Kinematics

This lesson covers kinematics — the study of motion without considering the forces that cause it. Kinematics is one of the foundational topics in A-Level Mechanics and provides the mathematical tools needed to describe and analyse the motion of objects in one and two dimensions.

Spec Mapping — AQA 7357 Sections P and Q (Quantities and units in mechanics; Kinematics). This lesson links SI units and dimensional consistency to constant-acceleration motion in a straight line and under gravity. Refer to the official AQA specification document for exact wording.

Key Quantities

Quantity	Symbol	Unit	Type
Displacement	$s$	m	Vector
Velocity	$v$	m/s	Vector
Speed	$	v	$
Acceleration	$a$	m/s $^2$	Vector
Time	$t$	s	Scalar

Displacement is the distance from a fixed point in a specified direction (a vector).
Velocity is the rate of change of displacement with respect to time.
Acceleration is the rate of change of velocity with respect to time.

Exam Tip: Always distinguish between distance (scalar) and displacement (vector), and between speed (scalar) and velocity (vector). Using the wrong term will lose marks.

The SUVAT Equations (Constant Acceleration)

When acceleration is constant, the following equations relate the five kinematic quantities:

Equation	Variables used
$v = u + at$	$v, u, a, t$
$s = ut + \frac{1}{2}at^2$	$s, u, a, t$
$s = vt - \frac{1}{2}at^2$	$s, v, a, t$
$v^2 = u^2 + 2as$	$v, u, a, s$
$s = \frac{(u + v)t}{2}$	$s, u, v, t$

Where: $s$ = displacement, $u$ = initial velocity, $v$ = final velocity, $a$ = acceleration, $t$ = time.

Example: A car accelerates from rest at 3 m/s $^2$ for 8 seconds. Find the final velocity and distance travelled.

$u = 0$ , $a = 3$ , $t = 8$

$v = u + at = 0 + 3 \times 8 = 24$ m/s

$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 3 \times 64 = 96$ m

Velocity-Time Graphs

A velocity-time graph provides a visual representation of an object's motion:

Feature	Interpretation
Gradient	Acceleration
Area under the curve	Displacement
Horizontal line	Constant velocity (zero acceleration)
Line sloping upwards	Positive acceleration
Line sloping downwards	Deceleration (negative acceleration)

Exam Tip: The area under a velocity-time graph gives displacement, not distance. If the graph goes below the time axis, the area below represents negative displacement (motion in the opposite direction).

Vertical Motion Under Gravity

For objects moving vertically under gravity (ignoring air resistance):

Take upwards as positive (or downwards — but state your convention clearly).
The acceleration due to gravity is $g = 9.8$ m/s $^2$ (downwards).
Use the SUVAT equations with $a = -g$ (if upwards is positive) or $a = g$ (if downwards is positive).

Example: A ball is thrown vertically upwards with speed 20 m/s. Find the maximum height.

At maximum height, $v = 0$ .

$v^2 = u^2 + 2as$

$0 = 20^2 + 2(-9.8)s$

$s = \frac{400}{19.6} = 20.4$ m

Displacement-Time Graphs

Feature	Interpretation
Gradient	Velocity
Horizontal line	Object is stationary
Straight line with positive gradient	Constant positive velocity
Curve	Changing velocity (acceleration)

Summary

Kinematics describes motion using displacement, velocity, acceleration, and time.
The SUVAT equations apply when acceleration is constant.
Velocity-time graphs: gradient = acceleration, area = displacement.
Displacement-time graphs: gradient = velocity.
For vertical motion under gravity, use $a = \pm g = \pm 9.8$ m/s $^2$ .
Always state your sign convention and use consistent units.

Exam Tip: Before applying SUVAT equations, list all known quantities and identify which equation to use. If three quantities are known, you can find the fourth. Always check your sign convention is consistent throughout the problem.

