Study Guides/Physics/A Biker Rides 700m North 300m East — Displacement
Study Guide · Physics

A Biker Rides 700m North then 300m East — Find Displacement

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.

Question (Click to Flip)

A biker rides 700m north and then 300m east. What is the displacement?

Answer

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.

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Key Facts

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).

Solution — Distance vs Displacement

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

Vector Diagram and Explanation

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

Key Concepts — Distance vs Displacement

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.

Questions and Answers

A biker rides 700m north and then 300m east. What is the displacement?+

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.

What is the total distance traveled by the biker who rides 700m north and 300m east?+

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.

Why is displacement less than distance in this problem?+

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.

How do you find the direction of displacement for 700m north and 300m east?+

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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