Question 7 of 18advanced🔧 ApplyNumerical5 marks

On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

Correct Answer

(At eighth turn) Displacement: 866 m (5003\sqrt{3} m) at 30° from initial direction. Path length: 4000 m. Comparison: Path length (4000 m) > Magnitude of displacement (866 m).

(At sixth turn) Displacement: Zero (motorist returns to starting point). Path length: 3000 m. Comparison: Path length (3000 m) > Magnitude of displacement (0 m).

(At third turn) Displacement: 1000 m at 60° from initial direction. Path length: 1500 m. Comparison: Path length (1500 m) > Magnitude of displacement (1000 m).

Exercise: EXERCISES | Q: 3.10 | (Page 21)
For More Understanding

Explanation

This problem involves understanding displacement vectors and path length. The motorist traces a regular hexagonal path since each turn of 60° after 6 turns completes 360°. The displacement depends only on the end points, while path length depends on the actual route taken. As stated in the Points to Ponder, path length is generally greater than magnitude of displacement when direction changes during motion.

Solution Steps

  1. Step 1: Understand the motion pattern. Each turn of 60° left after 500 m creates a regular hexagonal path. After 6 turns (6×606 \times 60° = 360°), the motorist returns to the starting point.

  2. Step 2: Calculate position after each turn using vector addition. Let initial direction be along positive x-axis.

  3. Step 3: At 3rd turn - Position = (500 m, 5003\sqrt{3} m). Displacement magnitude = √[(500)² + (5003\sqrt{3})²] = √[250000 + 750000] = 1000000\sqrt{1000000} = 1000 m. Direction = tan⁻¹(3\sqrt{3}) = 60° from initial direction. Path length = 3×5003 \times 500 = 1500 m.

  4. Step 4: At 6th turn - The motorist completes one full hexagon and returns to origin. Displacement = 0 m. Path length = 6×5006 \times 500 = 3000 m.

  5. Step 5: At 8th turn - Position after 8th turn = (750 m, 2503\sqrt{3} m). Displacement magnitude = √[(750)² + (2503\sqrt{3})²] = √[562500 + 187500] = 750000\sqrt{750000} = 5003\sqrt{3} ≈\approx 866 m. Direction = tan⁻¹(1/3\sqrt{3}) = 30° from initial direction. Path length = 8×5008 \times 500 = 4000 m.

  6. Step 6: Compare: In all cases, path length exceeds displacement magnitude because the motorist changes direction during motion.

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