Kinematics: Variable Acceleration with Calculus

This lesson covers kinematics when acceleration is not constant, using differentiation and integration. This is a key part of the Edexcel 9MA0 specification that links pure mathematics with mechanics.

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When SUVAT Does Not Apply

The SUVAT equations require constant acceleration. When acceleration varies with time, we must use calculus.

The fundamental relationships are:

• Velocity is the derivative of displacement with respect to time: v = ds/dt
• Acceleration is the derivative of velocity with respect to time: a = dv/dt = d²s/dt²
• Displacement is the integral of velocity: s = ∫v dt
• Velocity is the integral of acceleration: v = ∫a dt

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Differentiation: From Displacement to Velocity to Acceleration

If the displacement is given as a function of time, s = f(t), then:

• v = ds/dt = f'(t)
• a = dv/dt = f''(t)

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

v = ds/dt = 6t² - 18t + 12 a = dv/dt = 12t - 18

At t = 0: v = 12 m s⁻¹, a = -18 m s⁻². At t = 1: v = 6 - 18 + 12 = 0 m s⁻¹ (the particle is momentarily at rest). At t = 2: v = 24 - 36 + 12 = 0 m s⁻¹ (at rest again).

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Integration: From Acceleration to Velocity to Displacement

If the acceleration is given as a function of time, a = g(t), then:

v = ∫a dt + C₁

The constant C₁ is found from the initial condition (usually v at t = 0).

s = ∫v dt + C₂

The constant C₂ is found from the initial condition (usually s at t = 0).

Example: A particle has acceleration a = 6t - 4 m s⁻² and starts from rest at the origin.

v = ∫(6t - 4) dt = 3t² - 4t + C₁ At t = 0, v = 0: C₁ = 0, so v = 3t² - 4t.

s = ∫(3t² - 4t) dt = t³ - 2t² + C₂ At t = 0, s = 0: C₂ = 0, so s = t³ - 2t².

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Finding When a Particle is at Rest

A particle is at rest when v = 0. Set ds/dt = 0 and solve for t.

Example (continuing from above): v = 3t² - 4t = t(3t - 4) = 0 t = 0 or t = 4/3 seconds.

At t = 4/3: s = (4/3)³ - 2(4/3)² = 64/27 - 32/9 = 64/27 - 96/27 = -32/27 m.

The negative displacement indicates the particle has moved in the negative direction.

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Finding the Distance Travelled

Distance is always positive, but displacement can be negative. If the particle changes direction during the time interval, you must:

1. Find the times when v = 0 (direction changes).
2. Calculate the displacement for each sub-interval.
3. Add the absolute values of the displacements.

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Definite Integration for Displacement

The displacement between t = a and t = b is:

s = ∫ₐᵇ v dt

This gives the net displacement (it can be positive, negative or zero).

The distance travelled is:

distance = ∫ₐᵇ |v| dt

In practice, split the integral at the points where v = 0 and take the absolute value of each part.

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Maximum and Minimum Displacement

Maximum or minimum displacement occurs when v = ds/dt = 0. Use the second derivative (acceleration) to determine whether it is a maximum or minimum:

• If a = d²s/dt² < 0 at that point, it is a maximum displacement.
• If a = d²s/dt² > 0 at that point, it is a minimum displacement.

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Using Integration with Definite Limits

Example: v = 4t - t². Find the displacement between t = 0 and t = 5.

s = ∫₀⁵ (4t - t²) dt = [2t² - t³/3]₀⁵ = (50 - 125/3) - 0 = 150/3 - 125/3 = 25/3 m

But the particle changes direction when v = 0: 4t - t² = t(4 - t) = 0, so t = 0 or t = 4.

Distance = |∫₀⁴ (4t - t²) dt| + |∫₄⁵ (4t - t²) dt| = |[2t² - t³/3]₀⁴| + |[2t² - t³/3]₄⁵| = |32 - 64/3| + |25/3 - 32 + 64/3| = |32/3| + |-7/3| = 32/3 + 7/3 = 39/3 = 13 m

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Summary

• When acceleration is not constant, use calculus instead of SUVAT.
• Differentiate displacement to get velocity, and velocity to get acceleration.
• Integrate acceleration to get velocity, and velocity to get displacement.
• Always determine the constant of integration from initial conditions.
• For distance (not displacement), split at v = 0 and sum absolute values.

