Variable Acceleration

This lesson covers motion with variable acceleration, where acceleration changes over time rather than remaining constant. In these situations, the SUVAT equations cannot be used, and instead we use calculus (differentiation and integration) to relate displacement, velocity, and acceleration.

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The Calculus Relationships

Relationship	Calculus operation
Velocity from displacement	$v = \frac{ds}{dt}$v=dtds​ (differentiate $s$s with respect to $t$t)
Acceleration from velocity	$a = \frac{dv}{dt}$a=dtdv​ (differentiate $v$v with respect to $t$t)
Acceleration from displacement	$a = \frac{d^2s}{dt^2}$a=dt2d2s​ (differentiate $s$s twice)
Displacement from velocity	$s = \int v \, dt$s=∫vdt (integrate $v$v with respect to $t$t)
Velocity from acceleration	$v = \int a \, dt$v=∫adt (integrate $a$a with respect to $t$t)

Exam Tip: When the question says acceleration is "variable" or gives displacement/velocity as a function of time (e.g., $s = 3t^2 - 2t + 1$s=3t2−2t+1), you must use calculus. SUVAT equations only work for constant acceleration.

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Differentiation: Finding Velocity and Acceleration

Example: A particle moves along a straight line with displacement $s = 2t^3 - 9t^2 + 12t$s=2t3−9t2+12t metres, where $t$t is in seconds.

Velocity: $v = \frac{ds}{dt} = 6t^2 - 18t + 12$v=dtds​=6t2−18t+12

Acceleration: $a = \frac{dv}{dt} = 12t - 18$a=dtdv​=12t−18

Finding When the Particle is at Rest

The particle is at rest when $v = 0$v=0:

$6t^2 - 18t + 12 = 0$6t2−18t+12=0 $t^2 - 3t + 2 = 0$t2−3t+2=0 $(t - 1)(t - 2) = 0$(t−1)(t−2)=0

So $t = 1$t=1 s or $t = 2$t=2 s.

Finding When Acceleration is Zero

$12t - 18 = 0 \implies t = 1.5 \text{ s}$12t−18=0⟹t=1.5 s

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Integration: Finding Velocity and Displacement

Example: A particle has acceleration $a = 6t - 4$a=6t−4 m/s$^2$2. When $t = 0$t=0, $v = 3$v=3 m/s and $s = 0$s=0.

Velocity: $v = \int (6t - 4) \, dt = 3t^2 - 4t + C$v=∫(6t−4)dt=3t2−4t+C

Using $v = 3$v=3 when $t = 0$t=0: $3 = 0 - 0 + C$3=0−0+C, so $C = 3$C=3.

$v = 3t^2 - 4t + 3$v=3t2−4t+3

Displacement: $s = \int (3t^2 - 4t + 3) \, dt = t^3 - 2t^2 + 3t + D$s=∫(3t2−4t+3)dt=t3−2t2+3t+D

Using $s = 0$s=0 when $t = 0$t=0: $D = 0$D=0.

$s = t^3 - 2t^2 + 3t$s=t3−2t2+3t

Exam Tip: When integrating, always include the constant of integration and use the initial conditions (given values at $t = 0$t=0 or another specified time) to find it. Forgetting the constant is one of the most common errors.

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Using Definite Integration for Displacement

The displacement between times $t_1$t1​ and $t_2$t2​ is:

$s = \int_{t_1}^{t_2} v \, dt$s=∫t1​t2​​vdt

Important: If the velocity changes sign in the interval (the particle reverses direction), the integral gives the net displacement, not the total distance. To find the total distance, split the integral at the point(s) where $v = 0$v=0 and take the absolute values.

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Maximum and Minimum Displacement

The particle reaches maximum or minimum displacement when $v = 0$v=0 (the particle momentarily stops). Use the second derivative (acceleration) to determine which:

• If $a < 0$a<0 when $v = 0$v=0: maximum displacement
• If $a > 0$a>0 when $v = 0$v=0: minimum displacement

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Summary

• When acceleration is variable, use calculus instead of SUVAT equations.
• Differentiate displacement to find velocity; differentiate velocity to find acceleration.
• Integrate acceleration to find velocity; integrate velocity to find displacement.
• Always use initial conditions to determine constants of integration.
• The particle is at rest when $v = 0$v=0; acceleration is zero when $a = 0$a=0.
• For total distance, check whether velocity changes sign and split the integral accordingly.

Exam Tip: Read the question carefully: if displacement or velocity is given as a function of $t$t (e.g., $s = t^3 - 6t$s=t3−6t), use differentiation. If acceleration is given as a function of $t$t with initial conditions, use integration.

