What is variable acceleration in the context of Edexcel International A Levels Mechanics 2?

Variable acceleration refers to a situation where the acceleration of an object is not constant but changes with time or position. In the Edexcel International A Levels Mechanics 2 syllabus, students learn to analyze motion where acceleration is a function of time, velocity, or displacement, using calculus techniques to solve related problems.

How do you find velocity from variable acceleration in Mechanics 2?

To find velocity from variable acceleration in Mechanics 2, you integrate the acceleration function with respect to time. If acceleration a(t) is given as a function of time, the velocity v(t) can be found by integrating a(t) with respect to t, and applying initial conditions to determine the constant of integration.

What are common methods to solve problems involving variable acceleration in Edexcel Mechanics 2?

Common methods to solve variable acceleration problems in Edexcel Mechanics 2 include using calculus techniques such as differentiation and integration. Students often use these methods to derive expressions for velocity and displacement from a given acceleration function. Additionally, understanding the relationships between these quantities and applying initial conditions are crucial for solving these problems.

How does variable acceleration differ from constant acceleration in Mechanics 2?

Variable acceleration differs from constant acceleration in that the rate of change of velocity is not uniform over time. In constant acceleration, the acceleration value remains the same, simplifying the equations of motion. In contrast, variable acceleration requires the use of calculus to handle the changing acceleration, making the analysis more complex and requiring integration to find velocity and displacement.

Can you provide an example of a variable acceleration problem in Edexcel International A Levels Mechanics 2?

Certainly! Consider a problem where the acceleration of a particle is given by a(t) = 3t^2. To find the velocity function, integrate a(t) with respect to time: v(t) = ∫3t^2 dt = t^3 + C, where C is the constant of integration. If the initial velocity at t = 0 is 5 m/s, then C = 5, giving v(t) = t^3 + 5. This example illustrates how calculus is used to solve variable acceleration problems in the Edexcel International A Levels Mechanics 2 curriculum.

Why is "Applying suvat formulae when acceleration is variable" a common mistake in Variable Acceleration?

Why it happens: Suvat works for almost everything in M1, so students reach for it reflexively. How to avoid it: Suvat REQUIRES constant acceleration. If a is a function of t (or anything else), you MUST use calculus: v = a\,dt, s = v\,dt. Source: WME02/01 examiner reports — flagged annually

Why is "Forgetting the constant of integration" a common mistake in Variable Acceleration?

Why it happens: Definite integrals are common in pure maths; here we work with indefinites and need C. How to avoid it: Every time you integrate, write '+ C'. Then substitute the initial condition (usually given: 'when t = 0, v = ' or 'starts at the origin') to find C before going further.

Why is "Computing $\int v\,dt$ for 'total distance travelled' when $v$ changes sign" a common mistake in Variable Acceleration?

Why it happens: The two are equal only when v does not change sign. How to avoid it: Find when v = 0 in the interval. Split the integral. Take absolute values of each piece and add. Source: WME02/01 examiner reports — recurring

Why is "Calling the magnitude of velocity 'velocity' (not 'speed')" a common mistake in Variable Acceleration?

Why it happens: Loose everyday English — physics demands the distinction. How to avoid it: If the question says 'find the velocity at t = 2', give the VECTOR v(2) = 9i + 8j. If it says 'find the speed', give the scalar |v(2)| = √(145).

Why is "Omitting units on the final answer" a common mistake in Variable Acceleration?

Why it happens: Focus on the algebra. How to avoid it: Distance/displacement: m. Velocity/speed: m s^-1. Acceleration: m s^-2. Always quote them — B-marks for units are routine.

Mechanics 2Edexcel International A Levels Mathematics (XMA01-YMA01)

Variable Acceleration

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Variable Acceleration Study Notes — Edexcel IAL Mathematics M2 (WME02/01, 2018 spec — 2026 onwards)

Kinematics with non-constant acceleration: differentiate position to get velocity and acceleration; integrate acceleration to recover velocity and position. Same calculus in vector form for 2D motion. Suvat does NOT apply.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

M2 2.1 — Use differentiation to find velocity and acceleration from displacement as a function of time.
M2 2.2 — Use integration to find velocity from acceleration, and displacement from velocity, applying boundary conditions.
M2 2.3 — Apply calculus methods to vector kinematics in two dimensions.

The calculus link: position, velocity, acceleration

Differentiate to go 'down' (position → velocity → acceleration). Integrate to go 'up'.

