Kinematics: Constant Acceleration and SUVAT

This lesson covers straight-line kinematics with constant acceleration as required by the Edexcel 9MA0 A-Level Mathematics specification. You need to understand displacement-time and velocity-time graphs, and be fluent with the five SUVAT equations.

Key Quantities

Symbol	Quantity	SI Unit
s	Displacement	metres (m)
u	Initial velocity	m s⁻¹
v	Final velocity	m s⁻¹
a	Acceleration	m s⁻²
t	Time	seconds (s)

Important distinctions:

Distance is a scalar (always positive); displacement is a vector (can be negative — direction matters).
Speed is a scalar; velocity is a vector.
Acceleration is the rate of change of velocity. Negative acceleration (deceleration) means the object is slowing down in the positive direction.

The SUVAT Equations

These five equations apply only when acceleration is constant (uniform).

v = u + at
s = ut + ½at²
s = vt - ½at²
v² = u² + 2as
s = ½(u + v)t

Each equation links four of the five SUVAT variables. To solve a problem:

List the known quantities and the quantity you need to find.
Choose the equation that contains exactly those four variables.

Exam Tip: Always define the positive direction clearly at the start of a problem. This is especially important for vertical motion where you must decide whether upwards or downwards is positive.

Displacement-Time Graphs

On a displacement-time (s-t) graph:

The gradient at any point equals the velocity at that instant.
A straight line means constant velocity (zero acceleration).
A curve means the velocity is changing (non-zero acceleration).
A horizontal line means the object is stationary.

Velocity-Time Graphs

On a velocity-time (v-t) graph:

The gradient at any point equals the acceleration.
A horizontal line means constant velocity (zero acceleration).
A straight line with positive gradient means constant positive acceleration.
The area under the graph between two times gives the displacement during that interval.

For a trapezium under the graph: Area = ½(u + v) x t

This is consistent with SUVAT equation 5.

Vertical Motion Under Gravity

Objects falling freely near the Earth's surface have constant acceleration due to gravity:

a = g ≈ 9.8 m s⁻² (downwards)

Convention: If upwards is taken as positive, then a = -9.8 m s⁻².

Example: A ball is thrown vertically upwards at 14.7 m s⁻¹. Taking upwards as positive and g = 9.8 m s⁻²:

To find the time to reach the highest point (v = 0): v = u + at 0 = 14.7 + (-9.8)t t = 14.7/9.8 = 1.5 s

To find the maximum height: s = ut + ½at² s = 14.7(1.5) + ½(-9.8)(1.5²) s = 22.05 - 11.025 = 11.025 m

Multi-Stage Problems

Some problems involve two or more stages with different accelerations. For each stage:

Apply SUVAT separately.
The final velocity of one stage becomes the initial velocity of the next.
Keep track of the total displacement and total time.

Deriving the SUVAT Equations

From the definitions:

Acceleration: a = (v - u)/t, giving v = u + at.
Displacement is the area under the v-t graph. For constant acceleration, the v-t graph is a straight line from u to v over time t: s = ½(u + v)t
Substituting v = u + at into s = ½(u + v)t gives s = ut + ½at².
Eliminating t from v = u + at and s = ½(u + v)t gives v² = u² + 2as.

Summary

The five SUVAT equations apply only when acceleration is constant.
Always define the positive direction clearly.
On s-t graphs, gradient = velocity. On v-t graphs, gradient = acceleration and area = displacement.
For vertical motion, use a = ±g (depending on your sign convention).
For multi-stage problems, apply SUVAT separately to each stage.

