Kinematics Revision Notes — 9702 Paper 1

Distance, displacement, speed and velocity

Syllabus requirement · §2.1

“

define and use distance, displacement, speed, velocity and acceleration

”

Five words that carry the whole topic

Mechanics describes motion, and it needs five quantities to do it: distance, displacement, speed, velocity and acceleration. They come as two pairs plus one:

distance ↔ displacement,
speed ↔ velocity,
and acceleration, which is built from velocity.

In each pair one member is a scalar (size only) and the other a vector (size and direction). You met this split in the Physical Quantities and Units note; now we use it on motion. Paper 2 often opens with a one-mark "Define velocity" or "Define acceleration", marked against a fixed sentence. So we learn the definitions properly first, then calculate with them.

Distance versus displacement

Walk from your desk to the door by any route you like. The distance you travel is the total length of the path you actually took — every wobble counts. Your displacement is different: it is the straight-line arrow from start to finish, with its direction stated.

Distance: a scalar; the total path length travelled, however winding.
Displacement: a vector; the straight line from the starting point to the finishing point, in a stated direction.

Only if you walk a perfectly straight route do the two agree.

A walking path from start to finish: the distance is the length of the winding route; the displacement is the single straight arrow from start to finish.

Direction reverses the meaning

Walk $4\ \text{m}$ east, then $4\ \text{m}$ west. Distance travelled: $8\ \text{m}$ . Displacement: zero — you finished where you started, and displacement only depends on where you started and where you finished. Most of the traps in this section come from this one difference.

Speed versus velocity, average versus instantaneous

The same scalar/vector pairing repeats one level up:

Speed is distance travelled per unit time — a scalar.
Velocity is displacement per unit time — rate of change of displacement — a vector.

Each can be an average over a whole journey, or an instantaneous value — what a speedometer reads at one moment. The averages are:

\text{average speed} = \frac{\text{total distance travelled}}{\text{total time taken}} \qquad\qquad \text{average velocity} = \frac{\text{displacement}}{\text{time taken}}

Whenever the object reverses direction during the journey, the total distance and the displacement are different — so the average speed and the average velocity are different too. A runner completing one lap of a $400\ \text{m}$ track in $50\ \text{s}$ has an average speed of $8.0\ \text{m s}^{-1}$ but an average velocity of exactly zero, because her displacement (start to finish — same place) is zero.

Quantity	Definition (mark-scheme wording)	Unit	Scalar or vector?
distance	total path length travelled	$\text{m}$	scalar
displacement	distance travelled in a stated direction (from start to finish)	$\text{m}$	vector
speed	distance travelled per unit time	$\text{m s}^{-1}$	scalar
velocity	rate of change of displacement	$\text{m s}^{-1}$	vector
acceleration	rate of change of velocity	$\text{m s}^{-2}$	vector

Learn this table verbatim — the middle column is marked word-for-word. Every vector definition must name its vector partner: velocity is defined through displacement, never speed.

Acceleration: the rate of change of velocity

\text{acceleration} = \frac{\text{change in velocity}}{\text{time taken}} = \frac{\Delta v}{\Delta t}

Its unit follows from the definition: $\text{m s}^{-1}$ divided by $\text{s}$ gives $\text{m s}^{-2}$ — read aloud as "metres per second, per second". An acceleration of $2\ \text{m s}^{-2}$ means the velocity changes by $2\ \text{m s}^{-1}$ every second.

Because velocity is a vector, acceleration means any change in velocity — speeding up, slowing down, or changing direction. That last case feels strange but matters later: a ball at the top of its flight has zero vertical velocity yet still accelerates downwards at $9.81\ \text{m s}^{-2}$ , because its velocity is changing all the same. A negative acceleration (in whatever direction you have called positive) simply means the velocity is decreasing in that direction — the everyday word for that is deceleration, but in calculations we keep the minus sign.

Earning a definition mark

1
Write "[quantity] = rate of change of [its vector partner]". Velocity → rate of change of displacement. Acceleration → rate of change of velocity.
This sentence pattern is exactly what the B1 mark scheme line says. Learn it as a fixed phrase, not something to improvise.
2
Check that no scalar word slipped in. If your velocity definition contains the word "speed", or your acceleration definition contains "speed" where "velocity" belongs, the mark is lost.
Mark schemes explicitly distinguish velocity from speed in definitions — 'rate of change of speed' scores zero.
3
Alternative form: a quotient earns equal credit — "change in displacement ÷ time taken" is accepted for velocity.
Both published forms appear in mark schemes; pick whichever you will remember under pressure.

