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Difficulty: MediumDirection and Distance Test

A delivery supervisor starts from a central warehouse facing East. He drives 15 km15\text{ km} due East, turns 90∘90^\circ to his left, and drives 12 km12\text{ km}. He then turns 135∘135^\circ clockwise and drives 52 km5\sqrt{2}\text{ km}. Next, he turns 135∘135^\circ anti-clockwise and drives 8 km8\text{ km} due North. Finally, he turns 90∘90^\circ to his left and drives 20 km20\text{ km}. What is the shortest distance between his final position and the central warehouse, and in which direction is he located relative to the central warehouse?

  1. 15 km15\text{ km}, NorthAnswer
  2. B
    25 km25\text{ km}, North-East
  3. C
    15 km15\text{ km}, South
  4. D
    20 km20\text{ km}, North-West

Answer

15 km15\text{ km}, North
The net horizontal displacement cancels completely to 0 km0\text{ km} (15+5−20=015 + 5 - 20 = 0), while the net vertical displacement totals +15 km+15\text{ km} (12−5+8=1512 - 5 + 8 = 15). This places the final position exactly 15 km15\text{ km} North of the starting point.

Step-by-Step Solution

1
Set initial position and track the first two straight movements.
Position becomes (15,0)(15, 0) after moving East, then (15,12)(15, 12) after moving North.
Moving East adds to the x-coordinate; turning 90∘90^\circ left faces North and adds to the y-coordinate.
2
Apply the angular rotation of 135∘135^\circ clockwise.
New facing direction is South-East (−45∘-45^\circ).
Facing North (90∘90^\circ), a 135∘135^\circ clockwise rotation turns the direction to South-East.
3
Calculate displacement for 52 km5\sqrt{2}\text{ km} South-East.
Displacement is Δx=+5 km\Delta x = +5\text{ km} and Δy=−5 km\Delta y = -5\text{ km}, giving position (20,7)(20, 7).
52cos⁡(−45∘)=+55\sqrt{2} \cos(-45^\circ) = +5 and 52sin⁡(−45∘)=−55\sqrt{2} \sin(-45^\circ) = -5.
4
Apply 135∘135^\circ anti-clockwise turn and 8 km8\text{ km} movement North.
New position is (20,7+8)=(20,15)(20, 7 + 8) = (20, 15).
Rotating 135∘135^\circ anti-clockwise from South-East points North, increasing the y-coordinate by 88.
5
Apply final 90∘90^\circ left turn (West) and 20 km20\text{ km} movement.
Final position is (20−20,15)=(0,15)(20 - 20, 15) = (0, 15).
Turning left from North points West, decreasing the x-coordinate by 2020.
6
Calculate net distance and direction from origin (0,0)(0,0).
Shortest distance =02+152=15 km= \sqrt{0^2 + 15^2} = 15\text{ km}, direction is due North.
The x-offset is zero and y-offset is positive 1515.

Key Concept

Vector displacement in Cartesian coordinates with angular turns
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Topics

Direction and Distance Test

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