Synoptic Links

This topic connects to:

Differentiation (aqa-alevel-maths-pure-1 / differentiation) — instantaneous velocity and acceleration as derivatives of displacement.
Integration (aqa-alevel-maths-pure-1 / integration) — displacement obtained by integrating velocity-time relationships.
Variable acceleration (aqa-alevel-maths-mechanics / variable-acceleration) — SUVAT is the constant-acceleration special case of the calculus framework.

A-Level Deep Dive: Kinematics

Spec mapping

AQA 7357 specification, Paper 3 — Mechanics, Section T (Kinematics) covers the language of kinematics — position, displacement, distance travelled, velocity, speed, acceleration. Understand, use and derive the formulae for constant acceleration for motion in a straight line (refer to the official specification document for exact wording). This is the procedural foundation of Paper 3 and is examined synoptically with Section U (Forces and Newton's laws, where $F = ma$ converts forces into accelerations that feed SUVAT), Section V (Moments, where rigid-body equilibrium uses zero acceleration as a SUVAT special case) and indirectly with the calculus of motion in Section T's later sub-strand ( $v = \dfrac{ds}{dt}$ , $a = \dfrac{dv}{dt}$ ). The AQA formula booklet does list the four SUVAT equations, but the conditions under which they apply (constant acceleration, straight-line motion) must be recognised by the candidate — examiners will not flag a non-constant-acceleration scenario.

The four equations to memorise (and recognise the situations that select each):

$v = u + at$ — no displacement involved
$s = ut + \tfrac{1}{2}at^2$ — no final velocity
$v^2 = u^2 + 2as$ — no time
$s = \tfrac{1}{2}(u + v)t$ — no acceleration

In all UK A-Level mechanics, take $g = 9.8\,\text{m s}^{-2}$ downward unless the question states otherwise. Older textbooks use $g = 9.81$ ; AQA 7357 mark schemes are written for $g = 9.8$ .

Worked example with full mark scheme

Question (8 marks):

A small ball is projected vertically upward from a point $1.2\,\text{m}$ above horizontal ground with initial speed $14\,\text{m s}^{-1}$ . The ball moves freely under gravity ( $g = 9.8\,\text{m s}^{-2}$ ).

(a) Find the maximum height above the ground reached by the ball. (3)

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

Solution with mark scheme:

(a) Step 1 — choose a sign convention and identify the SUVAT variables.

Take upward as positive. At maximum height, $v = 0$ .

$u = +14\,\text{m s}^{-1}$
$v = 0$
$a = -9.8\,\text{m s}^{-2}$ (gravity acts downward, against the chosen positive direction)
$s = ?$ (displacement from projection point)
Equation needed: no $t$ — choose $v^2 = u^2 + 2as$ .

M1 — selecting $v^2 = u^2 + 2as$ with $v = 0$ and the correct sign on $a$ . A common error is leaving $a = +9.8$ , which gives a negative $s$ and a candidate who "knows it should be positive" then drops the minus by hand — examiners spot this and award no marks for a sign-fudged solution.

Step 2 — solve for $s$ .

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

A1 — $s = 10\,\text{m}$ above the projection point.

Step 3 — add the projection height.

Maximum height above ground $= 10 + 1.2 = 11.2\,\text{m}$ .

A1 — final answer $11.2\,\text{m}$ . Many candidates stop at $10\,\text{m}$ , missing the displacement-vs-position distinction. The question asks for height above the ground, not height gained.

(b) Step 1 — set up the displacement equation from projection to impact.

Taking upward positive and origin at the projection point, the ground is at $s = -1.2\,\text{m}$ .

$u = +14$
$a = -9.8$
$s = -1.2$
$t = ?$
Equation: $s = ut + \tfrac{1}{2}at^2$ .

M1 — correct setup with consistent signs (ground at negative displacement because the chosen origin is the launch point, which is above the ground).

Step 2 — substitute.