A clean demonstration

A delivery cyclist rides $300\ \text{m}$ east along a street, then turns round and rides $100\ \text{m}$ back west, taking $80\ \text{s}$ in total.

Total distance: the path length adds both legs:

d = 300 + 100 = 400\ \text{m}

Displacement: east is positive, so the second leg carries a minus sign:

s = (+300) + (-100) = +200\ \text{m},\ \text{i.e.}\ 200\ \text{m east of the start}

Average speed: total distance over total time:

\frac{400}{80} = 5.0\ \text{m s}^{-1}

Average velocity: displacement over time:

\frac{200}{80} = 2.5\ \text{m s}^{-1}\ \text{east}

The average velocity is only half the average speed, because the ride back west cancelled part of the displacement while still adding to the distance. Now an acceleration: if the cyclist speeds up uniformly from rest to $6.0\ \text{m s}^{-1}$ in $12\ \text{s}$ ,

a = \frac{\Delta v}{\Delta t} = \frac{6.0 - 0}{12} = 0.50\ \text{m s}^{-2}

and if she later brakes from $6.0\ \text{m s}^{-1}$ to rest in $4.0\ \text{s}$ , still facing east,

a = \frac{0 - 6.0}{4.0} = -1.5\ \text{m s}^{-2}

The minus sign is the answer doing its job: the velocity decreases along the positive (eastward) axis.

Define acceleration

9702/21 O/N 2025 Q1(a)1 mark

Define acceleration.

Show full working

1
Apply the pattern: acceleration is the rate of change of its own vector partner.
$\text{acceleration} = \text{the rate of change of velocity}$
The mark scheme pays B1 for exactly this phrase. Nothing more is needed — no formula, no unit, no example.
2
Self-check before moving on: does the sentence contain "speed"? No — velocity only ✓.
"Rate of change of speed" is the classic near-miss: it names the scalar, so the mark scheme refuses it.

Answer

Acceleration is the rate of change of velocity.

One-mark definitions are free marks if the sentence is learned verbatim. Never spend them back by paraphrasing.

Define velocity

9702/23 M/J 2025 Q1(a)1 mark

Define velocity.

Show full working

1
Velocity is defined through displacement:
$\text{velocity} = \text{the rate of change of displacement}$
Mark scheme: 'rate of change of displacement' — B1. The vector partner (displacement), not the scalar (distance).
2
An equally credited alternative spells out the quotient: velocity = change in displacement divided by the time taken.
Published mark schemes accept both phrasings — memorise whichever reads more naturally to you, but never mix in 'distance' or 'speed'.

Answer

Velocity is the rate of change of displacement (equivalently: change in displacement per unit time).

Velocity ↔ displacement, acceleration ↔ velocity: each definition climbs one rung up the vector ladder.

Common mistakes

"Velocity is the rate of change of speed."
Velocity is the rate of change of displacement.
Mark schemes explicitly distinguish velocity from speed in definitions. Naming the scalar loses the mark even though the sentence sounds right.
"Acceleration is the rate of change of distance."
Acceleration is the rate of change of velocity.
Each quantity is defined through its own vector partner — acceleration sits above velocity, not above distance.
Average velocity = total distance ÷ total time.
Average velocity = displacement ÷ time taken; total distance ÷ total time is average speed.
The pair diverge the moment direction reverses — an out-and-back trip has large average speed but small (even zero) average velocity.
"That car is accelerating, so its speed must be increasing."
Acceleration is a change of velocity, which can be speeding up, slowing down, or turning.
Slowing is acceleration too (negative along the chosen axis); so is circular motion at constant speed, because the direction changes.

Your turn

A definition drill, then two distance-versus-displacement calculations.

