AshPhys

Distance and displacement

AshPhys · Revision cheat sheet
AS Level9702 AS Ch.1Chapter 1: KinematicsLesson 1.2Sections 1.2

Goal: Tell distance from displacement and scalars from vectors, and find a displacement from two perpendicular moves.

01The big picture

Distance is the total length of the path travelled: a scalar. Displacement is the straight-line distance from the start to the finish in a stated direction: a vector.

Scalars have size only (distance, speed, time, mass, energy). Vectors have size and direction (displacement, velocity, acceleration, force), so a vector answer needs a direction.

Displacement [paraphrase]
The distance moved in a particular direction from a starting point.
Scalar and vector [paraphrase]
A scalar has magnitude only; a vector has magnitude and direction.
Fig. 1 · Distance or displacement
Distance or displacementA path 3.0 km east then 4.0 km north drawn in grey, and a black arrow from the start to the finish, the displacement, at angle theta north of east.3.0 km4.0 kms = 5.0 kmθstartN
The path (grey) is 7.0 km long; the displacement (black) is the 5.0 km straight line from start to finish.

02Key equations

Two perpendicular moves
s=x2+y2s = \sqrt{x^{2} + y^{2}}
Direction
tan⁡θ=yx\tan\theta = \frac{y}{x}
Symbols and units
  • sssize of the displacementm\mathrm{m}
  • x, yx,\ yperpendicular parts of the journeym\mathrm{m}
  • θ\thetaangle from the x direction∘^\circ

03Sample problems with solutions

Sample 1 · A walk east then north
A walker goes 3.0 km east and then 4.0 km north. Calculate the distance walked and the displacement.

  1. Distance adds the path lengths:

    d=3.0+4.0=7.0 kmd = 3.0 + 4.0 = 7.0\ \mathrm{km}

  2. Displacement is the straight line from start to finish:

    s=3.02+4.02=5.0 kms = \sqrt{3.0^{2} + 4.0^{2}} = 5.0\ \mathrm{km}

  3. Direction:

    θ=tan⁡−14.03.0=53∘ north of east\theta = \tan^{-1}\frac{4.0}{3.0} = 53^\circ \text{ north of east}
Answer: distance 7.0 km; displacement 5.0 km at 53° north of east

04Practice problems

  1. P1
    A runner completes one lap of a 400 m track. State the distance and the displacement.
    Final answer400 m; zero
    Answer: 400 m; zero
  2. P2
    Sort into scalars and vectors: mass, weight, time, velocity.
    Final answerscalars: mass, time; vectors: weight, velocity
    Answer: scalars: mass, time; vectors: weight, velocity
  3. P3
    A walker goes 6.0 m north and then 8.0 m west. Calculate the displacement.
    Final answer10 m at 53° west of north
    Answer: 10 m at 53° west of north
  4. P4
    A car drives 5.0 km east, then 2.0 km west. State the distance and the displacement.
    Final answer7.0 km; 3.0 km east
    Answer: 7.0 km; 3.0 km east

05Other useful information

Common mistakes

  • ✗ Wrong: Displacement can never be zero if you moved.

    ✓ Right: Back at the start, the displacement is zero; the distance is not.

Exam tips

  • Sketch the path; draw the displacement from start to end and give its direction.

What the syllabus asks

  • 9702 topics 1 and 2: scalars and vectors; define and use displacement. Coursebook 1.2.

Explore: Practice questions for this lesson