$-1.2 = 14t - 4.9 t^2$ $4.9 t^2 - 14 t - 1.2 = 0$

M1 — rearrangement to a quadratic in $t$ .

Step 3 — solve the quadratic.

Using the quadratic formula with $a = 4.9$ , $b = -14$ , $c = -1.2$ :

$t = \dfrac{14 \pm \sqrt{196 + 4 \cdot 4.9 \cdot 1.2}}{2 \cdot 4.9} = \dfrac{14 \pm \sqrt{196 + 23.52}}{9.8} = \dfrac{14 \pm \sqrt{219.52}}{9.8}$

$\sqrt{219.52} \approx 14.816$ , so $t \approx \dfrac{14 + 14.816}{9.8} \approx 2.94\,\text{s}$ or $t \approx \dfrac{14 - 14.816}{9.8} \approx -0.083\,\text{s}$ .

M1 — correct application of the quadratic formula.

A1 — selecting the positive root $t \approx 2.94\,\text{s}$ and rejecting the negative root explicitly with reasoning ("time after projection must be positive").

A1 — final answer $t \approx 2.94\,\text{s}$ (3 s.f.). Candidates who carry intermediate values to insufficient precision finish at $2.93\,\text{s}$ , which is on the boundary of acceptance. Carry at least 4 s.f. through working and round only at the end.

Total: 8 marks (M3 A5, distributed as shown).

Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Specimen question modelled on the AQA 7357 Paper 3 format

Question (6 marks): A car travels in a straight line. It accelerates uniformly from rest to $20\,\text{m s}^{-1}$ over a distance of $100\,\text{m}$ , then travels at constant speed for $T$ seconds before decelerating uniformly to rest at a deceleration of $2.5\,\text{m s}^{-2}$ .

(a) Find the acceleration during the first phase. (2)

(b) Given that the total distance travelled by the car is $420\,\text{m}$ , find $T$ . (4)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — selecting $v^2 = u^2 + 2as$ with $u = 0$ , $v = 20$ , $s = 100$ .
A1 (AO1.1b) — solving $400 = 200 a$ to obtain $a = 2\,\text{m s}^{-2}$ .

(b)

M1 (AO1.1b) — finding the deceleration-phase distance: $0 = 20^2 + 2(-2.5) s_3 \implies s_3 = 80\,\text{m}$ .
M1 (AO3.1a) — distance budgeting: middle-phase distance $= 420 - 100 - 80 = 240\,\text{m}$ .
M1 (AO1.1b) — applying $\text{distance} = \text{speed} \times \text{time}$ for constant-speed phase: $240 = 20 T$ .
A1 (AO1.1b) — $T = 12\,\text{s}$ .

Total: 6 marks split AO1 = 5, AO3 = 1. AQA mechanics often uses three-phase journeys to test that candidates can decompose a problem into SUVAT-friendly stages — the AO3 mark is for noticing that the three phases together must account for the total displacement.

Synoptic links

Connects to:

Section T (variable acceleration via calculus): when acceleration is not constant, SUVAT does not apply and candidates must integrate. $v = \int a\,dt$ , $s = \int v\,dt$ . The boundary case where $a$ happens to be constant recovers SUVAT — the equations $v = u + at$ and $s = ut + \tfrac{1}{2}at^2$ are the integrals of constant $a$ with respect to $t$ , with $u$ and $0$ as constants of integration. Recognising this connection earns AO2 marks on questions that switch between phases.
Section U (forces, $F = ma$ ): every SUVAT problem with non-zero $a$ implicitly involves a net force. A "ball under gravity" problem treats $g$ as the constant acceleration produced by the gravitational force $W = mg$ on a body of mass $m$ . Combined SUVAT-and-Newton questions are routine on Paper 3 and require the candidate to compute $a$ from forces before applying SUVAT.
Section T (vectors): displacement, velocity and acceleration are vector quantities. In two-dimensional problems (e.g. projectiles in the AS Mechanics paper, here applied vertically), candidates resolve into perpendicular components — typically horizontal (constant velocity) and vertical (SUVAT under gravity). The horizontal and vertical motions are independent and share only the time $t$ .
Projectile motion (AS Mechanics): the most common synoptic application. Horizontal: $x = (u\cos\theta) t$ . Vertical: $y = (u\sin\theta) t - \tfrac{1}{2} g t^2$ . Eliminating $t$ between these gives the trajectory parabola $y = x \tan\theta - \dfrac{g x^2}{2 u^2 \cos^2 \theta}$ .
Modelling assumptions: SUVAT under gravity tacitly assumes (a) air resistance is negligible, (b) $g$ is constant throughout the motion (true to within a fraction of a percent over A-Level scales), (c) the body is a particle (no rotation), (d) the ground is flat. Examiners reward candidates who name these assumptions when asked to "comment on the model"; "no air resistance" is by far the most marked-for assumption.

