Describing motion: distance, displacement, speed, velocity, acceleration

What you need to be able to do

The five quantities, in pairs

Kinematics describes how things move, without asking why. Everything is built from five quantities. They come in pairs: a scalar that only says how much, and a vector that also says which way.

Distance and displacement

Run once round a 400 m track and stop where you began: the distance is 400 m, but the displacement is 0. Walk 3 m east and then 4 m north: the distance is 7 m, the displacement is 5 m at 53° north of east (a 3-4-5 triangle). Displacement can also be negative: if east is positive, 20 m west is a displacement of −20 m.

Speed and velocity

Acceleration

• Unit: m s⁻¹ per second = m s⁻².
• Acceleration is a vector. A negative acceleration (a deceleration) points against the velocity: the object slows down. Acceleration and velocity do not have to point the same way (November 2023 report).
• Velocity can change in size or in direction. A car going round a bend at a steady 20 m s⁻¹ is accelerating, because its direction is changing.

Change in velocity: a vector subtraction

Change in velocity = final velocity − initial velocity: \(\Delta\mathbf{v} = \mathbf{v} - \mathbf{u}\). Subtract by adding the reversed initial velocity (from Subtracting vectors).

1. In a straight line, choose a positive direction. A ball hits a wall at 8 m s⁻¹ to the right and bounces back at 3 m s⁻¹ to the left. Right = +: \(\Delta v = (-3) - (+8) = -11\) m s⁻¹, i.e. 11 m s⁻¹ to the left 9702/12/M/J/24 Q9. Not 5: the direction reversed.
2. At an angle, draw \(\mathbf{v}\) then \(-\mathbf{u}\) tip to tail. 4.0 m s⁻¹ right changing to 3.0 m s⁻¹ down: \(\Delta\mathbf{v}\) = 3.0 down + 4.0 left = 5.0 m s⁻¹, pointing down and to the left.

Try it: distance vs displacement

Mark-scheme focus

Define velocity

Credited

"rate of change of displacement" B1, or "change in displacement / time (taken)" B1

Not credited

• "rate of change of distance"
• "speed in a particular direction"
• "change in displacement by time" (by = divided or multiplied?)
• "change in displacement in a unit time"

Why

Distance has no direction, so its rate of change is speed. A definition must be in terms of other defined quantities, which "speed in a direction" is not. "By" and "in a unit time" don't say what is divided by what.

Also accepted

Both wordings in the Credited box get the mark. Learn one of them word for word.

From: 9702/23/M/J/25 Q1(a), 9702/21/M/J/24 Q2(a) and the November 2021 and March 2017 examiner reports.

Define acceleration

Credited

"rate of change of velocity" B1, or "change in velocity / time (taken)" B1

Not credited

• "velocity divided by time"
• "rate of change of velocity per unit time"
• "rate of change of speed"
• "force per unit mass"

Why

Velocity ÷ time misses the change: an object at a steady 20 m s⁻¹ has no acceleration. "Rate of change" already means "per unit time", so adding "per unit time" says it twice. Speed has no direction, and a = F/m is Newton's second law, not the definition.

Also accepted

"Change in velocity per unit time" (the correct option in 9702/12/F/M/24 Q7, where "rate of change of speed" and "force per unit mass" were the wrong options).

From: 9702/21/M/J/25 Q1(a), 9702/22/F/M/16 Q2(a), 9702/21/O/N/12 Q1(a) and the November 2018, June 2017 and March 2016 examiner reports.

Define displacement

Credited

"distance (from the point) in a straight line in a given direction" B1

Not credited

• "the distance between two points"
• "the product of velocity and time"

Why

"Distance between two points" has no direction, so it is still a scalar. "Velocity × time" is a way to calculate it (for constant velocity only), not what it is.

Remember

Both parts score together: straight line (so it is not the path) and direction (so it is a vector).

From: 9702/23/M/J/24 Q2(a) and the November 2018 examiner report 9702/21/O/N/18 Q1(a)(i).

Command words you meet here

Worked examples

Example 1 · Speed or velocity round a curve (Paper 1, 1 mark) ★★ Medium

Why it trips people up: the June 2022 report says many chose D. D is the average speed; the question asks for the average velocity.

1 Read it with me what is actually happening?

A car drives three-quarters of the way round a circular road, from P to Q, in 4.0 s. The question wants its average VELOCITY - not its average speed.

