Kinematics: Uniform Acceleration

Kinematics is the branch of mechanics that describes motion without worrying about what causes it. Before we can understand forces and Newton’s laws, we need a solid grasp of how objects move — and the mathematical tools to describe that motion precisely.

In this lesson, we focus on uniformly accelerated motion in a straight line — the foundation of almost every mechanics problem at A-Level.

Key Definitions

Displacement (s) — the distance moved in a particular direction from a reference point. A vector quantity, measured in metres (m).

Velocity (v) — the rate of change of displacement. A vector quantity, measured in m/s.

Average velocity = total displacement / total time

Acceleration (a) — the rate of change of velocity. A vector quantity, measured in m/s².

Average acceleration = change in velocity / time taken = (v - u) / t

When acceleration is uniform (constant), we can use a special set of equations to solve problems.

The SUVAT Equations

For motion with constant acceleration in a straight line, five quantities are related by four equations. The five quantities are:

s — displacement (m)
u — initial velocity (m/s)
v — final velocity (m/s)
a — acceleration (m/s²)
t — time (s)

The four SUVAT equations:

Equation	Variables Used	Variable Missing
v = u + at	v, u, a, t	s
s = ut + (1/2)at²	s, u, a, t	v
v² = u² + 2as	v, u, a, s	t
s = (1/2)(u + v)t	s, u, v, t	a

Each equation connects four of the five SUVAT variables. The strategy for solving a problem is:

List the known quantities.
Identify the unknown you need to find.
Choose the equation that contains your three knowns and the one unknown.

flowchart TD
    A["Identify known SUVAT variables"] --> B{"Which variable\nis missing?"}
    B -->|"s missing"| C["Use v = u + at"]
    B -->|"v missing"| D["Use s = ut + ½ at²"]
    B -->|"t missing"| E["Use v² = u² + 2as"]
    B -->|"a missing"| F["Use s = ½(u + v)t"]
    B -->|"u missing"| G["Rearrange v = u + at\nor use v² = u² + 2as"]

Worked Example 1

A car accelerates uniformly from rest to 25 m/s in 10 s. Find the acceleration and the distance travelled.

Known: u = 0, v = 25 m/s, t = 10 s. Find: a, s.

Using v = u + at: 25 = 0 + a x 10 a = 25 / 10 = 2.5 m/s²

Using s = (1/2)(u + v)t: s = (1/2)(0 + 25)(10) = (1/2)(25)(10) = 125 m

Worked Example 2

A ball is thrown vertically upward at 20 m/s. How high does it rise before momentarily stopping? (Take g = 9.81 m/s²)

Known: u = 20 m/s (upward), v = 0 (at maximum height), a = -9.81 m/s² (downward). Find: s.

Using v² = u² + 2as: 0 = 20² + 2(-9.81)s 0 = 400 - 19.62s 19.62s = 400 s = 400 / 19.62 = 20.4 m

Worked Example 3

A train decelerates from 40 m/s at 2.0 m/s². How far does it travel before stopping?

Known: u = 40 m/s, v = 0, a = -2.0 m/s². Find: s.

Using v² = u² + 2as: 0 = 40² + 2(-2.0)s 0 = 1600 - 4.0s s = 1600 / 4.0 = 400 m

Time to stop: v = u + at, 0 = 40 + (-2.0)t, t = 20 s

Worked Example 4: Multi-Stage Problem

A car accelerates from rest at 3.0 m/s² for 5.0 s, then travels at constant velocity for 10 s, then decelerates at 2.0 m/s² to rest. Find the total distance.