Mark-scheme literacy

Kinematics questions on AQA 7357 Paper 3 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	55–70%	Selecting the right SUVAT equation, applying it with correct signs, executing the algebra, computing numerical answers to required precision
AO2 (reasoning / interpretation)	15–30%	Justifying the sign convention, distinguishing displacement from distance, explaining why a negative root is rejected, interpreting "maximum height above ground" vs "height gained"
AO3 (problem-solving / modelling)	10–25%	Decomposing multi-phase journeys, criticising modelling assumptions, choosing variables when more than one approach is possible

Examiner-rewarded phrasing: "taking upward as positive, so $a = -g = -9.8\,\text{m s}^{-2}$ "; "rejecting $t = -0.083\,\text{s}$ as time after projection must be non-negative"; "assuming air resistance is negligible". Phrases that lose marks: omitting units in a final answer; writing $g = 9.81$ when AQA convention is $9.8$ (rarely deducted but breaks consistency with the printed answer); silently switching sign conventions mid-solution.

A specific AQA pattern: questions that ask "find the speed" expect a positive scalar — even if your velocity calculation produces $v = -8.4\,\text{m s}^{-1}$ (downward in your sign convention), the speed is $8.4\,\text{m s}^{-1}$ . Stating the magnitude explicitly with a brief justification ("speed is the magnitude of velocity") earns the AO2 mark.

Grade-band model answers

3-mark question

Question: A stone is dropped from rest from a cliff and reaches the sea $3.0\,\text{s}$ later. Taking $g = 9.8\,\text{m s}^{-2}$ and ignoring air resistance, find the height of the cliff above the sea.

Grade C response (~150 words):

Use $s = ut + \tfrac{1}{2} a t^2$ with $u = 0$ , $a = 9.8$ , $t = 3.0$ :

$s = 0 + \tfrac{1}{2}(9.8)(3.0)^2 = \tfrac{1}{2}(9.8)(9) = 44.1\,\text{m}$ .

So the cliff is $44.1\,\text{m}$ high.

Examiner commentary: Full marks (3/3). The candidate selects the correct SUVAT equation given the absence of a final-velocity datum, treats downward as positive (so $a = +9.8$ ), and computes the answer to 3 s.f. with units. The answer is direct and correct. Many Grade C candidates lose a mark by writing $s = 44\,\text{m}$ without units, or $44.1$ with no unit suffix — examiners do deduct for missing units in mechanics, where dimensional clarity is graded.

Grade A response (~200 words):*

Take downward as positive, with origin at the top of the cliff. The stone is "dropped from rest" so $u = 0$ . Air resistance is negligible (stated), so the stone moves under gravity alone with $a = g = +9.8\,\text{m s}^{-2}$ . The displacement after $t = 3.0\,\text{s}$ is:

$s = ut + \tfrac{1}{2} a t^2 = 0 + \tfrac{1}{2}(9.8)(3.0)^2 = 44.1\,\text{m}$

Since the stone starts at the top of the cliff and ends at the sea, the cliff height equals this displacement: $h = 44.1\,\text{m}$ (3 s.f.).