A-Level Deep Dive: Kinematics — Constant Acceleration and SUVAT

Spec mapping

Edexcel 9MA0 Paper 3 — Statistics and Mechanics, Section 9: Kinematics. The constant-acceleration sub-strand requires fluency with the five SUVAT equations: $v = u + at$ , $s = ut + \tfrac{1}{2}at^2$ , $v^2 = u^2 + 2as$ , $s = \tfrac{1}{2}(u + v)t$ and $s = vt - \tfrac{1}{2}at^2$ . Vertical motion under gravity is examined using $g = 9.8\ \text{m/s}^2$ unless the question specifies otherwise. Although nominally a Mechanics topic, SUVAT is examined synoptically alongside vectors (Section 10), forces (Section 11) and variable acceleration (also Section 9, but via calculus). The Edexcel formula booklet does list the SUVAT equations — but the choice of which to apply is a candidate skill.

Worked example with full mark scheme

Question (8 marks):

A small ball is projected vertically upwards from a point $O$ at the edge of a cliff with initial speed $14\ \text{m/s}$ . The point $O$ is $24.5\ \text{m}$ above horizontal ground. Take $g = 9.8\ \text{m/s}^2$ and model the ball as a particle moving freely under gravity.

(a) Find the maximum height of the ball above $O$ . (3)

(b) Find the total time from projection until the ball strikes the ground. (5)

Solution with mark scheme:

(a) Step 1 — set up direction convention. Take upwards as positive. Then $u = 14$ , $a = -9.8$ , and at maximum height $v = 0$ .

M1 — using $v^2 = u^2 + 2as$ with the correct signed values. Candidates who silently drop the negative sign on $a$ get $s$ negative and lose both A marks.

$0 = 14^2 + 2(-9.8)s \implies 19.6 s = 196 \implies s = 10\ \text{m}$

A1 — $s = 10\ \text{m}$ .

A1 — answer stated with units, interpreted as "above $O$ ".

(b) Step 1 — define displacement. The ball strikes the ground when its displacement from $O$ is $s = -24.5\ \text{m}$ (below $O$ , with upwards positive).

M1 — correctly setting $s = -24.5$ rather than $+24.5$ .

Step 2 — apply $s = ut + \tfrac{1}{2}at^2$ .

$-24.5 = 14t + \tfrac{1}{2}(-9.8)t^2 \implies -24.5 = 14t - 4.9 t^2$

M1 — substituting correct signed values.

Rearranging: $4.9 t^2 - 14 t - 24.5 = 0$ , or dividing through by $4.9$ : $t^2 - \tfrac{20}{7}t - 5 = 0$ .

Step 3 — solve the quadratic.

$t = \dfrac{14 \pm \sqrt{196 + 4 \cdot 4.9 \cdot 24.5}}{2 \cdot 4.9} = \dfrac{14 \pm \sqrt{196 + 480.2}}{9.8} = \dfrac{14 \pm \sqrt{676.2}}{9.8}$

M1 — applying the quadratic formula or completing the square correctly.

$\sqrt{676.2} \approx 26.00$ , so $t \approx \dfrac{14 + 26.00}{9.8} \approx 4.08\ \text{s}$ (rejecting the negative root).

A1 — $t \approx 4.08\ \text{s}$ (3 s.f.).

A1 — explicit rejection of the negative root with a brief justification ("time must be positive").

Total: 8 marks (M4 A4).

Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Question (6 marks): A car travelling along a straight horizontal road accelerates uniformly from $8\ \text{m/s}$ to $20\ \text{m/s}$ over a distance of $112\ \text{m}$ .

(a) Find the acceleration of the car. (2)

(b) Find the time taken to cover this distance. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO3.1b) — selecting $v^2 = u^2 + 2as$ with $u = 8$ , $v = 20$ , $s = 112$ .
A1 (AO1.1b) — $a = \dfrac{400 - 64}{224} = 1.5\ \text{m/s}^2$ .

(b)

M1 (AO1.1b) — using $v = u + at$ or $s = \tfrac{1}{2}(u + v)t$ with consistent values.
A1 (AO1.1b) — $t = 8\ \text{s}$ .