Stage 1 (acceleration): u = 0, a = 3.0 m/s², t = 5.0 s v = u + at = 0 + 3.0 x 5.0 = 15 m/s s₁ = ut + (1/2)at² = 0 + (1/2)(3.0)(25) = 37.5 m

Stage 2 (constant velocity): v = 15 m/s, t = 10 s s₂ = vt = 15 x 10 = 150 m

Stage 3 (deceleration): u = 15 m/s, v = 0, a = -2.0 m/s² v² = u² + 2as: 0 = 225 + 2(-2.0)s₃, s₃ = 225/4.0 = 56.3 m

Total distance = 37.5 + 150 + 56.3 = 243.8 m

Displacement-Time Graphs

A displacement-time (s-t) graph plots displacement on the vertical axis against time on the horizontal axis.

Key features:

The gradient of an s-t graph equals velocity.
A straight line means constant velocity.
A horizontal line means the object is stationary (v = 0).
A curved line means the velocity is changing (the object is accelerating).
The steeper the line, the greater the velocity.
A line sloping downward means the object is moving back towards the starting point (negative displacement).

If the graph is curved, the instantaneous velocity at any point is found by drawing a tangent to the curve at that point and calculating its gradient.

Velocity-Time Graphs

A velocity-time (v-t) graph plots velocity on the vertical axis against time on the horizontal axis.

Key features:

The gradient of a v-t graph equals acceleration.
A horizontal line means constant velocity (zero acceleration).
A straight line with positive gradient means constant positive acceleration.
A straight line with negative gradient means constant deceleration (negative acceleration).
The area under a v-t graph equals displacement.

Area Under v-t Graph = Displacement

This is one of the most important results in kinematics. The area between the velocity-time line and the time axis gives the displacement:

Shape on v-t Graph	Area Formula	Situation
Rectangle	v x t	Constant velocity
Triangle	(1/2) x v x t	Uniform acceleration from rest
Trapezium	(1/2)(u + v) x t	Uniform acceleration with initial velocity

This last expression is exactly the SUVAT equation s = (1/2)(u + v)t — demonstrating that the equations and the graphs tell the same story.

Worked Example 5

A train accelerates uniformly from 10 m/s to 30 m/s over 40 s. Find: (a) the acceleration, (b) the displacement.

(a) Gradient of v-t graph = acceleration: a = (30 - 10) / 40 = 20 / 40 = 0.50 m/s²

(b) Area under v-t graph (trapezium): s = (1/2)(10 + 30)(40) = (1/2)(40)(40) = 800 m

Deriving SUVAT from Graphs

The SUVAT equations can be derived from the v-t graph for uniform acceleration:

Starting with a straight line from (0, u) to (t, v):

Gradient: a = (v - u)/t, which rearranges to v = u + at (equation 1).
Area (trapezium): s = (1/2)(u + v)t — equation 4.
Substituting v = u + at into equation 4: s = (1/2)(u + u + at)t = ut + (1/2)at² — equation 2.
From equation 1: t = (v - u)/a. Substituting into equation 4: s = (1/2)(u + v)(v - u)/a = (v² - u²)/(2a), which gives v² = u² + 2as (equation 3).

Understanding these derivations helps you see that all four SUVAT equations are simply different ways of extracting information from the same v-t graph.

Free-Fall and Vertical Motion

Objects in free fall near the Earth’s surface accelerate at g = 9.81 m/s² downward. Important special cases:

Situation	Initial Conditions	Key Result
Dropped from rest	u = 0, a = g downward	v = gt, s = (1/2)gt²
Thrown upward	u = positive upward, a = -g	Time to top: t = u/g; Max height: u²/(2g)
Thrown downward	u = positive downward, a = g	Both v and s increase faster than dropped

Worked Example 6: Symmetry of Vertical Throw

A ball is thrown vertically upward at 15 m/s. Find (a) the time to reach maximum height, (b) the maximum height, (c) the total time of flight, (d) the speed on return.

(a) At maximum height, v = 0: 0 = 15 + (-9.81)t, t = 15/9.81 = 1.53 s

(b) Using v² = u² + 2as: 0 = 225 - 19.62s, s = 225/19.62 = 11.5 m

(d) By symmetry, the ball returns at the same speed it was thrown: 15 m/s (but directed downward).

