Kinematics of Motion in a Straight Line Revision Notes — 9709 Paper 4

topic	marks/paper
Kinematics of Motion in a Straight Line	20.9
Forces and Equilibrium	13.8
Newton's Laws of Motion	13.2
Energy, Work and Power	12.9
Momentum	4.6

sub-topic	parts	marks	sections
Constant acceleration formulae (suvat)	102	359	05–08
Differentiation and integration (variable acceleration)	90	349	09–12
Displacement-time and velocity-time graphs	47	135	02–04
Scalars and vectors (distance, speed, displacement, …)	30	114	01, 11

Distance, displacement, speed, velocity and acceleration

Syllabus requirement · §4.2

“

understand the concepts of distance and speed as scalar quantities, and of displacement, velocity and acceleration as vector quantities (restricted to motion in one dimension only; the term 'deceleration' may sometimes be used in the context of decreasing speed).

”

Every question in this topic has the same set-up: a particle $P$ moves along a straight line, and there is a fixed point $O$ on that line to measure from. Before anything else, you choose which way along the line counts as positive — usually the direction the particle first moves in. After that, "which way" is carried entirely by a sign: $+$ means the positive direction, $-$ means the other way.

Five quantities describe the motion. Two of them only ever have a size (scalars); three have a size and a direction, shown by the sign (vectors):

Displacement $s$ (metres, m) — where the particle is: its position measured from $O$ along the line. $s = -4$ means 4 m from $O$ on the negative side.
Distance (m) — how much ground the particle has covered in total. It only ever grows, and it is never negative.
Velocity $v$ (metres per second, $\text{m s}^{-1}$ ) — how fast the displacement is changing, with its sign. $v = -3$ means moving at 3 m each second in the negative direction.
Speed ( $\text{m s}^{-1}$ ) — the size of the velocity, ignoring the sign: speed $= |v|$ .
Acceleration $a$ (metres per second per second, $\text{m s}^{-2}$ ) — how fast the velocity is changing, with its sign. $a = 2$ means the velocity increases by $2\text{ m s}^{-1}$ every second.

Quantity	Scalar / vector	What it measures	Can it be negative?
Distance	scalar	total length of path actually travelled	no — always $\geqslant 0$
Displacement $s$	vector	position relative to the fixed point $O$	yes — the sign says which side of $O$
Speed	scalar	how fast, whatever the direction: $\|v\|$	no — always $\geqslant 0$
Velocity $v$	vector	rate of change of displacement	yes — the sign says which way it is moving
Acceleration $a$	vector	rate of change of velocity	yes — the sign says which way the velocity is changing

The right-hand column is the whole distinction. A scalar answer with a minus sign in front of it means a scalar/vector mix-up has happened somewhere upstream.

A particle walks 8 m in the positive direction, then 3 m back: total distance 11 m, but displacement only +5 m.

Averages. Because distance and displacement differ, so do the two averages built from them:

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

Paper 4 asks for average speed far more often (it appeared in several 2021–2025 papers); average velocity is worth knowing mainly so you never use it by mistake. The two have the same size only if the particle never turns back.

Distance, displacement, average speed and average velocity

A particle $P$ moves along a straight line. Taking the direction of its initial motion as positive, $P$ travels $12\text{ m}$ in $4\text{ s}$ , then reverses direction and travels back $5\text{ m}$ in $3\text{ s}$ . Find (a) the total distance travelled, (b) the final displacement from the starting point, (c) the average speed, (d) the average velocity, for the whole $7\text{ s}$ of motion.

Show full working

1
Fix the positive direction. The question tells us: the direction of the initial motion is positive. So the first leg is $+12$ and the return leg is $-5$ .
Write the convention down before any number — it is what turns the words 'travels back 5 m' into the signed number −5.
2
(a) Distance is the total length of path covered, regardless of direction, so the two legs simply add: $\text{distance} = 12 + 5 = 17\text{ m}$
Distance never subtracts a 'backward' leg — it only ever accumulates.
3
(b) Displacement is signed, so the legs are added with their signs: $\text{displacement} = (+12) + (-5) = 7\text{ m}$
Displacement answers 'where is it now, relative to where it started?' — 7 m on the positive side.
4
(c) Average speed uses the scalar distance, divided by the total time: $\text{average speed} = \frac{17}{7} = 2.43\text{ m s}^{-1} \ (\text{3 s.f.})$
Average speed can never come out negative, because both distance and time are positive.
5
(d) Average velocity uses the signed displacement, divided by the same total time: $\text{average velocity} = \frac{7}{7} = 1\text{ m s}^{-1}$
Same time as (c), different numerator. If (c) and (d) had come out equal, that would only be because the particle never reversed.