The three quantities.

Quantity	Symbol	Unit
Displacement	$s(t)$	m
Velocity	$v(t) = \dfrac{ds}{dt}$	m s $^{-1}$
Acceleration	$a(t) = \dfrac{dv}{dt} = \dfrac{d^{2}s}{dt^{2}}$	m s $^{-2}$

To go the other way, integrate:

$v(t) = \int a(t)\,dt + C_{1}, \qquad s(t) = \int v(t)\,dt + C_{2}.$

The constants $C_{1}, C_{2}$ are determined by initial conditions — typically $v(0)$ and $s(0)$ , given in the question.

Why suvat fails. Suvat formulae ( $v = u + at$ , etc.) assume constant $a$ . When $a$ depends on $t$ , the formulae no longer hold. Calculus is the general tool.

Worked example — differentiation. $s = 2t^{3} - 9t^{2} + 12t + 5$ .

$v = ds/dt = 6t^{2} - 18t + 12$ .
$a = dv/dt = 12t - 18$ .

So $a(3) = 18$ m s $^{-2}$ .

Worked example — integration. Given $a = 6t - 4$ , $v(0) = 5$ , $s(0) = 0$ .

$v = \int (6t - 4)\,dt = 3t^{2} - 4t + C_{1}$ . Use $v(0) = 5$ : $C_{1} = 5$ . So $v = 3t^{2} - 4t + 5$ .
$s = \int v\,dt = t^{3} - 2t^{2} + 5t + C_{2}$ . Use $s(0) = 0$ : $C_{2} = 0$ . So $s = t^{3} - 2t^{2} + 5t$ .

Then $s(3) = 27 - 18 + 15 = 24$ m.

Differentiate: $s \to v \to a$ .
Integrate: $a \to v \to s$ (with constants of integration).
Boundary conditions pin down the constants.
Suvat does NOT apply; reach for calculus reflexively.

Instantaneously at rest, changing direction

Set $v(t) = 0$ to find when the particle is momentarily stationary.

'Instantaneously at rest' means $v(t) = 0$ at that instant. The particle may be reversing direction (if $a \neq 0$ at that instant), or genuinely starting/stopping.

Worked example. $v = t^{2} - 4t + 3 = (t - 1)(t - 3)$ .

$v(t) = 0$ at $t = 1$ and $t = 3$ .
Sign chart: $v > 0$ on $[0, 1)$ , $v < 0$ on $(1, 3)$ , $v > 0$ on $(3, \infty)$ .
The particle moves forward, comes to rest at $t = 1$ , reverses (moves backward), comes to rest at $t = 3$ , then moves forward again.

Maximum displacement, maximum speed. Both occur at stationary points:

Max/min of $s$ at $ds/dt = 0$ , i.e. $v = 0$ .
Max/min of $v$ at $dv/dt = 0$ , i.e. $a = 0$ .

Worked example. Find the maximum value of $|v|$ on $[0, 4]$ for $v = t^{2} - 4t + 3$ .

$a = 2t - 4 = 0$ at $t = 2$ , where $v(2) = 4 - 8 + 3 = -1$ .
Endpoints: $v(0) = 3$ , $v(4) = 3$ .
Maximum $|v| = \max(|3|, |-1|, |3|) = 3$ m/s.

At rest ⇔ $v = 0$ .
Max/min of $s$ at $v = 0$ ; max/min of $v$ at $a = 0$ .
Sign chart to know when the particle reverses.
Don't forget endpoints when finding extrema on an interval.

Distance vs displacement

Distance is total path length; displacement is net change in position. They differ when $v$ changes sign.

Displacement from $t = a$ to $t = b$ :

$\Delta s = s(b) - s(a) = \int_{a}^{b} v(t)\,dt.$

Distance travelled (a non-negative quantity):

$D = \int_{a}^{b} |v(t)|\,dt.$

These are equal only if $v$ does not change sign on $[a, b]$ . Otherwise $D > |\Delta s|$ .

Algorithm for distance travelled.

Solve $v(t) = 0$ in $[a, b]$ . Call the roots $t_{1} < t_{2} < \ldots$ .
Compute $s$ at $a, t_{1}, t_{2}, \ldots, b$ .
Add $|s(t_{k+1}) - s(t_{k})|$ across each piece.

