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Zorluk: Çok zorDirection and Distance Test

Match each autonomous robotic surveyor (P, Q, R, S) executing a multi-step navigation sequence with its exact shortest straight-line displacement from its initial starting point.

  • Robotic Surveyor P: Starts facing North, walks 15 m15\text{ m}, turns 135∘135^\circ clockwise and walks 102 m10\sqrt{2}\text{ m}, then turns 90∘90^\circ right and walks 52 m5\sqrt{2}\text{ m}, and finally turns North and walks 12 m12\text{ m}.13 m13\text{ m}
  • Robotic Surveyor Q: Starts facing East, walks 20 m20\text{ m}, turns 135∘135^\circ counter-clockwise and walks 82 m8\sqrt{2}\text{ m}, then turns 45∘45^\circ clockwise and walks 7 m7\text{ m}, and finally turns South and walks 6 m6\text{ m}.15 m15\text{ m}
  • Robotic Surveyor R: Starts facing West, walks 10 m10\text{ m}, turns 45∘45^\circ counter-clockwise and walks 62 m6\sqrt{2}\text{ m}, and finally turns 135∘135^\circ clockwise and walks 18 m18\text{ m}.20 m20\text{ m}
  • Robotic Surveyor S: Positioned at sunrise facing its own shadow, turns 90∘90^\circ right and walks 10 m10\text{ m}, then turns 45∘45^\circ right and walks 72 m7\sqrt{2}\text{ m}, and finally turns North and walks 7 m7\text{ m}.25 m25\text{ m}

Cevap

The correct matching pairs are: Robotic Surveyor P matches with 13 m13\text{ m}, Robotic Surveyor Q matches with 15 m15\text{ m}, Robotic Surveyor R matches with 20 m20\text{ m}, and Robotic Surveyor S matches with 25 m25\text{ m}.
Each robotic surveyor's final position is determined by establishing a Cartesian coordinate system with the starting point at (0,0)(0,0). By breaking each movement into orthogonal components (x,y)(x, y) using trigonometry for diagonal movements (45∘,135∘45^\circ, 135^\circ) and applying the Pythagorean theorem d=x2+y2d = \sqrt{x^2 + y^2}, we obtain exact straight-line displacements of 13 m13\text{ m} for P, 15 m15\text{ m} for Q, 20 m20\text{ m} for R, and 25 m25\text{ m} for S.

Adım Adım Çözüm

1
Calculate displacement coordinates for Robotic Surveyor P
Path vector components:
1. 15 m15\text{ m} North: (0,15)(0, 15)
2. Turn 135∘135^\circ CW (facing SE), walk 102 m10\sqrt{2}\text{ m}: Δx=102sin⁡(135∘)=10\Delta x = 10\sqrt{2}\sin(135^\circ) = 10, Δy=102cos⁡(135∘)=−10  ⟹  (10,5)\Delta y = 10\sqrt{2}\cos(135^\circ) = -10 \implies (10, 5)
3. Turn 90∘90^\circ right (facing SW), walk 52 m5\sqrt{2}\text{ m}: Δx=−5\Delta x = -5, Δy=−5  ⟹  (5,0)\Delta y = -5 \implies (5, 0)
4. Walk 12 m12\text{ m} North: Δx=0\Delta x = 0, Δy=12  ⟹  (5,12)\Delta y = 12 \implies (5, 12)
Final distance: d=52+122=25+144=13 md = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = 13\text{ m}.
Decomposing vectors into xx (East) and yy (North) components allows precise tracking of position.
2
Calculate displacement coordinates for Robotic Surveyor Q
Path vector components:
1. 20 m20\text{ m} East: (20,0)(20, 0)
2. Turn 135∘135^\circ CCW (facing NW), walk 82 m8\sqrt{2}\text{ m}: Δx=−8\Delta x = -8, Δy=8  ⟹  (12,8)\Delta y = 8 \implies (12, 8)
3. Turn 45∘45^\circ CW (facing North), walk 7 m7\text{ m}: Δx=0\Delta x = 0, Δy=7  ⟹  (12,15)\Delta y = 7 \implies (12, 15)
4. Walk 6 m6\text{ m} South: Δx=0\Delta x = 0, Δy=−6  ⟹  (12,9)\Delta y = -6 \implies (12, 9)
Final distance: d=122+92=144+81=15 md = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = 15\text{ m}.
Using trigonometry and coordinate tracking ensures correct evaluation of net displacement.
3
Calculate displacement coordinates for Robotic Surveyor R
Path vector components:
1. 10 m10\text{ m} West: (−10,0)(-10, 0)
2. Turn 45∘45^\circ CCW (facing SW), walk 62 m6\sqrt{2}\text{ m}: Δx=−6\Delta x = -6, Δy=−6  ⟹  (−16,−6)\Delta y = -6 \implies (-16, -6)
3. Turn 135∘135^\circ CW from SW (facing North), walk 18 m18\text{ m}: Δx=0\Delta x = 0, Δy=18  ⟹  (−16,12)\Delta y = 18 \implies (-16, 12)
Final distance: d=(−16)2+122=256+144=20 md = \sqrt{(-16)^2 + 12^2} = \sqrt{256 + 144} = 20\text{ m}.
Angular rotations must be measured relative to the current heading, not absolute cardinal directions.
4
Determine initial orientation and calculate displacement for Robotic Surveyor S
At sunrise, the Sun is in the East, so shadows fall toward the West. Facing its own shadow means S initially faces West.
1. Turn 90∘90^\circ right (facing North), walk 10 m10\text{ m}: (0,10)(0, 10)
2. Turn 45∘45^\circ right (facing NE), walk 72 m7\sqrt{2}\text{ m}: Δx=7\Delta x = 7, Δy=7  ⟹  (7,17)\Delta y = 7 \implies (7, 17)
3. Turn North, walk 7 m7\text{ m}: Δx=0\Delta x = 0, Δy=7  ⟹  (7,24)\Delta y = 7 \implies (7, 24)
Final distance: d=72+242=49+576=25 md = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = 25\text{ m}.
Shadow orientation establishes the initial cardinal direction from which relative turns are executed.

Anahtar Kavram

Vector Displacement, Angular Rotation, and Shadow Orientation
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Konular

Direction and Distance Test

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