A-Level Deep Dive: Variable Acceleration

Spec mapping

AQA 7357 specification, Paper 3 — Mechanics, Section T (Year 2) covers calculus in kinematics for motion in a straight line: $v = \dfrac{dr}{dt}$v=dtdr​, $a = \dfrac{dv}{dt} = \dfrac{d^2 r}{dt^2}$a=dtdv​=dt2d2r​, $r = \int v\,dt$r=∫vdt, $v = \int a\,dt$v=∫adt. Extension to two dimensions using vectors (refer to the official specification document for exact wording). Variable acceleration sits at the apex of the mechanics syllabus because it weaves together Pure (differentiation, integration, vectors) with the SUVAT framework students learnt in AS. The formula booklet does not list these calculus relationships — they must be memorised as definitions, not formulae.

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Worked example with full mark scheme

Question (8 marks):

A particle $P$P moves along the $x$x-axis. At time $t$t seconds ($t \geq 0$t≥0) its acceleration is $a = (6t - 4)$a=(6t−4) m s$^{-2}$−2. When $t = 0$t=0, $P$P is at the origin moving with velocity $3$3 m s$^{-1}$−1 in the positive $x$x-direction.

(a) Show that the velocity of $P$P at time $t$t is $v = 3t^2 - 4t + 3$v=3t2−4t+3 m s$^{-1}$−1. (3)

(b) Find the displacement of $P$P from the origin when $t = 4$t=4. (3)

(c) Determine whether $P$P ever comes to instantaneous rest, justifying your answer. (2)

Solution with mark scheme:

(a) Step 1 — integrate acceleration to obtain velocity.

$v = \int a\,dt = \int (6t - 4)\,dt = 3t^2 - 4t + C$v=∫adt=∫(6t−4)dt=3t2−4t+C

M1 — integrating the acceleration expression with the constant of integration explicit. Forgetting $+C$+C is the single most common error on this topic and forfeits both subsequent marks.

Step 2 — apply the boundary condition.

When $t = 0$t=0, $v = 3$v=3. Substituting: $3 = 0 - 0 + C$3=0−0+C, hence $C = 3$C=3.

M1 — substituting the initial condition to determine $C$C.

A1 — printed result $v = 3t^2 - 4t + 3$v=3t2−4t+3 obtained.

(b) Step 1 — integrate velocity to obtain displacement.

$s = \int v\,dt = \int (3t^2 - 4t + 3)\,dt = t^3 - 2t^2 + 3t + K$s=∫vdt=∫(3t2−4t+3)dt=t3−2t2+3t+K

M1 — integrating the velocity expression with a new constant $K$K (distinct from $C$C). Recycling the symbol $C$C is a presentation error but rarely penalised; reusing the value would be catastrophic.

Step 2 — apply the boundary condition.

When $t = 0$t=0, $s = 0$s=0 (origin). So $K = 0$K=0, giving $s = t^3 - 2t^2 + 3t$s=t3−2t2+3t.

M1 — second boundary condition correctly applied.

Step 3 — evaluate at $t = 4$t=4.

$s(4) = 64 - 32 + 12 = 44$s(4)=64−32+12=44 m.

A1 — exact answer with units.

(c) Step 1 — set $v = 0$v=0.

$3t^2 - 4t + 3 = 0$3t2−4t+3=0. The discriminant is $\Delta = 16 - 36 = -20 < 0$Δ=16−36=−20<0.

M1 — testing $v = 0$v=0 via the discriminant.

A1 — conclusion: since $\Delta < 0$Δ<0, the quadratic has no real roots, so $P$P never comes to instantaneous rest. The minimum speed (by completing the square) occurs at $t = 2/3$t=2/3 with $v_{\min} = 5/3$vmin​=5/3 m s$^{-1}$−1.

Total: 8 marks (M5 A3).

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Specimen question modelled on the AQA 7357 Paper 3 format

Question (6 marks): A particle moves in the $xy$xy-plane with position vector $\mathbf{r}(t) = (3t^2)\mathbf{i} + (4t - t^2)\mathbf{j}$r(t)=(3t2)i+(4t−t2)j metres at time $t$t seconds.

(a) Find the velocity vector at time $t$t. (2)

(b) Find the time at which the velocity is perpendicular to the initial velocity. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — differentiating component-wise: $\mathbf{v} = \dfrac{d\mathbf{r}}{dt} = (6t)\mathbf{i} + (4 - 2t)\mathbf{j}$v=dtdr​=(6t)i+(4−2t)j.
• A1 (AO1.1b) — fully correct vector form.