• average velocitydisplacement ÷ time: the straight line from P to Q
• magnitudesize only
• anticlockwisetells you the path (¾ of the circle) - but only the start and end points matter for velocity
• radius 12 mP and Q are both 12 m from the centre

2 What is it asking for? the target, part by part

• size of the average velocity P → Qm s⁻¹ · MCQ

3 What do we know? pull every number out, in SI units

• r = 12 mcircle
• t = 4.0 sP to Q
• 90°angle between OP and OQ (from the figure)

4 Plan the route work backwards from what we want

FindPhysics ideaAlready have itWork out firstTrap

• Findaverage velocity
  • Usedisplacement ÷ time
    • Work out firstdisplacement PQ: Pythagoras on two 12 m radii at 90°
    • Trapthe arc length gives average SPEED (option D)

Does it matter which way round the car went?

Not for velocity. Displacement depends only on where it started and finished - the straight line P to Q.

Which should be bigger: average speed or average velocity?

Average speed. The arc is always at least as long as the straight line.

5 Now solve it have a go on paper first, then open the worked solution

1. Anticlockwise from P to Q is three-quarters of the circle. That is only needed for the distance.
2. Displacement: P and Q are both 12 m from the centre, at right angles, so \(PQ = \sqrt{12^2 + 12^2} = 17.0\) m.
3. Average velocity = displacement ÷ time = 17.0 / 4.0 = 4.2 m s⁻¹ → A 1

D comes from the arc: distance = ¾ × 2π × 12 = 56.5 m, and 56.5 / 4.0 = 14 m s⁻¹ is the average speed.

The straight line is shorter than the arc, so the average velocity must be smaller than the average speed: 4.2 < 14.

Example 2 · Half a lap and a whole lap (Paper 1, 1 mark) ★★ Medium

1 Read it with me what is actually happening?

A toy car goes round a circular track at a steady 0.50 m s⁻¹; one lap takes 40 s. We want the average velocity after half a lap (20 s) and after a whole lap (40 s).

• constant speedthe SPEED never changes - but the velocity's direction does
• average velocitydisplacement ÷ time
• once around the trackafter 40 s it is back where it started: displacement zero

2 What is it asking for? the target, part by part

• v₂₀ and v₄₀m s⁻¹ · MCQ

3 What do we know? pull every number out, in SI units

• v = 0.50 m s⁻¹speed
• T = 40 sone lap

• circumference = 0.50 × 40 = 20 m, so the radius is 20 ÷ 2π

4 Plan the route work backwards from what we want

FindPhysics ideaAlready have itWork out firstTrap

• Findv₂₀
  • Usedisplacement after half a lap = the diameter
    • Work out firstradius from the circumference
• Findv₄₀
  • Useback at the start: displacement 0 → v₄₀ = 0

After half a lap, how far is the car from where it started, in a straight line?

Right across the circle: one diameter.

Can an object moving the whole time have zero average velocity?

Yes - if it ends where it started. Its average SPEED is still 0.50 m s⁻¹.

5 Now solve it have a go on paper first, then open the worked solution

1. Circumference = speed × time for one lap = 0.50 × 40 = 20 m, so the radius is \(20 / 2\pi = 3.18\) m.
2. After 20 s the car is half way round: displacement = the diameter = 6.37 m. \(v_{20} = 6.37 / 20 = 0.32\) m s⁻¹.
3. After 40 s it is back at the start: displacement = 0, so \(v_{40} = 0\) → A 1

The November 2022 report: many candidates worked out the average speed after 20 s and 40 s instead of the average velocity. The speed is 0.50 m s⁻¹ the whole time (option D); the average velocity is not.

6 Check it does the answer make sense?

v₂₀ (0.32) is less than the speed (0.50), since the diameter is shorter than the half-circle. v₄₀ = 0 because the displacement is zero.

Example 3 · Change in velocity when the direction changes (Paper 1, 1 mark) ★★★ Hard

Why it is hard: the June 2022 report says over half of the candidates chose A, the sum of the two velocities. The question asks for the change.

1 Read it with me what is actually happening?

An object starts moving right at 4.0 m s⁻¹ and ends moving down at 3.0 m s⁻¹. Which arrow shows how much its velocity CHANGED?