A-Level Deep Dive: Kinematics — Variable Acceleration with Calculus

Spec mapping

Edexcel 9MA0-03 specification section 9 — Kinematics, sub-strands 9.1, 9.2 and 9.3 covers calculus in kinematics for motion in a straight line: $v = \dfrac{ds}{dt}$v=dtds​, $a = \dfrac{dv}{dt} = \dfrac{d^2 s}{dt^2}$a=dtdv​=dt2d2s​, $s = \int v\,dt$s=∫vdt, $v = \int a\,dt$v=∫adt. Extend to 2D using vectors $\mathbf{r}(t)$r(t) with $\mathbf{v} = \dot{\mathbf{r}}$v=r˙ and $\mathbf{a} = \ddot{\mathbf{r}}$a=r¨ (refer to the official specification document for exact wording). Variable-acceleration kinematics is the Year 2 mechanics topic that distinguishes 9MA0 from AS-level 8MA0: where AS uses the SUVAT equations (constant acceleration only), Year 2 demands the full calculus treatment whenever $a$a depends on $t$t. It is examined on Paper 3 — Statistics and Mechanics, section 9, and synoptically pulls in section 7 (Differentiation) and section 8 (Integration) of Paper 1/2 Pure.

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Worked example with full mark scheme

Question (8 marks): A particle $P$P moves in a straight line. Its acceleration at time $t$t seconds ($t \geq 0$t≥0) is given by $a = 6 - 2t$a=6−2t m s$^{-2}$−2. At $t = 0$t=0, $P$P has velocity $1$1 m s$^{-1}$−1 and is at the origin.

(a) Find an expression for the velocity $v$v of $P$P at time $t$t. (3)

(b) Find the maximum velocity of $P$P for $t \geq 0$t≥0, justifying that it is a maximum. (2)

(c) Find the displacement of $P$P from the origin at the instant the velocity is maximum. (3)

Solution with mark scheme:

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

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

M1 — integrating $a$a with respect to $t$t and including a constant of integration. Forgetting $+C$+C is the single most common mark-loser on this topic; integration without a constant is mathematically incomplete.

Step 2 — apply the initial condition.

At $t = 0$t=0, $v = 1$v=1, so $1 = 0 - 0 + C \implies C = 1$1=0−0+C⟹C=1.

M1 — substituting the boundary condition to evaluate the constant.

A1 — $v = 6t - t^2 + 1$v=6t−t2+1.

(b) Step 1 — locate the stationary point of $v(t)$v(t).

$\dfrac{dv}{dt} = a = 6 - 2t = 0 \implies t = 3$dtdv​=a=6−2t=0⟹t=3.

M1 — setting $a = 0$a=0 (equivalently $\frac{dv}{dt} = 0$dtdv​=0) to find the time at which velocity is stationary. Note: at maximum velocity, acceleration is zero, not velocity.

Step 2 — evaluate and justify maximum.

$v(3) = 18 - 9 + 1 = 10$v(3)=18−9+1=10 m s$^{-1}$−1. Since $\dfrac{d^2 v}{dt^2} = -2 < 0$dt2d2v​=−2<0, this is indeed a maximum.

A1 — maximum velocity $10$10 m s$^{-1}$−1 with second-derivative justification (or sign-change argument on $a$a).

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

$s = \int v\,dt = \int (6t - t^2 + 1)\,dt = 3t^2 - \dfrac{t^3}{3} + t + D$s=∫vdt=∫(6t−t2+1)dt=3t2−3t3​+t+D

M1 — integrating $v$v with a new constant $D$D.

Step 2 — apply the second initial condition.

At $t = 0$t=0, $s = 0$s=0, so $D = 0$D=0.

M1 — using the displacement boundary condition.

Step 3 — evaluate at $t = 3$t=3.

$s(3) = 27 - 9 + 3 = 21$s(3)=27−9+3=21 m.

A1 — displacement $21$21 m from the origin.

Total: 8 marks (M5 A3).

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Specimen question modelled on the Edexcel 9MA0 Paper 3 format

Question (6 marks): A particle moves on the $x$x-axis. Its velocity at time $t \geq 0$t≥0 seconds is $v = t^3 - 6t^2 + 8t$v=t3−6t2+8t m s$^{-1}$−1.

(a) Find the times at which the particle is instantaneously at rest. (2)

(b) Find the total distance travelled in the first $4$4 seconds. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — setting $v = 0$v=0: $t(t^2 - 6t + 8) = 0$t(t2−6t+8)=0, factorising to $t(t - 2)(t - 4) = 0$t(t−2)(t−4)=0.
• A1 (AO1.1b) — $t = 0, 2, 4$t=0,2,4 seconds.

