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Zorluk: ZorDirection and Distance Test

At sunset, a surveyor facing the setting sun launches a drone from a fixed control pad. The drone executes the following sequence of movements:
1. It travels 40 m40\text{ m} straight ahead in the direction the surveyor is facing.
2. It turns 135∘135^\circ clockwise and flies 302 m30\sqrt{2}\text{ m}.
3. It makes a 90∘90^\circ right turn and flies 102 m10\sqrt{2}\text{ m}.
4. Finally, it turns 135∘135^\circ counter-clockwise and flies 15 m15\text{ m}.

What is the shortest straight-line distance of the drone from the control pad, and in which direction is it located relative to the starting point?

  1. 35 m35\text{ m}, NorthCevap
  2. B
    35 m35\text{ m}, South-West
  3. C
    25 m25\text{ m}, North-West
  4. D
    95 m95\text{ m}, North

Cevap

The drone is 35 m35\text{ m} away from the control pad in the North direction.
At sunset, facing the setting sun means facing West. The drone travels 40 m40\text{ m} West (x=−40,y=0x = -40, y = 0). Turning 135∘135^\circ clockwise reorients it to North-East; flying 302 m30\sqrt{2}\text{ m} changes position to x=−10,y=30x = -10, y = 30. Turning 90∘90^\circ right reorients it to South-East; flying 102 m10\sqrt{2}\text{ m} shifts position to x=0,y=20x = 0, y = 20. Turning 135∘135^\circ counter-clockwise reorients it North; flying 15 m15\text{ m} brings it to x=0,y=35x = 0, y = 35. The drone is 35 m35\text{ m} North of the origin.

Adım Adım Çözüm

1
Determine initial facing direction and first displacement vector
The surveyor faces West at sunset. Moving 40 m40\text{ m} straight ahead places the drone at coordinates (−40,0)(-40, 0) relative to origin (0,0)(0,0). Current heading: West.
Sun sets in the West, establishing the initial cardinal reference.
2
Apply 135∘135^\circ clockwise turn and calculate second displacement
Turning 135∘135^\circ clockwise from West reorients the drone to North-East. Flying 302 m30\sqrt{2}\text{ m} adds +30 m+30\text{ m} East and +30 m+30\text{ m} North. Position becomes (−40+30,0+30)=(−10,30)(-40 + 30, 0 + 30) = (-10, 30).
A 135∘135^\circ clockwise shift from West (270∘270^\circ) leads to North-East (45∘45^\circ). Vector components are 302cos⁡(45∘)=3030\sqrt{2}\cos(45^\circ) = 30 and 302sin⁡(45∘)=3030\sqrt{2}\sin(45^\circ) = 30.
3
Apply 90∘90^\circ right turn and calculate third displacement
Turning 90∘90^\circ right from North-East reorients heading to South-East. Flying 102 m10\sqrt{2}\text{ m} adds +10 m+10\text{ m} East and −10 m-10\text{ m} South. Position becomes (−10+10,30−10)=(0,20)(-10 + 10, 30 - 10) = (0, 20).
A 90∘90^\circ right turn rotates North-East to South-East. Vector components are +10 m+10\text{ m} along East and −10 m-10\text{ m} along North.
4
Apply 135∘135^\circ counter-clockwise turn and calculate final displacement
Turning 135∘135^\circ counter-clockwise from South-East reorients heading to North. Flying 15 m15\text{ m} North advances position to (0,20+15)=(0,35)(0, 20 + 15) = (0, 35).
Turning 135∘135^\circ counter-clockwise from South-East points directly North.
5
Calculate net displacement magnitude and cardinal direction
Net coordinates are (0,35)(0, 35). Distance = 02+352=35 m\sqrt{0^2 + 35^2} = 35\text{ m} due North.
X-component is 0 m0\text{ m} and Y-component is +35 m+35\text{ m}.

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