(c)

B1 (AO3.5a) — naming a valid assumption (e.g. "the car is modelled as a particle" or "acceleration is constant" or "no air resistance").
B1 (AO3.5b) — explaining its effect (e.g. "in reality acceleration would vary with engine response, so the calculated time is an approximation").

Total: 6 marks split AO1 = 3, AO3 = 3. AO3 weighting is high because the modelling sub-part is examined explicitly — this is typical of Section 9 Mechanics questions.

Synoptic links

Connects to:

Variable acceleration via calculus (Section 9): when acceleration is not constant, SUVAT fails. Instead $v = \dfrac{ds}{dt}$ and $a = \dfrac{dv}{dt} = \dfrac{d^2 s}{dt^2}$ , with displacement recovered by integration. The transition between constant and variable contexts is a frequent A* discriminator.
Newton's second law (Section 11, Forces): $F = ma$ links kinematics to dynamics. Once a constant net force is established, acceleration is constant and SUVAT applies — every dynamics problem reduces to kinematics after the force analysis.
Vectors in mechanics (Section 10): in two dimensions, position, velocity and acceleration become vectors $\mathbf{r}$ , $\mathbf{v}$ , $\mathbf{a}$ . The SUVAT equations generalise componentwise: $\mathbf{v} = \mathbf{u} + \mathbf{a} t$ and $\mathbf{r} = \mathbf{u} t + \tfrac{1}{2} \mathbf{a} t^2$ .
Projectile motion (Section 9 / 10 boundary): treats horizontal motion (constant velocity) and vertical motion (constant $-g$ ) independently. SUVAT is applied separately to each component and the results combined parametrically.
Modelling assumptions (Section 8 overview): "particle", "smooth", "light", "inextensible" are technical terms in 9MA0. Each assumption simplifies the equations; A* candidates state which assumptions were used and how relaxing them would change the answer.

Mark-scheme literacy

Constant-acceleration kinematics on 9MA0 Paper 3 splits AO marks across all three objectives:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Choosing the correct SUVAT equation, substituting signed values, solving the resulting linear or quadratic equation
AO2 (reasoning / interpretation)	15–25%	Justifying choice of positive direction, rejecting non-physical roots, interpreting answers in context with units
AO3 (problem-solving / modelling)	20–30%	Identifying modelling assumptions, evaluating their effect, multi-stage problems requiring coordination of phases

Examiner-rewarded phrasing: "taking upwards as positive, so $a = -g = -9.8\ \text{m/s}^2$ "; "since time must be positive, we reject $t = -1.78$ "; "modelling the ball as a particle ignores its size and air resistance, so the calculated maximum height overestimates reality slightly". Phrases that lose marks: writing $g = 10$ when the question specifies $9.8$ ; omitting units on a final answer; failing to state which root was rejected; "acceleration = gravity" without a sign.

Grade-band model answers

3-mark question

Question: A particle moves in a straight line with initial velocity $3\ \text{m/s}$ and constant acceleration $2\ \text{m/s}^2$ . Find its velocity after $5\ \text{s}$ .

Grade C response (~150 words):

Using $v = u + at$ with $u = 3$ , $a = 2$ , $t = 5$ :

$v = 3 + 2 \times 5 = 13\ \text{m/s}$ .

Examiner commentary: Full marks (3/3). The candidate identifies the right equation, substitutes correctly and gives units. Brief but complete. The C/B borderline on this kind of question is whether the candidate states units — many drop the $\text{m/s}$ and lose the final A1.

Grade A response (~190 words):*

Taking the direction of motion as positive, the particle has $u = 3\ \text{m/s}$ , $a = 2\ \text{m/s}^2$ and $t = 5\ \text{s}$ . Acceleration is constant, so SUVAT applies.

Using $v = u + at$ :

$v = 3 + 2 \times 5 = 13\ \text{m/s}$

The velocity after $5\ \text{s}$ is $13\ \text{m/s}$ in the direction of initial motion.