Answer

Distance $= 17\text{ m}$ ; displacement $= +7\text{ m}$ ; average speed $\approx 2.43\text{ m s}^{-1}$ ; average velocity $= 1\text{ m s}^{-1}$ .

A particle that ends where it started has displacement 0 and average velocity 0 — but its distance and average speed are not zero. Examiners like exactly that contrast.

Constant speed. When a particle moves at a constant speed, the distance it covers grows at the same rate every second, so

\text{distance} = \text{speed} \times \text{time}, \qquad \text{time} = \frac{\text{distance}}{\text{speed}}, \qquad \text{speed} = \frac{\text{distance}}{\text{time}}

This is the simplest motion there is, but it turns up constantly: a stage of a journey "at constant speed", a particle sliding on a smooth horizontal plane after a collision, and every catching-up problem in §08.

A journey in three legs

A runner runs $400\text{ m}$ along a straight path at a constant $5\text{ m s}^{-1}$ , stops for $20\text{ s}$ , then runs $200\text{ m}$ back towards the start at a constant $4\text{ m s}^{-1}$ . Find the average speed and the average velocity for the whole journey.

Show full working

1
Time for the first leg, using time $=$ distance ÷ speed: $\frac{400}{5} = 80\text{ s}$
Each leg is at constant speed, so each gets its own time from the same rule.
2
Time for the rest: given as $20\text{ s}$ (no distance is covered).
Rest still counts in the total time — leaving it out is the most common slip in an average-speed question.
3
Time for the return leg: $\frac{200}{4} = 50\text{ s}$
Speed is always positive, so time comes out positive even though the runner is going backwards.
4
Total time and total distance: $80 + 20 + 50 = 150\text{ s}, \qquad 400 + 200 = 600\text{ m}$
Distance adds the legs as positive lengths — the direction of the return leg is irrelevant here.
5
Average speed: $\frac{600}{150} = 4\text{ m s}^{-1}$
Total distance over total time — not the average of 5 and 4, which would ignore how long was spent at each speed and the rest.
6
Displacement, with the outward direction positive: $(+400) + (-200) = 200\text{ m}$
The runner finishes 200 m from the start, on the outward side.
7
Average velocity: $\frac{200}{150} = 1.33\text{ m s}^{-1} \ (\text{3 s.f.}) \text{ in the outward direction}$
Much smaller than the average speed, because half of the outward distance was undone by the return leg.

Answer

Average speed $= 4\text{ m s}^{-1}$ ; average velocity $\approx 1.33\text{ m s}^{-1}$ in the outward direction.

Constant speed on a real paper

9709/42 M/J 2025 Q1(a)1 mark

A crate is being pushed in a straight line along a horizontal surface by a force of magnitude $25\text{ N}$ inclined at $20^\circ$ above the horizontal. The crate moves a distance of $12\text{ m}$ in $8$ seconds with constant speed.

Find the constant speed of the crate.

Show full working

1
Pick out what this part needs. Constant speed, distance $12\text{ m}$ , time $8\text{ s}$ . The force and its angle are for the later parts (work done and power) and play no role here.
Mechanics stems carry information for every part at once. Deciding which numbers a part actually needs is a skill in itself.
2
Use speed $=$ distance ÷ time: $\text{speed} = \frac{12}{8} = 1.5\text{ m s}^{-1}$
Constant speed is exactly the situation where this formula is valid — for changing speed it would only give the average.

Answer

$1.5\text{ m s}^{-1}$ .

"Deceleration" does not always mean "negative acceleration"

The syllabus allows deceleration to mean decreasing speed. Whether that makes $a$ positive or negative depends on which way the particle is moving:

moving in the positive direction and slowing down: $v > 0$ is shrinking towards 0, so $a < 0$ ;
moving in the negative direction and slowing down: $v < 0$ is climbing back up towards 0, so $a > 0$ .