Examiner commentary: Full marks (3/3). Beyond procedural correctness, the candidate states the sign convention, names the modelling assumption (air resistance negligible), and connects the SUVAT displacement to the question's "height" via the geometry of the setup. The "(3 s.f.)" suffix is examination craft. This level of articulation earns the AO2 mark on harder questions and signals the candidate is ready to face multi-phase problems where sloppy sign work would be fatal.

6-mark question

Question: A particle moves in a straight line with constant acceleration. It passes through points $A$ , $B$ and $C$ in that order, with $AB = 24\,\text{m}$ and $BC = 40\,\text{m}$ . The particle takes $2.0\,\text{s}$ to travel from $A$ to $B$ and a further $2.0\,\text{s}$ to travel from $B$ to $C$ . Find (a) the acceleration, (b) the speed at $A$ .

Grade B response (~250 words):

Using $s = ut + \tfrac{1}{2} a t^2$ from $A$ over the first 2 s: $24 = 2u + 2a$ , so $u + a = 12$ .

Over 4 s from $A$ to $C$ , total $s = 64$ : $64 = 4u + 8a$ , so $u + 2a = 16$ .

Subtract: $a = 4\,\text{m s}^{-2}$ . Then $u = 12 - 4 = 8\,\text{m s}^{-1}$ .

So acceleration $= 4\,\text{m s}^{-2}$ and speed at $A$ is $8\,\text{m s}^{-1}$ .

Examiner commentary: Full marks (6/6) — but the working is uncomfortably terse. The candidate sets up two simultaneous equations correctly and solves them, but does not explicitly identify " $u$ " as the speed at $A$ (the question asks for speed, which is a scalar; the working uses $u$ as a velocity). On a question with a sign-sensitive setup this elision would cost the AO2 mark. The Grade B label fits because the algebra is right but the presentation lacks the discipline that would survive a harder question.

Grade A response (~300 words):*

Let $u$ be the velocity at $A$ (taking the direction $A \to C$ as positive) and $a$ the constant acceleration.

For the leg $A \to B$ : $s = 24$ , $t = 2$ , so by $s = ut + \tfrac{1}{2} a t^2$ :

$24 = 2u + 2a \implies u + a = 12 \quad (1)$

For the leg $A \to C$ : $s = 24 + 40 = 64$ , $t = 4$ :

$64 = 4u + 8a \implies u + 2a = 16 \quad (2)$

Subtracting (1) from (2):

$a = 4\,\text{m s}^{-2}$

Substituting into (1): $u = 12 - 4 = 8\,\text{m s}^{-1}$ .

Since $u > 0$ , the velocity is in the $A \to C$ direction and the speed at $A$ equals the magnitude: $|u| = 8\,\text{m s}^{-1}$ .

Examiner commentary: Full marks (6/6). The candidate names the sign convention, distinguishes velocity from speed at the end (the question asks for speed), and presents the simultaneous-equations approach cleanly with labelled equations (1) and (2). The phrase "since $u > 0$ , the velocity is in the $A \to C$ direction" is precisely the AO2 reasoning examiners reward — it shows the candidate has understood that "speed" is the magnitude of "velocity" and treated them differently.

9-mark question

Question: A ball is projected vertically upward from ground level with speed $U\,\text{m s}^{-1}$ . After the ball has been in flight for $T\,\text{s}$ , a second ball is projected vertically upward from the same point with speed $U\,\text{m s}^{-1}$ . Show that the two balls collide at height $h = \dfrac{U^2}{2g} - \dfrac{g T^2}{8}$ above the ground, and state the condition on $U$ and $T$ for a collision to occur.

Grade A response (~370 words):*

Take upward as positive; ground level as origin. Both balls move under gravity alone with $a = -g$ .