(b)

• M1 (AO3.1a) — evaluate initial velocity: $\mathbf{v}(0) = 0\mathbf{i} + 4\mathbf{j}$v(0)=0i+4j.
• M1 (AO1.1b) — perpendicularity condition via dot product: $\mathbf{v}(t) \cdot \mathbf{v}(0) = 0$v(t)⋅v(0)=0, giving $(6t)(0) + (4 - 2t)(4) = 0$(6t)(0)+(4−2t)(4)=0.
• M1 (AO1.1b) — solving $16 - 8t = 0$16−8t=0.
• A1 (AO2.1) — conclusion $t = 2$t=2 s with reasoning visible.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. AQA mechanics calculus questions tilt heavier toward AO3 than the corresponding pure papers because applying calculus to a physical setup is itself problem-solving.

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Synoptic links

Connects to:

1. Pure section H — Differentiation: $v = ds/dt$v=ds/dt and $a = dv/dt$a=dv/dt are direct applications of the derivative as a rate of change. Every technique from Pure differentiation (chain rule, product rule, implicit differentiation) is fair game when the displacement is given as a complicated function of time.

2. Pure section I — Integration: the inverse relationships $s = \int v\,dt$s=∫vdt and $v = \int a\,dt$v=∫adt require the full integration toolkit, including substitution and parts in harder questions. The constant of integration is physical — it is the initial position or velocity.

3. Mechanics section S — SUVAT: the SUVAT equations are the special case of variable-acceleration calculus where $a$a is constant. Integrating $a$a (constant) gives $v = u + at$v=u+at; integrating again gives $s = ut + \tfrac{1}{2}at^2$s=ut+21​at2. Recognising this is examiner-rewarded synoptic reasoning.

4. Pure section J — Vectors: in two-dimensional motion, position, velocity and acceleration are all vectors. Differentiation and integration act component-wise, and the dot product handles perpendicularity and angle questions.

5. Pure section H/I — Differential equations (Year 2): Newton's second law $F = ma$F=ma becomes the ODE $m\dfrac{dv}{dt} = F(v, t)$mdtdv​=F(v,t) when the force depends on velocity (e.g. air resistance). Variable-acceleration kinematics is the bridge from descriptive mechanics into ODE-based modelling.

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Mark-scheme literacy

Variable-acceleration questions on AQA 7357 Paper 3 split AO marks more evenly than Pure topics:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Differentiating and integrating polynomials, applying boundary conditions, evaluating at specified times
AO2 (reasoning / interpretation)	20–30%	Justifying choice of method, distinguishing speed from velocity, interpreting sign of $v$v as direction
AO3 (problem-solving / modelling)	15–25%	Setting up the calculus relationship from a physical description, choosing the right boundary condition, deciding when SUVAT does not apply

Examiner-rewarded phrasing: "since acceleration is not constant, SUVAT does not apply; we must integrate"; "the constant of integration represents the initial velocity"; "the particle changes direction when $v = 0$v=0 and $a \neq 0$a=0 at that instant". Phrases that lose marks: writing "speed" when "velocity" is required (or vice versa); omitting units on a final numerical answer; failing to state the domain restriction $t \geq 0$t≥0 when relevant.

A specific AQA pattern: questions phrased "find the distance travelled between $t = 0$t=0 and $t = T$t=T" require splitting the integral at every sign change of $v$v, then summing absolute values. $\int_0^T v\,dt$∫0T​vdt alone gives net displacement, not total distance — a single-mark difference that A* candidates spot instantly.

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Grade-band model answers

3-mark question

Question: A particle has velocity $v = 4t - t^2$v=4t−t2 m s$^{-1}$−1 at time $t$t seconds. Find its acceleration when $t = 3$t=3.

Grade C response (~120 words):

Acceleration is the derivative of velocity: $a = \dfrac{dv}{dt} = 4 - 2t$a=dtdv​=4−2t.

When $t = 3$t=3: $a = 4 - 6 = -2$a=4−6=−2 m s$^{-2}$−2.

Examiner commentary: Full marks (3/3). The candidate correctly identifies $a = dv/dt$a=dv/dt, differentiates accurately, and substitutes the specified time. Units are included. The negative sign is preserved, indicating deceleration in the positive direction. This is the standard Grade C response for a procedural question — efficient and correct. A common slip would be to write $a = 2 - 2t$a=2−2t (mis-differentiating the constant term) or to omit units.

Grade A response (~190 words):*

Acceleration is defined as the rate of change of velocity:

$a = \dfrac{dv}{dt} = \dfrac{d}{dt}(4t - t^2) = 4 - 2t$a=dtdv​=dtd​(4t−t2)=4−2t

Substituting $t = 3$t=3: $a = 4 - 2(3) = -2$a=4−2(3)=−2 m s$^{-2}$−2.

The negative value indicates that, at $t = 3$t=3, the particle is decelerating (if travelling in the positive direction) or, equivalently, accelerating in the negative $x$x-direction.