Examiner commentary: Full marks (3/3). Beyond procedural correctness, the candidate states the direction convention, justifies the use of SUVAT (constant acceleration) and gives the answer with both magnitude and direction. The directional commentary matters because velocity is a vector — Grade A* candidates routinely note this even on simple questions.

6-mark question

Question: A stone is dropped from rest from the top of a cliff and falls freely under gravity. It strikes the ground after $3\ \text{s}$ . Take $g = 9.8\ \text{m/s}^2$ .

(a) Find the height of the cliff. (3)

(b) Find the speed of the stone as it strikes the ground. (3)

Grade B response (~210 words):

(a) Using $s = ut + \tfrac{1}{2}at^2$ with $u = 0$ , $a = 9.8$ , $t = 3$ :

$s = 0 + \tfrac{1}{2} \times 9.8 \times 9 = 44.1\ \text{m}$ .

(b) Using $v = u + at$ : $v = 0 + 9.8 \times 3 = 29.4\ \text{m/s}$ .

Examiner commentary: Full marks (6/6) for the maths. The candidate has implicitly taken downwards as positive, which works because the stone is dropped (initial velocity zero) and falls in one direction. A pickier examiner might want the convention stated, but it is unambiguous here. This is solid Grade B work — correct, efficient, but lacking the polish that distinguishes A*.

Grade A response (~245 words):*

Take downwards as positive, so $a = +9.8\ \text{m/s}^2$ . The stone is dropped, so $u = 0$ .

(a) Using $s = ut + \tfrac{1}{2}at^2$ :

$s = 0 \times 3 + \tfrac{1}{2}(9.8)(3)^2 = 4.9 \times 9 = 44.1\ \text{m}$

The cliff is $44.1\ \text{m}$ high.

(b) Using $v = u + at$ :

$v = 0 + 9.8 \times 3 = 29.4\ \text{m/s}$

The speed at impact is $29.4\ \text{m/s}$ .

Examiner commentary: Full marks (6/6). The candidate states the direction convention, distinguishes "dropped" ( $u = 0$ ) from "thrown", and gives both numerical answers with units. The display-math formatting and explicit working make every step verifiable. Note that "speed" was asked for, not "velocity" — speed is the magnitude, so giving $29.4\ \text{m/s}$ without a direction is correct here. An A* candidate distinguishes these terms precisely.

9-mark question

Question: A particle $P$ moves in a straight line with constant acceleration. At time $t = 0$ it passes through point $A$ with velocity $5\ \text{m/s}$ . At time $t = 4\ \text{s}$ it passes through point $B$ with velocity $13\ \text{m/s}$ .

(a) Find the acceleration of $P$ . (2)

(b) Find the distance $AB$ . (2)

(c) After passing through $B$ , $P$ continues with the same acceleration for a further $2\ \text{s}$ before the acceleration reverses to deceleration of equal magnitude. Find the total distance from $A$ when $P$ first comes to rest. (5)

Grade A response (~340 words):*

(a) Using $v = u + at$ with $u = 5$ , $v = 13$ , $t = 4$ :

$13 = 5 + 4a \implies a = 2\ \text{m/s}^2$ .

(b) Using $s = \tfrac{1}{2}(u + v)t$ :

$AB = \tfrac{1}{2}(5 + 13)(4) = 36\ \text{m}$ .

Velocity at C: $v = 13 + 2 \times 2 = 17\ \text{m/s}$ . Distance BC: $s = 13 \times 2 + \tfrac{1}{2}(2)(4) = 26 + 4 = 30\ \text{m}$ .

Stage 3 (C onwards, $a = -2$ , initial velocity $17\ \text{m/s}$ , final velocity $0\ \text{m/s}$ ):

Using $v^2 = u^2 + 2as$ : $0 = 17^2 + 2(-2)s \implies s = \dfrac{289}{4} = 72.25\ \text{m}$ .

Total distance from A: $36 + 30 + 72.25 = 138.25\ \text{m}$ .