Worked example. $v = t^{2} - 4t + 3$ , $s(0) = 0$ , distance from $t = 0$ to $t = 4$ .

Roots of $v$ : $t = 1, 3$ .
$s(t) = t^{3}/3 - 2t^{2} + 3t$ . Compute: $s(0) = 0$ , $s(1) = 4/3$ , $s(3) = 0$ , $s(4) = 4/3$ .
Distance $= |4/3 - 0| + |0 - 4/3| + |4/3 - 0| = 4$ m.

The displacement over the same interval is $s(4) - s(0) = 4/3$ m — only one third of the distance travelled.

Displacement = $\int v\,dt$ (signed).
Distance = $\int |v|\,dt$ (always positive).
Split at sign changes of $v$ .
Add absolute values of pieces.

Vector kinematics in 2D

Same calculus, applied component-by-component to the position vector.

Setup. Position vector $\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}$ in metres.

Differentiation. Component-by-component:

$\mathbf{v}(t) = \dfrac{d\mathbf{r}}{dt} = \dfrac{dx}{dt}\mathbf{i} + \dfrac{dy}{dt}\mathbf{j}, \qquad \mathbf{a}(t) = \dfrac{d\mathbf{v}}{dt} = \dfrac{d^{2}x}{dt^{2}}\mathbf{i} + \dfrac{d^{2}y}{dt^{2}}\mathbf{j}.$

Integration. Component-by-component, with a vector constant of integration:

$\mathbf{v}(t) = \int \mathbf{a}(t)\,dt + \mathbf{C}_{1}, \qquad \mathbf{r}(t) = \int \mathbf{v}(t)\,dt + \mathbf{C}_{2}.$

Speed. The magnitude of the velocity vector:

$\text{speed} = |\mathbf{v}| = \sqrt{v_x^{2} + v_y^{2}}.$

Direction. Angle to $\mathbf{i}$ -axis: $\theta = \arctan(v_y / v_x)$ — pay attention to the quadrant.

Worked example. $\mathbf{r} = (t^{3} - 3t)\mathbf{i} + (2t^{2} + 1)\mathbf{j}$ .

$\mathbf{v} = (3t^{2} - 3)\mathbf{i} + 4t\,\mathbf{j}$ .
$\mathbf{a} = 6t\,\mathbf{i} + 4\mathbf{j}$ .
At $t = 2$ : $\mathbf{v}(2) = 9\mathbf{i} + 8\mathbf{j}$ , $|\mathbf{v}(2)| = \sqrt{145} \approx 12.04$ m/s.

Common question variants.

"Find $t$ when the velocity is perpendicular to $\mathbf{i}$ ": $v_x = 0$ .
"Find $t$ when the velocity is parallel to $\mathbf{i} + \mathbf{j}$ ": $v_y = v_x$ .
"Find $t$ when speed is minimum": differentiate $|\mathbf{v}|^{2} = v_x^{2} + v_y^{2}$ and set to zero. (Easier than differentiating $|\mathbf{v}|$ .)

Differentiate / integrate component-by-component.
Speed = $|\mathbf{v}|$ via Pythagoras.
Vector constants of integration in 2D problems.
Velocity perpendicular / parallel conditions reduce to component equations.

Typical M2 question structure

What examiners ask and how marks are distributed.

Variable acceleration questions on WME02/01 are usually 6-10 marks, often Q1 or Q2 (i.e. early in the paper, used as an entry-level question). Typical structures:

1. Forward direction (differentiate).

Given $s(t)$ or $\mathbf{r}(t)$ .
Find $v$ or $\mathbf{v}$ at a specific time, and $a$ or $\mathbf{a}$ .
Find time of zero velocity / acceleration.

2. Reverse direction (integrate).

Given $a(t)$ or $\mathbf{a}(t)$ + initial conditions.
Find $v(t)$ , $s(t)$ , then evaluate at a given time.

3. Distance vs displacement.

Given $v(t)$ with at least one sign change in the interval.
Find both displacement and total distance — students who confuse these lose 3+ marks.

4. Vector with perpendicular / parallel direction conditions.

"Find $t$ when velocity is perpendicular to position vector": dot product $\mathbf{r} \cdot \mathbf{v} = 0$ .
"Find $t$ when velocity is parallel to $\mathbf{i}$ ": $v_y = 0$ .

Marking pattern.

M-marks for setting up the differentiation / integration.
A-marks for the correct expressions.
B-marks for boundary conditions and units.
ECF carries through generously.