• change in velocityfinal − initial: Δv = v − u, a vector subtraction
• which arrowonly the DIRECTION matters here

2 What is it asking for? the target, part by part

• the arrow for ΔvMCQ · 1 mark

3 What do we know? pull every number out, in SI units

• u = 4.0 m s⁻¹right
• v = 3.0 m s⁻¹down

4 Plan the route work backwards from what we want

FindPhysics ideaAlready have itWork out firstTrap

• FindΔv
  • UseΔv = v + (−u)
    • Step 1reverse u: 4.0 to the LEFT
    • Step 2add tip to tail: 3.0 down, then 4.0 left
    • Trapu + v (down-right) is the sum, not the change

How do you subtract a vector?

Add its reverse. v − u is v plus an arrow the same size as u pointing the opposite way.

Starting right and ending down, which way must the 'push' have been?

It had to cancel the rightward motion (push left) and create downward motion (push down): down and to the left.

5 Now solve it have a go on paper first, then open the worked solution

1. Change in velocity = final − initial: \(\Delta\mathbf{v} = \mathbf{v} - \mathbf{u} = \mathbf{v} + (-\mathbf{u})\).
2. Reverse the initial velocity: −u = 4.0 m s⁻¹ to the left.
3. Add tip to tail: 3.0 down, then 4.0 left. The result points down and to the left (5.0 m s⁻¹) → B 1

Option	Direction	What it is
A	down-right	u + v: the sum, not the change. The most popular wrong answer.
B	down-left	v − u ✓
C	up-right	u − v: initial minus final, the wrong way round.
D	up-left	−(u + v): the sum reversed.

6 Check it does the answer make sense?

Δv must point partly left - it has to undo the rightward velocity. Only B and D do, and it must point down too, to create the downward velocity: B.

Example 4 · Average speed and average velocity on two legs (Paper 2, 4 marks) ★★ Medium

1 Read it with me what is actually happening?

A girl runs north, then east. We sketch her path and find her average speed (whole path) and average velocity (straight line from start to finish, with its direction).

• Sketch a vector diagramarrows tip to tail, then R from start to finish
• average speedtotal DISTANCE ÷ total time
• average velocity … and its angleDISPLACEMENT ÷ time, plus a direction measured from north

2 What is it asking for? the target, part by part

• (i)vector diagram with Rsketch · 1 mark
• (ii)1average speedm s⁻¹ · 1 mark
• (ii)2average velocity: size and anglem s⁻¹ + ° · 3 marks

3 What do we know? pull every number out, in SI units

• 120 m northin 15 s
• 80 m eastin 12 s

• total time is 27 s for both averages

4 Plan the route work backwards from what we want

FindPhysics ideaAlready have itWork out firstTrap

• (ii)1 Findaverage speed
  • Use200 m ÷ 27 s
    • Trapnot the mean of the two speeds
• (ii)2 Findaverage velocity
  • UseR ÷ 27 s
    • R√(120² + 80²)
    • Angletan⁻¹(80 ÷ 120) from north

Why can't we average the two speeds (8.0 and 6.7)?

She spends different times on each leg. Average speed must be total distance over total time.

For the angle 'with respect to the initial path', which side is opposite the angle?

The initial path is north (120 m). The angle from north has 80 m (east) opposite it: tan θ = 80 ÷ 120.

5 Now solve it have a go on paper first, then open the worked solution

1. (i) Draw 120 m north, then 80 m east from its tip, both with arrows; R joins the start to the finish, labelled R. B1
2. (ii) 1. Total distance = 200 m, total time = 15 + 12 = 27 s. Average speed = 200 / 27 = 7.4 m s⁻¹ A1 Not (8.0 + 6.7)/2: the June 2016 report says that method gets no credit.
3. (ii) 2. Displacement \(R = \sqrt{120^2 + 80^2} = 144\) m C1
4. Average velocity = 144 / 27 = 5.3 m s⁻¹ A1
5. Angle from the initial path (north): \(\tan^{-1}(80/120) =\) 34° (east of north) A1

average speed = 200 / 27 = 7.4 m s⁻¹; R = √(120² + 80²) = 144 m; v = 144 / 27 = 5.3 m s⁻¹ at 34° to the initial direction (east of north)

5.3 < 7.4: average velocity can never be bigger than average speed.

Summary

• Distance and speed are scalars; displacement, velocity and acceleration are vectors.
• Velocity = rate of change of displacement. Acceleration = rate of change of velocity = change in velocity ÷ time taken.
• Average speed = total distance ÷ total time. Average velocity = displacement ÷ total time, with a direction.
• Change in velocity = final − initial, done as vectors.

Can I do this?

Quiz