Examiner commentary: Full marks (9/9). The candidate splits the motion into three phases and applies SUVAT separately to each — the standard technique for multi-stage problems. The intermediate velocity at $C$ is computed first (acting as the initial velocity for stage 3), and the deceleration stage uses $v = 0$ as the final condition. Display math, units throughout, and a clean total. The decomposition into named stages (B to C, C onwards) is the Grade A* organisational trick: it makes the working auditable for the examiner and prevents sign errors when acceleration changes direction.

A-Level-depth misconceptions

The errors that distinguish A from A* on SUVAT questions:

Sign of $g$ . Whether $g = +9.8$ or $g = -9.8$ depends entirely on the chosen positive direction. Taking upwards as positive gives $a = -9.8$ on the way up and on the way down (gravity always acts downwards). Switching signs mid-problem because "the ball is now falling" is the classic error.
Choosing the right equation. SUVAT has five equations and five variables $(s, u, v, a, t)$ ; each equation omits exactly one. Identify the variable you don't have and the variable you don't need — that pins down the equation. Picking randomly wastes time and introduces errors.
Vector vs scalar quantities. Displacement, velocity and acceleration are vectors (signed); distance and speed are scalars (always positive). A particle that travels $10\ \text{m}$ then returns has displacement $0$ but distance $20\ \text{m}$ . Using "distance" when "displacement" is meant breaks SUVAT.
Free fall $\neq$ "dropped". "Free fall" means moving under gravity alone (so $a = \pm g$ ). "Dropped" additionally implies $u = 0$ . "Thrown downwards" is free fall with $u \neq 0$ . Don't conflate these.
$s$ is displacement, not always height. When a ball is thrown up from a cliff and lands below the launch point, $s$ at impact is negative (with up positive), not the absolute height fallen. Treating $s$ as always positive yields nonsense quadratics.
Average velocity formula $s = \tfrac{1}{2}(u + v)t$ requires constant $a$ . It looks like a "kinematic average" but only holds when acceleration is constant. Apply it to variable-acceleration motion and you'll be wrong.
Quadratic root rejection. Solving $s = ut + \tfrac{1}{2}at^2$ for $t$ often gives two roots — one positive, one negative. The negative root corresponds to time before the motion started (the projectile's hypothetical past trajectory). State explicitly that you reject it.

Common errors and mark-loss patterns

Three patterns repeatedly cost candidates marks on Section 9 SUVAT questions. They are about discipline, not technique.

Sign-convention drift. Candidates state "take up as positive" then write $a = +9.8$ partway through. Cure: write the convention at the top of the page and check every signed value against it before substituting.
Forgetting units. SUVAT answers are physical quantities. $s = 44.1$ is not a complete answer — $s = 44.1\ \text{m}$ is. The final A1 is regularly withheld for missing units, especially on multi-part questions where the candidate dashes the final answer.
Failing to state modelling assumptions. Section 9 questions often include a "state one assumption" sub-part worth two marks. Candidates either skip it or write something vague like "assume gravity is constant". Examiner-credited assumptions are: "particle" (no rotation, point mass), "no air resistance", "constant acceleration", "smooth surface". Naming and explaining one of these earns both marks.

Going further — university and academic signposting

Constant-acceleration kinematics is the gateway to several physics and applied-mathematics undergraduate trajectories:

Galilean relativity (Year 1 mechanics): SUVAT is invariant under Galilean transformations — observers moving at constant relative velocity agree on accelerations and forces. This symmetry, formalised by Galileo, is the historical and conceptual root of inertial frames and Newton's first law.
Lagrangian mechanics (Year 2 / 3): instead of forces and SUVAT, motion is derived from a single scalar function (the Lagrangian $L = T - V$ ) via the Euler-Lagrange equations. Constant-acceleration motion under gravity falls out as the simplest non-trivial example, and the same machinery handles pendulums, orbits and field theories.
Relativistic kinematics (Year 2 / 3): at speeds approaching $c$ , SUVAT breaks down — velocities don't add linearly and "constant acceleration" requires reinterpretation (proper acceleration). Special relativity replaces SUVAT with four-vector equations involving $\gamma = 1/\sqrt{1 - v^2/c^2}$ .
Physics undergraduate degree: every first-year mechanics course revisits projectile motion as a worked example before generalising to forces, energy and momentum. The Edexcel SUVAT toolkit is the minimum fluency expected on day one.