A tip.* Always write the function you're differentiating / integrating in standard polynomial form (descending powers) before applying calculus. It costs a line of working and prevents algebra slips.

Q1/Q2 entry-level: 5-8 marks.
Forward (differentiate) vs reverse (integrate) direction.
Distance ≠ displacement trap appears yearly.
Vector perpendicular = dot product zero.

How it’s examined

Variable acceleration appears on every WME02/01 paper, almost always as a short question (Q1, Q2 or Q3), worth 5-10 marks. The 1D version (scalar) is most common; the 2D vector version appears in roughly half of all sittings. Examiners frequently include a part-(b) that distinguishes 'distance travelled' from 'displacement' — students who confuse these lose 3+ marks per question. ECF (error carried forward) is applied cleanly: a single integration slip in the antiderivative typically costs only one A-mark. A* candidates should target full marks here — the material is straightforward calculus.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WME02/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.

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Step-by-step worked examples — Variable Acceleration

Step-by-step solutions to past-paper-style questions on variable acceleration, written exactly the way a tutor would explain them at the board.

1Find displacement from velocity (5 marks, M2)
Core• Adapted from WME02/01 January 2024 Q1• variable acceleration, integration, kinematics
Question
A particle $P$ moves on a straight line so that at time $t$ seconds its velocity is $v = 3t^{2} - 12t + 9$ m s $^{-1}$ (for $t \geq 0$ ). The particle is at the origin when $t = 0$ . Find the displacement of $P$ from the origin when $t = 4$ s.
Step-by-step solution
1. Step 1
  Displacement is the integral of velocity: $s = \int v\,dt$ .
  $s(t) = \int (3t^{2} - 12t + 9)\,dt = t^{3} - 6t^{2} + 9t + C$
2. Step 2
  Apply the initial condition $s(0) = 0$ : $C = 0$ .
  $s(t) = t^{3} - 6t^{2} + 9t$
3. Step 3
  Evaluate at $t = 4$ :
  $s(4) = 64 - 96 + 36 = 4\,\text{m}$
Answer
Displacement from the origin is $4$ m.
Examiner tip
M1 for integrating $v$ . A1 for the correct antiderivative. M1 for using the initial condition $s(0) = 0$ (constant of integration). M1 for substituting $t = 4$ . A1 for $4$ m. This is the asked-for displacement, NOT the distance travelled — distance would require splitting where $v$ changes sign and integrating $|v|$ piecewise.
2Acceleration at a given instant (4 marks, M2)
Core• Adapted from WME02/01 June 2024 Q1• variable acceleration, differentiation
Question
A particle moves so that its displacement from the origin is $s = 2t^{3} - 9t^{2} + 12t + 5$ metres at time $t$ seconds. Find the acceleration when $t = 3$ s.
Step-by-step solution
1. Step 1
  Differentiate $s$ to get $v$ :
  $v = \dfrac{ds}{dt} = 6t^{2} - 18t + 12$
2. Step 2
  Differentiate $v$ to get $a$ :
  $a = \dfrac{dv}{dt} = 12t - 18$
3. Step 3
  Evaluate at $t = 3$ :
  $a(3) = 36 - 18 = 18\,\text{m s}^{-2}$
Answer
Acceleration at $t = 3$ is $18$ m s $^{-2}$ .
Examiner tip
M1 for differentiating once. A1 for $v(t)$ . M1 for differentiating again. A1 for $18$ m s $^{-2}$ . Units MUST be quoted on the final answer at IAL.
3Find when particle is instantaneously at rest, and distance travelled (8 marks, M2)
Extended• Adapted from WME02/01 January 2023 Q2• variable acceleration, distance travelled
Question
A particle moves on a straight line. Its velocity is $v = t^{2} - 4t + 3$ m s $^{-1}$ for $t \geq 0$ . The particle starts at the origin. (a) Find the values of $t$ at which the particle is instantaneously at rest. (b) Find the total distance travelled by the particle in the first $4$ seconds.
Step-by-step solution
1. Step 1
  (a) At rest: $v = 0$ . Factorise $t^{2} - 4t + 3 = (t - 1)(t - 3) = 0$ , so $t = 1$ s and $t = 3$ s.
2. Step 2