Let $t$ denote time measured from the launch of the first ball. The first ball's height: $y_1(t) = U t - \tfrac{1}{2} g t^2$ .

The second ball is launched at $t = T$ , so its height at time $t \geq T$ is:

$y_2(t) = U(t - T) - \tfrac{1}{2} g (t - T)^2$

For collision: $y_1(t) = y_2(t)$ at some $t^* > T$ :

$U t - \tfrac{1}{2} g t^2 = U(t - T) - \tfrac{1}{2} g (t - T)^2$

Expanding the right-hand side:

$U t - U T - \tfrac{1}{2} g (t^2 - 2 t T + T^2) = U t - U T - \tfrac{1}{2} g t^2 + g t T - \tfrac{1}{2} g T^2$

Equate to LHS $U t - \tfrac{1}{2} g t^2$ and cancel:

$0 = -U T + g t T - \tfrac{1}{2} g T^2$

Divide by $T$ ( $T \neq 0$ ) and solve for $t$ :

$t^* = \dfrac{U}{g} + \dfrac{T}{2}$

Substitute into $y_1$ :

$h = U t^* - \tfrac{1}{2} g (t^*)^2 = U \left(\dfrac{U}{g} + \dfrac{T}{2}\right) - \tfrac{1}{2} g \left(\dfrac{U}{g} + \dfrac{T}{2}\right)^2$

Expanding and simplifying (algebra omitted for brevity):

$h = \dfrac{U^2}{2 g} - \dfrac{g T^2}{8}$

For the collision to occur in flight, $h > 0$ and $t^*$ must be before either ball returns to ground. The first condition $h > 0$ gives $U^2 > \dfrac{g^2 T^2}{4}$ , i.e. $U > \dfrac{g T}{2}$ (taking positive roots).

Examiner commentary: Full marks (9/9). The candidate sets up parallel SUVAT expressions with consistent time origins, solves the collision-time equation, substitutes back, and states the validity condition without prompting. The condition $U > gT/2$ is the AO3 mark — it identifies that the algebraic result is only physical when the first ball has not already landed before the second is launched. This is graduate-level mechanical reasoning expressed at A-Level.

A-Level-depth misconceptions

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

Sign of $g$ . When upward is positive, $a = -g = -9.8$ . When downward is positive (e.g. a dropped object), $a = +g = +9.8$ . The single most common A-Level error is mixing conventions mid-solution — using $+g$ for the upward leg and $-g$ after the apex, when the convention should have been fixed at the start.
Choosing the wrong SUVAT equation. Each SUVAT equation omits one of the five variables. Identify which variable the question does not mention and select accordingly. Candidates who memorise only $v = u + at$ and $s = ut + \tfrac{1}{2}at^2$ end up doing two-step calculations where one step would have sufficed — and accumulate rounding error along the way.
Vector vs scalar slip. "Velocity" is a vector (signed in 1D); "speed" is its magnitude (always non-negative); "displacement" is signed; "distance travelled" is the path length and is non-negative. A particle that moves $5\,\text{m}$ right then $3\,\text{m}$ left has displacement $+2\,\text{m}$ but distance $8\,\text{m}$ . Confusing these on multi-phase questions costs A1s consistently.
$u = 0$ vs $u$ unstated. "Dropped from rest" gives $u = 0$ . "Released from rest" gives $u = 0$ . "Projected" with no speed given does not imply $u = 0$ — the question expects you to use $u$ as an unknown or to re-read for the missing datum.
Apex velocity vs apex acceleration. At the maximum height of a vertical projectile, the velocity is momentarily zero but the acceleration is still $-g$ . Candidates who write "at the apex $a = 0$ " lose every mark on the descending-leg calculation.
Air resistance "of course". The phrase "ignoring air resistance" should be stated explicitly when the question asks for an assumption critique. A candidate who writes "we assume no air resistance" earns the AO3 mark; one who silently uses SUVAT does not.
Quadratic-in- $t$ root selection. When the SUVAT $s = ut + \tfrac{1}{2}at^2$ produces two positive roots (e.g. a ball that passes a target on the way up and on the way down), the question's wording selects which one. "First reaches" means the smaller $t$ ; "lands" means the larger. Selecting the wrong root costs the A1 even when the algebra is flawless.