The rule that always works: a particle is slowing down when $v$ and $a$ have opposite signs, and speeding up when they have the same sign. A "deceleration of $2\text{ m s}^{-2}$ " is a positive size; turn it into a signed acceleration only after you know the direction of motion.

Slowing down means the acceleration points against the motion. Moving right and slowing: a is negative. Moving left and slowing: a is positive.

Reading the sign of an acceleration

A particle moves along a straight line, with the positive direction to the right. At $t = 0$ its velocity is $-8\text{ m s}^{-1}$ and at $t = 3$ its velocity is $-2\text{ m s}^{-1}$ , the velocity changing at a constant rate. Find its acceleration, and state whether the particle is speeding up or slowing down.

Show full working

1
Acceleration is the change in velocity per second. With the velocity changing at a constant rate: $a = \frac{\text{change in velocity}}{\text{time taken}} = \frac{v_{\text{end}} - v_{\text{start}}}{3}$
This is just the definition of acceleration from the list above: how many m s⁻¹ the velocity changes by each second.
2
Substitute the signed velocities, end minus start: $a = \frac{(-2) - (-8)}{3} = \frac{6}{3}$
Subtracting a negative is where a sign goes wrong — (−2) − (−8) = −2 + 8 = +6.
3
Evaluate: $a = +2\text{ m s}^{-2}$
The acceleration is positive even though the particle is moving in the negative direction the whole time.
4
Speeding up or slowing down? The speed goes from $|-8| = 8$ to $|-2| = 2$ , so the particle is slowing down. This matches the sign rule: $v < 0$ and $a > 0$ have opposite signs.
So the particle has a deceleration of 2 m s⁻² — and a signed acceleration of +2 m s⁻². Both statements describe the same motion.

Answer

$a = +2\text{ m s}^{-2}$ ; the particle is slowing down (a deceleration of $2\text{ m s}^{-2}$ ).

Common mistakes

Computing "average velocity" or "average speed" as the mean of the speeds in each stage
Average speed $=$ total distance ÷ total time, always
Stages last different lengths of time (and rests count in the time), so a simple mean of the speeds is almost never right.
Treating "decelerating" as automatically meaning " $a$ is negative"
Deceleration means $a$ acts against the current direction of motion — $v$ and $a$ have opposite signs
A particle moving in the negative direction that is decelerating has a positive acceleration.
Giving a speed as a negative number, e.g. "speed $= -5.07\text{ m s}^{-1}$ "
Speed $= |v|$ : if $v = -5.07$ , the speed is $5.07\text{ m s}^{-1}$
Mark schemes routinely say 'must be positive' on a speed answer; a negative value is the velocity, not the speed.
Assigning signs without first stating a positive direction
Make "positive direction $= \dots$ " the first line of working, and keep it for the whole question
An unstated convention is invisible to the marker and easy to contradict by accident three lines later.

Your turn

State your positive direction before assigning any sign.