Oxbridge interview prompt: "A ball is thrown vertically upwards with speed $u$ . Ignoring air resistance, the time to reach maximum height equals the time to fall back. Now include linear air resistance. Which is longer — the time up or the time down? Justify physically."

Additional A* practice: two-stage motion — projectile thrown up, find total time including downward leg

A common A* trap on Paper 3 is to set a question that seems to require splitting motion into "up" and "down" phases, when in fact a single SUVAT application — chosen carefully — solves it in one step.

Worked example: A ball is thrown vertically upwards from ground level with speed $21\ \text{m/s}$ . Find the total time until it returns to ground level. Take $g = 9.8\ \text{m/s}^2$ .

Naive two-stage method. Stage 1 (up): use $v = u + at$ with $u = 21$ , $v = 0$ , $a = -9.8$ to get $t_1 = 21/9.8 \approx 2.143\ \text{s}$ . By symmetry, stage 2 (down) takes the same time, so total $\approx 4.286\ \text{s}$ .

Single-equation method. Take upwards as positive. The ball returns to ground when $s = 0$ (its starting height). Use $s = ut + \tfrac{1}{2}at^2$ :

$0 = 21 t + \tfrac{1}{2}(-9.8) t^2 = 21 t - 4.9 t^2 = t(21 - 4.9 t)$

So $t = 0$ (the launch instant) or $t = 21/4.9 \approx 4.286\ \text{s}$ .

Why A candidates prefer this:* the single equation handles both phases simultaneously because SUVAT is valid throughout the motion (constant acceleration $-9.8\ \text{m/s}^2$ ). Splitting into stages is unnecessary and risks sign errors at the apex.

A subtlety: if the ball were thrown from a cliff (start point above ground), the symmetry argument breaks. Then the two-stage method overcounts (the downward leg is longer than the upward leg) and the single-equation method, with $s = -h$ at impact, is the only safe approach. This generalisation — recognising when symmetry helps and when it doesn't — is the synoptic insight Edexcel rewards on harder Section 9 questions.

Edexcel A-Level alignment footer

This content is aligned with the Pearson Edexcel GCE A Level Mathematics (9MA0) specification, Paper 3 — Statistics and Mechanics, Section 9: Kinematics. For the most accurate and up-to-date information, please refer to the official Pearson Edexcel specification document.

Visual summary

graph TD
    A["Constant-acceleration<br/>motion"] --> B{"Which variable<br/>is missing?"}
    B -->|"No s"| C["v = u + at"]
    B -->|"No v"| D["s = ut + ½at²"]
    B -->|"No t"| E["v² = u² + 2as"]
    C --> F["Substitute signed<br/>values, solve,<br/>state units"]
    D --> F
    E --> F

    style A fill:#27ae60,color:#fff
    style F fill:#3498db,color:#fff

Kinematics: Constant Acceleration and SUVAT

Kinematics: Constant Acceleration and SUVAT

Key Quantities

The SUVAT Equations

Displacement-Time Graphs

Velocity-Time Graphs

Vertical Motion Under Gravity

Multi-Stage Problems

Deriving the SUVAT Equations

Summary

A-Level Deep Dive: Kinematics — Constant Acceleration and SUVAT

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

6-mark question

9-mark question

A-Level-depth misconceptions

Common errors and mark-loss patterns

Going further — university and academic signposting

Additional A* practice: two-stage motion — projectile thrown up, find total time including downward leg

Edexcel A-Level alignment footer

Visual summary

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