  (b) Sign of $v$ : $v > 0$ on $[0, 1)$ and $(3, 4]$ , $v < 0$ on $(1, 3)$ . Distance = $\int_{0}^{1} v\,dt - \int_{1}^{3} v\,dt + \int_{3}^{4} v\,dt$ (or equivalently $\int_{0}^{4} |v|\,dt$ ).
3. Step 3
  Antiderivative: $\int v\,dt = \dfrac{t^{3}}{3} - 2t^{2} + 3t$ .
4. Step 4
  Position at $t = 0, 1, 3, 4$ :
  $s(0) = 0,\;\; s(1) = \tfrac{1}{3} - 2 + 3 = \tfrac{4}{3},\;\; s(3) = 9 - 18 + 9 = 0,\;\; s(4) = \tfrac{64}{3} - 32 + 12 = \tfrac{4}{3}$
5. Step 5
  Distance travelled = $|s(1) - s(0)| + |s(3) - s(1)| + |s(4) - s(3)| = \tfrac{4}{3} + \tfrac{4}{3} + \tfrac{4}{3} = 4$ m.
Answer
(a) $t = 1$ s and $t = 3$ s. (b) Total distance $= 4$ m.
Examiner tip
(a) M1 for setting $v = 0$ . A1 for both times. (b) M1 for recognising distance $\neq$ displacement and identifying sign changes. M1 for splitting the integral. A1 for the antiderivative. A1 for the three pieces. A1 for $4$ m. A common error is computing $\int_{0}^{4} v\,dt = 4/3$ (the displacement) and quoting it as the distance.
42D vector motion: velocity, acceleration, speed (8 marks, M2)
Extended• Adapted from WME02/01 June 2023 Q3• variable acceleration, vectors
Question
A particle moves in a plane so that its position vector at time $t$ seconds is $\mathbf{r} = (t^{3} - 3t)\mathbf{i} + (2t^{2} + 1)\mathbf{j}$ metres. Find (a) the velocity vector $\mathbf{v}$ when $t = 2$ , (b) the acceleration vector $\mathbf{a}$ at $t = 2$ , (c) the speed at $t = 2$ .
Step-by-step solution
1. Step 1
  Differentiate $\mathbf{r}$ component-by-component:
  $\mathbf{v} = \dfrac{d\mathbf{r}}{dt} = (3t^{2} - 3)\mathbf{i} + 4t\,\mathbf{j}$
2. Step 2
  At $t = 2$ : $\mathbf{v}(2) = (12 - 3)\mathbf{i} + 8\mathbf{j} = 9\mathbf{i} + 8\mathbf{j}$ m s $^{-1}$ .
3. Step 3
  Differentiate $\mathbf{v}$ :
  $\mathbf{a} = \dfrac{d\mathbf{v}}{dt} = 6t\,\mathbf{i} + 4\mathbf{j}$
4. Step 4
  At $t = 2$ : $\mathbf{a}(2) = 12\mathbf{i} + 4\mathbf{j}$ m s $^{-2}$ .
5. Step 5
  Speed = $|\mathbf{v}(2)| = \sqrt{9^{2} + 8^{2}} = \sqrt{81 + 64} = \sqrt{145} \approx 12.04$ m s $^{-1}$ .
Answer
(a) $\mathbf{v}(2) = 9\mathbf{i} + 8\mathbf{j}$ m s $^{-1}$ . (b) $\mathbf{a}(2) = 12\mathbf{i} + 4\mathbf{j}$ m s $^{-2}$ . (c) Speed $= \sqrt{145} \approx 12.04$ m s $^{-1}$ .
Examiner tip
(a) M1 for differentiating both components. A1 for $\mathbf{v}(2)$ . (b) M1 for second derivative. A1 for $\mathbf{a}(2)$ . (c) M1 for $|\mathbf{v}|$ using Pythagoras. A1 for the speed. B1 for units. Note: 'velocity' is the vector $\mathbf{v}$ , 'speed' is the scalar $|\mathbf{v}|$ — these are distinct in IAL.
5Find velocity and position from acceleration (10 marks, M2)
Extended• Adapted from WME02/01 June 2024 Q5• variable acceleration, integration, initial conditions
Question
A particle moves on a straight line with acceleration $a = 6t - 4$ m s $^{-2}$ at time $t$ seconds, for $t \geq 0$ . When $t = 0$ , the particle is at the origin with velocity $5$ m s $^{-1}$ . Find (a) $v$ in terms of $t$ , (b) $s$ in terms of $t$ , (c) the displacement when $t = 3$ s.
Step-by-step solution
1. Step 1
  (a) Integrate $a$ to get $v$ :
  $v = \int (6t - 4)\,dt = 3t^{2} - 4t + C_{1}$
2. Step 2
  Apply $v(0) = 5$ : $C_{1} = 5$ . So $v = 3t^{2} - 4t + 5$ .
3. Step 3
  (b) Integrate $v$ to get $s$ :
  $s = \int (3t^{2} - 4t + 5)\,dt = t^{3} - 2t^{2} + 5t + C_{2}$
4. Step 4
  Apply $s(0) = 0$ : $C_{2} = 0$ . So $s = t^{3} - 2t^{2} + 5t$ .
5. Step 5
  (c) Substitute $t = 3$ : $s(3) = 27 - 18 + 15 = 24$ m.
Answer
(a) $v = 3t^{2} - 4t + 5$ m s $^{-1}$ . (b) $s = t^{3} - 2t^{2} + 5t$ m. (c) $s(3) = 24$ m.
Examiner tip
(a) M1 for integrating $a$ . A1 for the form. M1 for using $v(0) = 5$ . A1 for $v(t)$ . (b) M1 for integrating $v$ . A1 for the form. M1 for $s(0) = 0$ . A1 for $s(t)$ . (c) B1 for $24$ m with units. Forgetting the constants of integration (or using them lazily without explicit initial-condition statements) loses two M-marks reliably.