1
(a) State which quantity is defined as "the rate of change of displacement". (b) A student writes: "Acceleration is how much the speed changes." Explain why this does not earn the mark, and write a correct definition.
Stuck? Show hint
For (b): two separate defects — the wrong vector partner, and no sense of a rate.
Show solution
1. 1
  (a) Rate of change of displacement is the definition of velocity.
  Displacement is the vector partner of velocity, one rung below it.
2. 2
  (b) The student's sentence fails twice: it defines acceleration through speed (the scalar) instead of velocity, and "how much" gives a change, not a rate (no time involved).
  Spotting both defects is the habit that keeps your own definitions clean: right partner, and a rate.
3. 3
  Correct definition: acceleration is the rate of change of velocity.
  "Change in velocity ÷ time taken" earns the same mark; both name velocity and include time.
Answer
(a) Velocity. (b) It uses speed rather than velocity, and omits any timescale; acceleration is the rate of change of velocity.
2
A hiker walks $3.0\ \text{m}$ due east, then $4.0\ \text{m}$ due north. Calculate (a) the distance walked, (b) the magnitude and direction of the hiker's displacement.
Stuck? Show hint
East and north are perpendicular — Pythagoras for the size of the displacement, tangent for its direction.
Show solution
1. 1
  (a) Distance is the total path length — the two legs simply add, since it is a scalar:
  $d = 3.0 + 4.0 = 7.0\ \text{m}$
  Direction does not matter for distance: every metre walked counts.
2. 2
  (b) Displacement is the single straight arrow from start to finish. The legs are perpendicular, so its magnitude comes from Pythagoras:
  $|\vec{s}| = \sqrt{3.0^2 + 4.0^2}$
  The two legs and the displacement form a right-angled triangle, with the displacement as the hypotenuse.
3. 3
  Evaluate: $|\vec{s}| = \sqrt{9.0 + 16} = \sqrt{25} = 5.0\ \text{m}$
  The 3–4–5 triangle again — spotting it saves the square-rooting entirely.
4. 4
  Direction — measure the angle $\theta$ east of north (the adjacent side is the northerly $4.0$ ):
  $\tan\theta = \frac{3.0}{4.0} = 0.75 \quad\Longrightarrow\quad \theta = 37^\circ$
  State the reference direction: 'at 37° east of north' is complete; a bare angle is not.
Answer
(a) $7.0\ \text{m}$ . (b) Displacement $5.0\ \text{m}$ at $37^\circ$ east of north.
3
A tram travels from stop A to stop B, $600\ \text{m}$ apart in a straight line, in $60\ \text{s}$ . It waits at B for $60\ \text{s}$ , then returns to A, taking $75\ \text{s}$ . Calculate (a) the average speed and (b) the average velocity for the whole outing.
Stuck? Show hint
(b): where does the tram end up relative to where it started?
Show solution
1. 1
  Total distance, adding the two legs: $600 + 600 = 1200\ \text{m}$
  Distance is a scalar, so the return leg adds on; it does not cancel.
2. 2
  Total time, including the wait at B: $60 + 60 + 75 = 195\ \text{s}$
  The waiting time is still part of the time taken. Leaving it out is the usual slip here.
3. 3
  (a) Average speed = total distance ÷ total time: $\frac{1200}{195} = 6.15 \approx 6.2\ \text{m s}^{-1}$
  Round only at the end, to 2 s.f. to match the data.
4. 4
  (b) The tram finishes where it began, so its total displacement is zero — hence
  $\text{average velocity} = \frac{0}{195} = 0$
  Zero average velocity next to a clearly non-zero average speed: the two quantities really are different.
Answer
(a) $6.2\ \text{m s}^{-1}$ . (b) Zero — the displacement for the return trip cancels the outward displacement.

Paper 1 · MCQReal past-paper questions · Distance, displacement, speed, velocity, acceleration Paper 2 · TheoryReal past-paper questions · Distance, displacement, speed, velocity, acceleration

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Can you do all of these?

Define velocity and acceleration using the vector words — 'rate of change of displacement', 'rate of change of velocity' — with no 'speed' or 'distance' in sight
Tell distance from displacement and average speed from average velocity, especially when the motion reverses
Read gradients: s–t gradient = velocity, v–t gradient = acceleration; draw a long tangent for the value at one instant on a curve
Read areas: displacement from the area under a v–t graph, counting regions below the axis as negative; change in velocity from the area under an a–t graph
Sketch motion graphs deliberately: shape, starting value, end value, zero-crossings and labels each earn their own mark
Derive the four suvat equations from the v–t graph; list s, u, v, a, t and pick the equation that omits the unwanted quantity; convert to SI units first
Choose one positive direction for vertical motion, substitute signs consistently, and read a negative answer as motion in the other direction
Use the up-and-over symmetries of free fall: equal speeds at equal heights, equal times up and down
Describe the electromagnet–trapdoor measurement of g (including the s against t² graph), correct its ball-diameter systematic error, and combine uncertainties with Δg/g = Δs/s + 2Δt/t
Resolve a projectile launch into u cos θ (horizontal, constant) and u sin θ (vertical, suvat with a = −g); find the time from whichever list has enough data; combine components with Pythagoras and tan for the landing velocity