

Coordinate Geometry and Vector Resolution in Direction and Distance Problems
Estimated Time:2m 0s
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Topics

Direction and Distance Test

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