Common errors and mark-loss patterns on SUVAT questions

Three patterns repeatedly cost candidates marks on Paper 3 kinematics questions. They are all about discipline, not technique.

Sign-convention drift. Candidates declare "upward positive" at the start, then partway through write $g = +9.8$ as if downward were now positive, producing an incorrect signed result that they then "fix" by hand. The cure: write the chosen positive direction next to every signed quantity at first use, and never change convention mid-question.
Premature rounding. Carrying $\sqrt{219.52}$ as $14.8$ (3 s.f.) instead of $14.816$ (5 s.f.) into a subsequent division produces a final answer that fails the "3 s.f." accuracy check. The cure: keep at least one extra significant figure in intermediate steps and round only at the final answer.
Missing or inconsistent units. AQA mark schemes do deduct for missing or wrong units in mechanics. $10$ is not the same as $10\,\text{m}$ or $10\,\text{m s}^{-1}$ . The cure: write units on every numerical answer, including intermediate results that have physical meaning.

This pattern is endemic to Paper 3: candidates know SUVAT, lose marks on presentation discipline.

Going further — university and academic signposting

Constant-acceleration kinematics points directly toward several undergraduate trajectories:

Galilean relativity (Year 1 mechanics): the SUVAT equations are invariant under a change of inertial frame. If observer $O'$ moves at constant velocity $V$ relative to $O$ , then displacements and velocities transform as $s' = s - V t$ , $v' = v - V$ , but the acceleration is the same in both frames ( $a' = a$ ). This is the principle that motivated Galileo's relativity and underpins Newtonian mechanics.
Lagrangian mechanics (Year 2/3): the trajectory $s(t) = ut + \tfrac{1}{2} a t^2$ for a freely falling body is the path that minimises the action integral $S = \int (T - V)\,dt$ where $T$ is kinetic and $V$ potential energy. SUVAT is the easiest case of a much deeper variational principle that generalises to all of physics.
Relativistic kinematics (Year 3): at speeds approaching $c$ , the constant-acceleration formulae break down. The relativistic analogue of $v = u + at$ is the velocity-addition formula $v = \dfrac{u + a t}{1 + u a t / c^2}$ for a "constant proper acceleration" — and $v < c$ always, no matter how long $a$ is applied. SUVAT is the $c \to \infty$ limit.
Numerical ODE solvers (Year 2 numerical methods): when acceleration is not constant, kinematics becomes a differential equation $\ddot{s} = f(s, \dot{s}, t)$ . Computing trajectories then requires numerical integration (Euler, RK4). SUVAT is the closed-form special case that benchmarks the solver.

Oxbridge interview prompt: "A ball is dropped from height $h$ onto a perfectly elastic floor. It bounces forever. What is the total time before the ball is permanently at rest? What is the total distance travelled? Are these the same kind of infinity?"

Additional A* practice: two-stage motion under different accelerations

A common A* trap on AQA 7357 Paper 3 is to give a journey with two different constant accelerations and require the candidate to stitch the SUVAT phases together. The technique is to use the final velocity of phase 1 as the initial velocity of phase 2, and to budget displacement and time across phases.

Worked example: A train accelerates uniformly from rest at $0.5\,\text{m s}^{-2}$ for $30\,\text{s}$ , then immediately decelerates uniformly at $0.75\,\text{m s}^{-2}$ until it comes to rest. Find (a) the total distance travelled, (b) the total time of the journey.

Phase 1 (acceleration): $u_1 = 0$ , $a_1 = 0.5$ , $t_1 = 30$ .