Common Pitfalls

Sign conventions: Choose a positive direction and stick to it. If upward is positive, then g = -9.81 m/s² for a falling object.
SUVAT only applies to uniform acceleration. If acceleration changes, you must use graphs or split the problem into stages.
“From rest” means u = 0. “Comes to rest” means v = 0.
“Freely falling” means the only force is gravity — so a = g = 9.81 m/s² downward (or -9.81 if upward is positive).
Confusing velocity and speed: A ball thrown upward has zero velocity at the top but non-zero acceleration. Zero velocity does not mean zero acceleration.
Using the wrong equation: If you only know three SUVAT variables, there is exactly one equation you can use. Don’t try to use an equation that requires a variable you haven’t found yet.
Rounding too early: Keep intermediate values to at least 4 significant figures; only round your final answer to match the question data (usually 2 or 3 s.f.).

Exam Tips for Edexcel Physics

In "show that" questions, you must arrive at the given answer using clear algebraic steps. Do not work backwards from the answer.
When drawing or interpreting v-t graphs, label axes with quantities and units.
If a question involves multiple stages of motion (e.g. acceleration then constant speed then braking), draw a v-t graph first — it makes the problem much clearer.
Always state which SUVAT equation you are using before substituting values. This earns method marks even if arithmetic goes wrong.
For vertical motion questions, always define your positive direction at the start of your answer (e.g. "Taking upward as positive").

Key Points

Five SUVAT variables: s, u, v, a, t. Four equations, each linking four of the five.
The gradient of an s-t graph gives velocity; the gradient of a v-t graph gives acceleration.
The area under a v-t graph gives displacement.
Always define a positive direction before starting a calculation.
SUVAT equations are only valid for constant acceleration.
For multi-stage problems, apply SUVAT separately to each stage where acceleration is constant.

A-Level Deep Dive: Kinematics: Uniform Acceleration

Spec mapping

Edexcel 9PH0 specification, Topic 2 — Mechanics, sub-topic 2.2 (Motion) develops the formal description of motion in a straight line under constant acceleration: definitions of displacement, velocity and acceleration as vector quantities; interpretation of displacement-time and velocity-time graphs (gradients and areas); and use of the SUVAT equations for uniformly accelerated motion (refer to the official Pearson Edexcel specification document for exact wording). Although this is a Year-1 topic introduced early in the AS course, kinematics underpins almost every later mechanics question. The SUVAT toolkit is reused in Topic 2 sub-topic on projectile motion when horizontal and vertical components are decoupled, in Topic 6 (Further mechanics) when momentum and impulse are reasoned about over a finite acceleration phase, in Topic 5 (Waves) when oscillator displacement is differentiated to recover velocity, and in Topic 8 (Nuclear and particle physics) for collision recoil reasoning. The Edexcel data, formulae and relationships booklet does list the four SUVAT equations explicitly, but it does not state the sign convention — selecting an axis and reporting which direction is positive is left entirely to the candidate.

Worked example with full mark scheme

Question (8 marks):

A car of mass 1200 kg starts from rest and accelerates uniformly to a speed of 24.0 m s⁻¹ over a time of 8.00 s. The driver then maintains this speed for 2.00 s before applying the brakes, which produce a uniform deceleration of magnitude 6.00 m s⁻². The car comes to rest.

(a) Sketch a labelled velocity-time graph for the entire motion. (2)

(b) Calculate the total distance travelled from start to rest. (4)

(c) The driver claims the average speed for the whole journey was 18.0 m s⁻¹. Determine, with working, whether this claim is correct. (2)

Solution with mark scheme:

(a) B1 — three straight-line segments correctly drawn: a positive-gradient line from (0, 0) to (8.00, 24.0); a horizontal line from (8.00, 24.0) to (10.00, 24.0); a negative-gradient line from (10.00, 24.0) to the time axis at 14.00 s. B1 — axes labelled with quantities ("velocity / m s⁻¹", "time / s") and key values marked.

