A biker rides 700 m north and then 300 m east. The total distance traveled is 1000 m (700 + 300). However, displacement — the straight-line distance from start to finish — is 100√58 m ≈ 761.6 m, directed at an angle of approximately 23.2° east of north. This is a standard application of the Pythagoras theorem to vector problems in physics.
Distance traveled = 700 + 300 = 1000 m.
Displacement = √(700² + 300²) = √580,000 = 100√58 ≈ 761.6 m.
Direction of displacement: tan⁻¹(300/700) = tan⁻¹(3/7) ≈ 23.2° East of North.
North and East directions are perpendicular (90°) — Pythagoras theorem applies.
Displacement is always ≤ distance; here displacement < distance because path is L-shaped.
Distance is scalar; displacement is vector (has direction).
Given: • The biker first travels 700 m North • Then travels 300 m East
Part 1 — Total Distance: Distance = sum of all path lengths Distance = 700 + 300 = 1000 m
Part 2 — Displacement (magnitude): The two paths are perpendicular (North ⊥ East). Using Pythagoras theorem: Displacement² = (700)² + (300)² = 490,000 + 90,000 = 580,000
Displacement = √580,000 = √(10,000 × 58) = 100√58 ≈ 100 × 7.616 ≈ 761.6 m
Part 3 — Direction of Displacement: The displacement vector points from Start to End. tan θ = (East component) / (North component) tan θ = 300 / 700 = 3/7 ≈ 0.4286 θ = tan⁻¹(3/7) ≈ 23.2°
Direction: 23.2° East of North (i.e., N 23.2° E)
Final Answer: • Distance = 1000 m • Displacement = 100√58 m ≈ 761.6 m at 23.2° East of North
Visualising the path:
(End point)
|
700m | (North)
|
(Start) ───────── (After going east) 300m (East)
Actually the correct diagram:
Start → 700m North → Turn right → 300m East → End point
The displacement is the straight diagonal from Start to End:
End point is: • 700 m North of start • 300 m East of start
The displacement vector (diagonal) has: • Magnitude = √(700² + 300²) = 100√58 ≈ 761.6 m • Direction = tan⁻¹(300/700) ≈ 23.2° from North towards East
Key concepts: • Distance: total path length (scalar) = 1000 m • Displacement: straight-line vector from initial to final position = 761.6 m at 23.2° E of N • Distance ≥ Displacement always • Displacement = Distance only when motion is in a straight line
Distance vs Displacement:
Distance: • Total path length traveled • Scalar (magnitude only, no direction) • Always positive • Here: 700 + 300 = 1000 m
Displacement: • Shortest straight-line path from initial to final position • Vector (has both magnitude and direction) • Can be less than distance, equal to it (straight-line motion), or zero (return to start) • Here: 100√58 ≈ 761.6 m, at 23.2° East of North
Why displacement < distance here: • The biker changes direction (turns from North to East) • The actual path is L-shaped • The shortest path (displacement) cuts across the corner
Pythagoras theorem application: Whenever two displacement vectors are perpendicular (90°), the resultant displacement is given by: R = √(A² + B²) This is directly from the Pythagorean theorem.
The displacement is 100√58 m ≈ 761.6 m at 23.2° East of North. Using Pythagoras theorem: displacement = √(700² + 300²) = √(490000 + 90000) = √580000 = 100√58 ≈ 761.6 m. The direction is tan⁻¹(300/700) = tan⁻¹(3/7) ≈ 23.2° East of North.
Total distance = 700 + 300 = 1000 m. Distance is the total path length regardless of direction. The displacement (straight-line from start to end) is different: ≈761.6 m at 23.2° East of North.
Because the biker changes direction — the path is L-shaped (first north, then east). The displacement (straight-line from start to finish) cuts across the corner and is shorter than the total L-shaped path. Displacement equals distance only when motion is in a straight line.
The direction angle θ from North = tan⁻¹(East component / North component) = tan⁻¹(300/700) = tan⁻¹(3/7) ≈ 23.2°. So the displacement is directed 23.2° East of North (written as N 23.2° E).
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