Key Formulae — Variable Acceleration

The formulae you need to memorise for variable acceleration on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Differentiation: position → velocity → acceleration
$v = \dfrac{ds}{dt}, \quad a = \dfrac{dv}{dt} = \dfrac{d^{2}s}{dt^{2}}$
$s$
displacement (m), function of $t$
$v$
velocity (m s $^{-1}$ )
$a$
acceleration (m s $^{-2}$ )
When to use
When you are given displacement (or position vector) and need velocity or acceleration. NOT in the formula booklet — but the calculus is core P1 material.
Example
$s = 2t^{3} - 9t^{2} + 12t$ : $v = 6t^{2} - 18t + 12$ , $a = 12t - 18$ .
Integration: acceleration → velocity → displacement
$v = \int a\,dt + C_{1}, \quad s = \int v\,dt + C_{2}$
$C_{1}$
constant of integration — find from $v(0)$
$C_{2}$
constant of integration — find from $s(0)$
When to use
When acceleration (or velocity) is given as a function of $t$ and you need to recover the others. Always state the initial condition used. NOT in the formula booklet.
Example
$a = 6t - 4$ , $v(0) = 5$ : $v = 3t^{2} - 4t + 5$ .
Vector kinematics
$\mathbf{v} = \dfrac{d\mathbf{r}}{dt}, \quad \mathbf{a} = \dfrac{d\mathbf{v}}{dt} = \dfrac{d^{2}\mathbf{r}}{dt^{2}}$
$\mathbf{r}$
position vector (m)
$\mathbf{v}$
velocity vector (m s $^{-1}$ )
$\mathbf{a}$
acceleration vector (m s $^{-2}$ )
When to use
Two-dimensional motion in the $\mathbf{i}, \mathbf{j}$ plane. Differentiate / integrate component-by-component. NOT in the formula booklet.
Example
$\mathbf{r} = t^{2}\mathbf{i} + (2t + 1)\mathbf{j}$ : $\mathbf{v} = 2t\mathbf{i} + 2\mathbf{j}$ , $\mathbf{a} = 2\mathbf{i}$ .
Distance travelled (not displacement)
$\text{distance} = \int_{t_{1}}^{t_{2}} |v|\,dt$
$|v|$
speed (always non-negative)
$[t_{1}, t_{2}]$
interval of motion
When to use
When the velocity changes sign in the interval — the particle reverses direction. Split into pieces where $v$ keeps its sign and add the absolute values. NOT in the formula booklet.
Example
$v = (t - 1)(t - 3)$ on $[0, 4]$ : split at $t = 1$ and $t = 3$ .
Speed from velocity vector
$|\mathbf{v}| = \sqrt{v_x^{2} + v_y^{2}}$
$v_x, v_y$
$\mathbf{i}$ and $\mathbf{j}$ components of $\mathbf{v}$
When to use
Whenever 'speed' is asked for in a 2D vector problem. Note: 'velocity' is the vector itself. NOT in the formula booklet.
Example
$\mathbf{v} = 9\mathbf{i} + 8\mathbf{j}$ : $|\mathbf{v}| = \sqrt{145} \approx 12.04$ m/s.

