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Difficulty: MediumDirection and Distance Test

A cyclist starts from point PP facing East. She rides 15 km15\text{ km} due East to reach checkpoint QQ. At checkpoint QQ, she turns 135∘135^\circ to her left and rides 102 km10\sqrt{2}\text{ km} to reach checkpoint RR. From checkpoint RR, she turns 135∘135^\circ to her right and rides 7 km7\text{ km} due East to reach checkpoint SS. Finally, she turns 90∘90^\circ to her right and rides 5 km5\text{ km} to reach destination TT.

What is the shortest distance between her starting point PP and final destination TT, and in which direction is destination TT relative to starting point PP?

  1. 13 km13\text{ km}, North-EastAnswer
  2. B
    13 km13\text{ km}, South-East
  3. C
    17 km17\text{ km}, North-East
  4. D
    13 km13\text{ km}, South-West

Answer

13 km13\text{ km}, North-East
By placing the start point at the origin (0,0)(0,0), the movement sequence produces the following coordinates: Q(15,0)→R(5,10)→S(12,10)→T(12,5)Q(15,0) \rightarrow R(5,10) \rightarrow S(12,10) \rightarrow T(12,5). The Euclidean distance between (0,0)(0,0) and (12,5)(12,5) is 122+52=13 km\sqrt{12^2 + 5^2} = 13\text{ km}. Since both x and y coordinates of TT are positive relative to PP, the direction of TT from PP is North-East.

Step-by-Step Solution

1
Set up coordinate axis with point PP at origin (0,0)(0,0).
Initial position P=(0,0)P = (0,0). Initial facing direction = East (0∘0^\circ).
Establishing a standard Cartesian system makes vector tracking systematic.
2
Calculate position after moving 15 km15\text{ km} East to QQ.
Position of Q=(15,0)Q = (15, 0), facing East.
Displacement is purely along the positive x-axis.
3
Calculate position after turning 135∘135^\circ left (North-West) and riding 102 km10\sqrt{2}\text{ km} to RR.
Horizontal change Δx=102cos⁡(135∘)=−10 km\Delta x = 10\sqrt{2} \cos(135^\circ) = -10\text{ km}; Vertical change Δy=102sin⁡(135∘)=+10 km\Delta y = 10\sqrt{2} \sin(135^\circ) = +10\text{ km}. Position of R=(15−10,0+10)=(5,10)R = (15 - 10, 0 + 10) = (5, 10), facing North-West.
Turning 135∘135^\circ left from East points towards North-West (135∘135^\circ standard angle).
4
Calculate position after turning 135∘135^\circ right (East) and riding 7 km7\text{ km} to SS.
Position of S=(5+7,10)=(12,10)S = (5 + 7, 10) = (12, 10), facing East.
Turning 135∘135^\circ right from North-West re-orients facing direction due East.
5
Calculate position after turning 90∘90^\circ right (South) and riding 5 km5\text{ km} to destination TT.
Position of T=(12,10−5)=(12,5)T = (12, 10 - 5) = (12, 5).
Turning 90∘90^\circ right from East faces South (negative y-direction).
6
Calculate shortest distance PTPT and relative direction of TT from P(0,0)P(0,0).
Distance PT=(12−0)2+(5−0)2=144+25=169=13 kmPT = \sqrt{(12 - 0)^2 + (5 - 0)^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ km}. Since x=+12x = +12 and y=+5y = +5, TT lies in the North-East quadrant from PP.
Applying the Pythagorean theorem gives the straight-line displacement.

Key Concept

Direction and Distance Test - Multi-Turn Vector Displacement and Pythagorean Theorem
Estimated Time:1m 30s
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Topics

Direction and Distance Test

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