(b) Phase 1 — acceleration from rest. Take the direction of motion as positive throughout.

Known: $u = 0$ , $v = 24.0$ m s⁻¹, $t = 8.00$ s. Using $s = \tfrac{1}{2}(u + v)t$ :

$s_1 = \tfrac{1}{2}(0 + 24.0)(8.00) = 96.0 \text{ m}$

M1 — correct SUVAT equation chosen with the variable missing being $a$ . A1 — $s_1 = 96.0$ m.

Phase 2 — constant velocity. $s_2 = vt = 24.0 \times 2.00 = 48.0$ m.

Phase 3 — braking to rest. $u = 24.0$ m s⁻¹, $v = 0$ , $a = -6.00$ m s⁻². Using $v^2 = u^2 + 2as$ :

$0 = (24.0)^2 + 2(-6.00)s_3 \implies s_3 = \dfrac{576}{12.0} = 48.0 \text{ m}$

M1 — correct phase-3 equation, with the negative sign on $a$ explicitly shown.

A1 — total distance $s = 96.0 + 48.0 + 48.0 = 192$ m (3 sf).

Average speed $= 192 / 14.0 = 13.7$ m s⁻¹ (3 sf).

M1 — average speed computed as total distance over total time (not the arithmetic mean of phase speeds, a common error).

A1 — explicit conclusion that $13.7 \neq 18.0$ , so the driver's claim is incorrect; the value 18.0 m s⁻¹ would arise only from $\tfrac{1}{2}(0 + 24.0 + 24.0)/...$ style miscalculations or from using the mean of $u$ and $v$ for phase 1 alone.

Total: 8 marks (B2 M3 A3).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A cyclist passes the start of a long, straight slope travelling at 4.00 m s⁻¹. The cyclist then accelerates uniformly down the slope at 1.20 m s⁻². A point P is marked 50.0 m down the slope from the start.

(a) Calculate the speed of the cyclist as they pass P. (2)

(b) Determine the time taken to travel from the start to P. (2)

(c) The cyclist's measured time over the 50.0 m was 7.5 s. Discuss, with reference to your answer to (b), one physical reason why the measured time exceeds the calculated time. (2)

Mark scheme decomposition by AO:

Part	AO weighting	Earned by
(a) M1	AO1.2	Selecting $v^2 = u^2 + 2as$ with the variable missing being $t$ .
(a) A1	AO1.1b	$v = \sqrt{16.0 + 120} = \sqrt{136} = 11.7$ m s⁻¹ (3 sf).
(b) M1	AO2.1	Choosing $v = u + at$ and rearranging $t = (v - u)/a$ .
(b) A1	AO1.1b	$t = (11.7 - 4.00)/1.20 = 6.39$ s (3 sf).
(c)	AO3.2a	Identifies a physical effect omitted from the model — typically air resistance or rolling friction with the road — that opposes motion and reduces the effective acceleration.
(c)	AO3.2b	Concludes that the actual acceleration is therefore lower than 1.20 m s⁻², so the cyclist takes longer to reach P; alternatively, that the SUVAT analysis assumes uniform acceleration throughout, which the real situation does not satisfy.

Total: 6 marks split AO1 = 3, AO2 = 1, AO3 = 2. This is a typical Paper 1 SUVAT-with-evaluation item: a routine calculation followed by an AO3 prompt that rewards candidates who can compare an idealised model with a real measurement and name the missing physics rather than invoking generic "experimental error".

Synoptic links

Dynamics — Newton's second law (sub-topic 2.3): SUVAT yields the acceleration; $F = ma$ then yields the resultant force. The standard exam construction is "calculate the acceleration from the kinematics, then deduce the braking force on a 1200 kg car" — two topics in one question. Constant acceleration in kinematics corresponds exactly to constant resultant force in dynamics.