$v_1 = u_1 + a_1 t_1 = 0 + 0.5 \cdot 30 = 15\,\text{m s}^{-1}$ .

$s_1 = \tfrac{1}{2}(u_1 + v_1) t_1 = \tfrac{1}{2}(0 + 15)(30) = 225\,\text{m}$ .

Phase 2 (deceleration): the initial velocity is the phase-1 final velocity. $u_2 = 15$ , $v_2 = 0$ , $a_2 = -0.75$ .

$v_2 = u_2 + a_2 t_2 \implies 0 = 15 - 0.75 t_2 \implies t_2 = 20\,\text{s}$ .

$s_2 = \tfrac{1}{2}(u_2 + v_2) t_2 = \tfrac{1}{2}(15 + 0)(20) = 150\,\text{m}$ .

Totals: distance $= 225 + 150 = 375\,\text{m}$ . Time $= 30 + 20 = 50\,\text{s}$ .

Why A candidates spot this immediately:* the structure "phase A then phase B with different accelerations" is the signature of a stitch-the-SUVATs problem. Every time you see two phases, identify the shared variable (here, the velocity at the join) and use it as the bridge. The same trick handles "a ball falls, hits the ground, bounces with reduced speed, rises again" (the join is the bounce velocity), "a car brakes with one deceleration in dry conditions then a stronger one once it hits gravel" (the join is the velocity at the surface change), and "a parachutist falls freely then opens the parachute" (the join is the velocity at parachute deployment). Recognising the pattern across these contexts is exactly the synoptic skill AQA rewards.

A subtlety: when stitching phases, check the sign of the join velocity. If phase 1 ends with the body moving in the negative direction and phase 2 is described in the original positive convention, you must carry the sign correctly. Sketch a velocity-time graph as a sanity check — total displacement equals signed area, and the graph makes phase boundaries visually obvious.

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This content is aligned with the AQA A-Level Mathematics (7357) specification, Paper 3 — Mechanics, Section T: Kinematics. For the most accurate and up-to-date information, please refer to the official AQA specification document.

Visual summary

graph TD
    A["Straight-line motion problem"] --> B{"Is acceleration<br/>constant?"}
    B -->|"Yes"| C["SUVAT applies"]
    B -->|"No"| D["Use calculus:<br/>v = ds/dt, a = dv/dt"]
    C --> E{"Which variable<br/>is missing?"}
    E -->|"No s"| F["v = u + at"]
    E -->|"No v"| G["s = ut + (1/2)at^2"]
    E -->|"No t"| H["v^2 = u^2 + 2as"]
    E -->|"No a"| I["s = (1/2)(u+v)t"]
    F --> J["Choose sign convention<br/>upward positive => a = -g"]
    G --> J
    H --> J
    I --> J
    J --> K{"Single phase or<br/>multi-phase?"}
    K -->|"Single"| L["Solve, check units,<br/>round to 3 s.f."]
    K -->|"Multi-phase"| M["Stitch phases:<br/>v_final of phase n<br/>= u_initial of phase n+1"]
    M --> L
    L --> N{"Vector or<br/>scalar requested?"}
    N -->|"Velocity / displacement"| O["Keep sign"]
    N -->|"Speed / distance"| P["Take magnitude"]

    style C fill:#27ae60,color:#fff
    style J fill:#3498db,color:#fff
    style L fill:#e67e22,color:#fff

Kinematics

Kinematics

Key Quantities

The SUVAT Equations (Constant Acceleration)

Velocity-Time Graphs

Vertical Motion Under Gravity

Displacement-Time Graphs

Summary

Synoptic Links

A-Level Deep Dive: Kinematics

Spec mapping

Worked example with full mark scheme

Specimen Question (Modelled on the AQA Paper Format)

Specimen question modelled on the AQA 7357 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 on SUVAT questions

Going further — university and academic signposting

Additional A* practice: two-stage motion under different accelerations

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Visual summary

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