1
A cyclist rides $200\text{ m}$ east, then turns around and rides $120\text{ m}$ west, taking $80\text{ s}$ in total. Taking east as positive, find the distance travelled, the final displacement, the average speed and the average velocity.
Stuck? Show hint
Distance adds the two legs; displacement adds them with signs.
Show solution
1. 1
  Distance: $200 + 120 = 320\text{ m}$
  Both legs count as positive lengths.
2. 2
  Displacement, east positive: $(+200) + (-120) = 80\text{ m}$
  The west leg is −120 because west is the negative direction.
3. 3
  Average speed: $\frac{320}{80} = 4\text{ m s}^{-1}$
  Total distance over total time.
4. 4
  Average velocity: $\frac{80}{80} = 1\text{ m s}^{-1} \text{ east}$
  Displacement over total time; the answer carries a direction.
Answer
Distance $320\text{ m}$ ; displacement $80\text{ m}$ east; average speed $4\text{ m s}^{-1}$ ; average velocity $1\text{ m s}^{-1}$ east.
2
A particle moves on a straight line with the positive direction to the right. Its velocity changes at a constant rate from $6\text{ m s}^{-1}$ at $t = 0$ to $-4\text{ m s}^{-1}$ at $t = 5$ . Find its acceleration. Describe what happens to its speed during the 5 seconds.
Stuck? Show hint
Change in velocity is end minus start, with signs. The velocity passes through 0 on the way.
Show solution
1. 1
  Acceleration $=$ change in velocity ÷ time: $a = \frac{(-4) - 6}{5} = \frac{-10}{5} = -2\text{ m s}^{-2}$
  End minus start, keeping both signs.
2. 2
  When is the velocity zero? It falls by $2\text{ m s}^{-1}$ each second from 6, so it reaches 0 after $\frac{6}{2} = 3\text{ s}$
  The sign of v changes at this instant — the particle stops and turns round.
3. 3
  Describe the speed. For $0 \leqslant t < 3$ : $v > 0$ and $a < 0$ have opposite signs, so the particle slows down, from $6$ to $0$ . For $3 < t \leqslant 5$ : $v < 0$ and $a < 0$ have the same sign, so it speeds up again, from $0$ to $4\text{ m s}^{-1}$ , now moving left.
  One constant acceleration can mean slowing down and then speeding up — it depends on the sign of v at each moment.
Answer
$a = -2\text{ m s}^{-2}$ . The speed falls from $6$ to $0$ (at $t = 3$ , where the particle turns round), then rises to $4\text{ m s}^{-1}$ with the particle moving in the negative direction.

Practise distance, displacement, speed and velocityReal past-paper questions · Scalar quantities (distance, speed) and vector quantities (displacement, velocity, acceleration)

Checking your access…

Can you do all of these?

Fix a positive direction before assigning any sign, and keep it for the whole question
Distance and speed are never negative; displacement, velocity and acceleration carry signs
Average speed = total distance ÷ total time (rests included) — never the mean of the stage speeds
Slowing down means v and a have opposite signs; a 'deceleration of 2' is a positive size
Gradient of an s-t graph is the velocity; a flat s-t graph means at rest
Gradient of a v-t graph is the acceleration; the area under it is the displacement
For distance from a v-t graph, split at every axis crossing and add the sizes of the areas
'Returns to its starting point' means area above the axis = area below it
Sketch a v-t graph from words: one straight segment per stage, key times and speeds labelled on the axes
For an unknown on a v-t graph, write each stage's time as Δv ÷ a and each area in terms of the unknown, then use the total time or distance
Check both roots of any quadratic against the question (positive speeds, positive times, stages shorter than the journey) and say why one is rejected
Choose the suvat formula that leaves out the letter you neither know nor want
Use suvat only when the acceleration is constant; in a multi-stage journey give each stage its own s, u, v, a, t, linked by the speed at each join
The distance in the nth second is s(n) − s(n − 1)
In vertical motion use a = −10 (upwards positive) for the whole flight — at the top v = 0 but a is still −10
Measure s from the point of projection; landing below it means s is negative
When a string goes slack, the rising particle switches from the system's acceleration to −g
With two particles, use one origin, one clock (t and t − 1 for a late starter), and one equation from the meeting condition
For a vertical collision, check it happens before either particle lands
The least (or greatest) distance between two particles comes where their velocities are equal — complete the square on d(t) = s_A − s_B
Given s, v or a as a function of t, use calculus, never suvat
Differentiate to go s → v → a; never write a = v/t
Maximum or minimum velocity: solve a = 0 and substitute into v
Integrate to go a → v → s, finding each constant from a condition; never write s = vt
A constant of integration is only 0 if the condition makes it 0 — brackets like (t + 1)ⁿ usually give a non-zero constant
For total distance, solve v = 0, keep the roots inside the interval where v changes sign, split there and add the sizes
For piecewise motion, the second piece's constant comes from continuity of velocity at the join
Except at an impact (a barrier, the ground): there the velocity does change instantly, and each formula gives its own value

Now do the questions

221 real Paper 4 parts from 2021–2025, sorted by difficulty, with mark schemes

Forces and Equilibrium

Momentum

Kinematics of Motion in a Straight Line

Distance, displacement, speed, velocity and acceleration

Displacement-time graphs: the gradient is the velocity

Velocity-time graphs: gradient is acceleration, area is displacement

Can you do all of these?

Distance, displacement, speed, velocity and acceleration

The rest of this note

Can you do all of these?