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Difficulty: HardDirection and Distance Test

Just before sunset, Ananya starts walking from point PP directly towards her shadow. After walking 10 m10\text{ m} to reach point QQ, she turns 135∘135^\circ anti-clockwise and walks 102 m10\sqrt{2}\text{ m} to point RR. From point RR, she turns 90∘90^\circ to her right and walks 152 m15\sqrt{2}\text{ m} to point SS. Finally, she turns to face South and walks 5 m5\text{ m} to reach point TT. What is the shortest straight-line distance between point PP and point TT, and in which direction is point TT located relative to point PP?

  1. 25 m25\text{ m}, North-EastAnswer
  2. B
    25 m25\text{ m}, North-West
  3. C
    35 m35\text{ m}, North-East
  4. D
    15 m15\text{ m}, South-East

Answer

The shortest straight-line distance is 25 m25\text{ m}, and point TT is located North-East relative to starting point PP.
The correct answer correctly resolves the implicit sunset shadow direction (East) and resolves each subsequent angular displacement vector on a Cartesian grid to arrive at coordinates (15,20)(15, 20). The straight-line distance is calculated as 152+202=25 m\sqrt{15^2 + 20^2} = 25\text{ m}, positioned in the North-East quadrant relative to the origin.

Step-by-Step Solution

1
Determine initial direction using sunset shadow orientation.
At sunset, the Sun is in the West, so shadows fall toward the East. Walking directly towards her shadow means Ananya initially walks East from P(0,0)P(0,0) to Q(10,0)Q(10,0).
Shadow orientation provides the initial cardinal vector.
2
Calculate position of point RR after 135∘135^\circ anti-clockwise turn.
Facing East (0∘0^\circ), an anti-clockwise turn of 135∘135^\circ reorients her toward North-West (135∘135^\circ). Walking 102 m10\sqrt{2}\text{ m} gives 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}. Thus, R=(10−10,0+10)=(0,10)R = (10 - 10, 0 + 10) = (0, 10).
Decompose angular displacement vector into orthogonal components.
3
Calculate position of point SS after 90∘90^\circ right turn.
Facing North-West (135∘135^\circ), turning 90∘90^\circ right places her facing North-East (45∘45^\circ). Walking 152 m15\sqrt{2}\text{ m} gives Δx=+15 m\Delta x = +15\text{ m} and Δy=+15 m\Delta y = +15\text{ m}. Thus, S=(0+15,10+15)=(15,25)S = (0 + 15, 10 + 15) = (15, 25).
Re-orient facing direction and resolve the North-East vector.
4
Calculate final position TT after walking South.
Walking 5 m5\text{ m} South from S(15,25)S(15, 25) changes the y-coordinate to 25−5=2025 - 5 = 20. Thus, T=(15,20)T = (15, 20).
Apply pure vertical displacement southward.
5
Compute shortest straight-line distance and direction from P(0,0)P(0,0) to T(15,20)T(15,20).
Distance PT=152+202=225+400=625=25 mPT = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ m}. Since both coordinates (15,20)(15, 20) are positive, TT lies in the North-East direction relative to PP.
Apply the Pythagorean theorem and quadrant analysis.

Key Concept

Direction and Distance Test - Multi-step Vector Resolution and Angular Turns
Estimated Time:2m 0s
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Topics

Direction and Distance Test

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