Key Definitions and Keywords — Variable Acceleration

Definitions to memorise and the exact keywords mark schemes credit for variable acceleration answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Variable acceleration
Examiner keyword
Acceleration that depends on time (or on position/velocity, but in M2 only time-dependence is examined). Distinguished from 'constant acceleration', for which the suvat formulae apply.
Velocity vs. speed
Examiner keyword
Velocity is a vector: magnitude AND direction. Speed is the scalar magnitude of velocity. In 1D, velocity is positive or negative; speed is $|v|$ .
Displacement vs. distance travelled
Examiner keyword
Displacement is the net change in position $s(t_{2}) - s(t_{1})$ (signed). Distance travelled is the total path length $\int |v|\,dt$ (unsigned, always $\geq$ displacement).
Instantaneously at rest
Examiner keyword
At the instant when $v = 0$ . The particle may be momentarily stationary while reversing direction. Set $v(t) = 0$ and solve for $t$ .
Position vector $\mathbf{r}$
Vector from a fixed origin to the particle. In 2D: $\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}$ . Velocity $\mathbf{v} = d\mathbf{r}/dt$ ; acceleration $\mathbf{a} = d^{2}\mathbf{r}/dt^{2}$ .

Common Mistakes and Misconceptions — Variable Acceleration

The traps other students keep falling into on variable acceleration questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Applying suvat formulae when acceleration is variable
WME02/01 examiner reports — flagged annually
Why it happens
Suvat works for almost everything in M1, so students reach for it reflexively.
How to avoid it
Suvat REQUIRES constant acceleration. If $a$ is a function of $t$ (or anything else), you MUST use calculus: $v = \int a\,dt$ , $s = \int v\,dt$ .
✕Forgetting the constant of integration
Why it happens
Definite integrals are common in pure maths; here we work with indefinites and need $C$ .
How to avoid it
Every time you integrate, write ' $+ C$ '. Then substitute the initial condition (usually given: 'when $t = 0$ , $v = \ldots$ ' or 'starts at the origin') to find $C$ before going further.
✕Computing $\int v\,dt$ for 'total distance travelled' when $v$ changes sign
WME02/01 examiner reports — recurring
Why it happens
The two are equal only when $v$ does not change sign.
How to avoid it
Find when $v = 0$ in the interval. Split the integral. Take absolute values of each piece and add.
✕Calling the magnitude of velocity 'velocity' (not 'speed')
Why it happens
Loose everyday English — physics demands the distinction.
How to avoid it
If the question says 'find the velocity at $t = 2$ ', give the VECTOR $\mathbf{v}(2) = 9\mathbf{i} + 8\mathbf{j}$ . If it says 'find the speed', give the scalar $|\mathbf{v}(2)| = \sqrt{145}$ .
✕Omitting units on the final answer
Why it happens
Focus on the algebra.
How to avoid it
Distance/displacement: m. Velocity/speed: m s $^{-1}$ . Acceleration: m s $^{-2}$ . Always quote them — B-marks for units are routine.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Variable Acceleration — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Variable Acceleration

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Find displacement from velocity (5 marks, M2)

2Acceleration at a given instant (4 marks, M2)

3Find when particle is instantaneously at rest, and distance travelled (8 marks, M2)

42D vector motion: velocity, acceleration, speed (8 marks, M2)

5Find velocity and position from acceleration (10 marks, M2)

Differentiation: position → velocity → acceleration

Integration: acceleration → velocity → displacement

Vector kinematics

Distance travelled (not displacement)

Speed from velocity vector

Variable acceleration

Velocity vs. speed

Displacement vs. distance travelled

Instantaneously at rest

Position vector r\mathbf{r}r

✕Applying suvat formulae when acceleration is variable

✕Forgetting the constant of integration

✕Computing ∫v dt\int v\,dt∫vdt for 'total distance travelled' when vvv changes sign

✕Calling the magnitude of velocity 'velocity' (not 'speed')

✕Omitting units on the final answer

Practice questions

Past paper style quiz

Video lesson

Variable Acceleration — frequently asked questions

Position vector $\mathbf{r}$

✕Computing $\int v\,dt$ for 'total distance travelled' when $v$ changes sign