(b)

• M1 (AO3.1b) — recognising that distance $\neq$= $|\int v\,dt|$∣∫vdt∣ when velocity changes sign; the integral must be split at the rest instants.
• M1 (AO1.1b) — computing $\left|\int_0^2 v\,dt\right|$​∫02​vdt​ and $\left|\int_2^4 v\,dt\right|$​∫24​vdt​ separately.
• A1 (AO1.1b) — $\int_0^2 (t^3 - 6t^2 + 8t)\,dt = \left[\tfrac{t^4}{4} - 2t^3 + 4t^2\right]_0^2 = 4 - 16 + 16 = 4$∫02​(t3−6t2+8t)dt=[4t4​−2t3+4t2]02​=4−16+16=4 m, and $\int_2^4 v\,dt = (64 - 128 + 64) - 4 = -4$∫24​vdt=(64−128+64)−4=−4 m.
• A1 (AO2.5) — total distance $= |4| + |-4| = 8$=∣4∣+∣−4∣=8 m, distinguishing from net displacement of $0$0 m.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. The AO3 mark is the synoptic insight that signed integrals give displacement while modulus of each segment gives distance — a hallmark Year 2 mechanics test.

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Synoptic links

Connects to:

1. Section 8 — Integration (Pure): every reverse-process kinematics step is an integration. $s = \int v\,dt$s=∫vdt requires the same antiderivatives, the same $+C$+C discipline and the same definite-integral evaluation as pure integration questions. Trigonometric, exponential and rational $v(t)$v(t) all appear.
2. Section 7 — Differentiation (Pure): $v = \dfrac{ds}{dt}$v=dtds​ and $a = \dfrac{dv}{dt}$a=dtdv​ apply the power rule, chain rule and product rule routinely. Maximum-velocity questions are differentiation-and-stationary-points questions in disguise.
3. Section 9 — SUVAT as a special case: the constant-acceleration equations ($v = u + at$v=u+at, $s = ut + \tfrac{1}{2}at^2$s=ut+21​at2) are derived by integrating $a = \text{const}$a=const. They are not a separate topic — they are the $a(t) = k$a(t)=k specialisation of the calculus framework. Using SUVAT when $a$a depends on $t$t scores zero method marks.
4. Section 10 — Vectors in mechanics: 2D variable-acceleration extends to $\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}$r(t)=x(t)i+y(t)j, with $\mathbf{v} = \dot{x}\mathbf{i} + \dot{y}\mathbf{j}$v=x˙i+y˙​j and $|\mathbf{v}| = \sqrt{\dot{x}^2 + \dot{y}^2}$∣v∣=x˙2+y˙​2​. Each component is an independent 1D kinematics problem.
5. Section 8 (Pure, Year 2) — Differential equations: writing $\dfrac{dv}{dt} = f(v, t)$dtdv​=f(v,t) converts kinematics into a first-order ODE. Newton's second law $F = ma$F=ma frequently produces separable or linear ODEs that integrate by methods learned in Pure.

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Mark-scheme literacy

Variable-acceleration questions on 9MA0 Paper 3 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Differentiating or integrating $s(t)$s(t), $v(t)$v(t), $a(t)$a(t); applying initial conditions; evaluating definite integrals
AO2 (reasoning / interpretation)	25–35%	Distinguishing displacement / distance / speed / velocity; justifying maxima with second-derivative or sign analysis; commenting on physical meaning of constants
AO3 (problem-solving / modelling)	10–20%	Setting up the model from a worded scenario; choosing whether to differentiate or integrate; recognising when SUVAT does not apply

Examiner-rewarded phrasing: "since acceleration depends on $t$t, the SUVAT equations cannot be used"; "applying the initial condition $v(0) = u$v(0)=u to find $C$C"; "since $\dfrac{d^2 v}{dt^2} < 0$dt2d2v​<0 at this point, $v$v is maximum". Phrases that lose marks: "$v = u + at$v=u+at" written when $a$a is variable (immediate zero); "the area under the velocity-time graph is the distance" without checking sign of $v$v; missing constants of integration.

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Grade-band model answers

3-mark question

Question: A particle moves in a straight line with velocity $v = 4t - t^2$v=4t−t2 m s$^{-1}$−1. Find its acceleration at $t = 3$t=3.

Grade C response (~180 words):

To find acceleration, differentiate velocity:

$a = \dfrac{dv}{dt} = 4 - 2t$a=dtdv​=4−2t.