Master Variable Acceleration (Exam Tips)

🚀 Master Variable Acceleration (Exam Tips)

Variable acceleration is where the tidy SUVAT world stops being tidy. Speeds change unevenly. Forces aren’t constant. Real motion creeps in. And examiners love it because it tests whether you can choose the right model, not just plug numbers into a memorised formula.

You’ll see it across AQA, Edexcel, OCR. Often a 6–9 mark build, sometimes with a small graph, sometimes a differential equation. If you can move confidently between:

calculus links (a = dv/dt, v = ds/dt, and the chain rule a = v dv/ds), and
graph reading (areas and gradients on v–t or a–t),

you’re in a very strong place. Don’t worry if the algebra looks messy at times — that’s normal. We keep it organised and the marks follow.

🔙 Previous topic:

“Review pulley systems before studying acceleration.”

🧠 What “variable acceleration” actually means

Acceleration is how fast velocity changes. If that change isn’t constant, it’s variable acceleration. The key is how it varies:

with time: a = a(t)
with velocity: a = a(v)
with position: a = a(s)

Three glue rules to remember (they’re the whole game):

v = ds/dt
a = dv/dt
a = v dv/ds (this is just the chain rule in disguise)

Pick the description that matches the question, then either integrate up (a → v → s) or differentiate down (s → v → a). Choose once, stick with it.

Teacher aside: write your positive direction at the top of the page. One small arrow avoids half the sign mistakes.

🔧 Core methods you’ll actually use

Keep this short list in your notes. No LaTeX, just working text.

If a = a(t): integrate a with respect to t to get v(t). Then integrate v to get s(t). Add constants, use initial conditions immediately.
If a = a(v): write dv/dt = a(v), separate variables → ∫ dv/a(v) = ∫ dt. Solve for v(t).
If a = a(s): use a = v dv/ds, so v dv = a(s) ds. Integrate both sides to link v and s directly.
Graphs:
- v–t area = displacement; gradient = acceleration
- a–t area = change in velocity

Quick habit: pin the constants as soon as you integrate. “At t = 0, v = …” fixes C before you forget what C was.

🧪 Quick pictures (so it feels real)

Car away from lights: big acceleration, then it tails off. That’s a = a(t) or a(v).
Parachutist: acceleration starts near g, then drops as air resistance grows; speed tends to a limit. That’s a = g − k v.
Particle with a spring: acceleration grows with stretch; that’s a = a(s). Chain rule time.

None of these need fancy tricks. Just choose the model and push it through calmly.

🧮 Worked Example 1 — acceleration depends on velocity

A particle moves along a line. dv/dt = 6 − 0.5 v. At t = 0, v = 0.
(i) Find v(t). (ii) How long to reach v = 10 m/s?

Separate: dv / (6 − 0.5 v) = dt
Integrate: ∫ dv/(6 − 0.5 v) = ∫ dt → −2 ln|6 − 0.5 v| = t + C
Initial condition: at t = 0, v = 0 → −2 ln 6 = C
So −2 ln|6 − 0.5 v| = t − 2 ln 6

Solve for v:
ln|6 − 0.5 v| = −t/2 + ln 6
6 − 0.5 v = 6 e^(−t/2)
v(t) = 12 (1 − e^(−t/2))

Time to hit 10:
10 = 12 (1 − e^(−t/2)) → e^(−t/2) = 1/6 → t = 2 ln 6 (about 3.58 s)

Sense-check: as t → ∞, v → 12. Looks right. The curve rises fast, then flattens.

🧮 Worked Example 2 — acceleration depends on position

A particle moves so that a = 0.4 s, where s is displacement from O (metres). When s = 0, v = 2 m/s.
Find v as a function of s, then the speed at s = 10 m.

Use chain rule form: a = v dv/ds = 0.4 s
Separate: v dv = 0.4 s ds
Integrate: 0.5 v^2 = 0.2 s^2 + C
Initial condition: s = 0, v = 2 → C = 2
So v^2 = 0.4 s^2 + 4 → at s = 10, v^2 = 44 → v ≈ 6.63 m/s

Short. Clean. Done. If you see a(s), think “v dv = a(s) ds” before anything else.

📈 Graph questions (easy marks if you don’t rush)

Examiners like to drop a small v–t or a–t graph into the mix.

On a v–t graph, the area under the curve is displacement. If it drops below the axis, that bit subtracts. The gradient is acceleration.
On an a–t graph, the area is the change in velocity.
Piecewise behaviour (accelerate, cruise, brake): treat each section separately, add the results, label your working.

If it looks fussy, write one sentence: “From 0 to 3 s acceleration is constant and positive; from 3 to 5 s it is zero.” That line quietly earns method credit.

🧭 A reliable method you can use every time

Underline the givens. Units, initial conditions, what they want at the end.
Choose the model. a(t)? a(v)? a(s)? Or start from the graph.
Write the governing equation (dv/dt = … or v dv/ds = …).
Separate and integrate (or differentiate, if they gave s(t)).
Use the initial condition immediately to fix constants.
Finish the task. If they want time, don’t stop at v; if they want displacement, integrate once more.
Sense-check (limits, signs, units). If it says a runner covers 3 km in 2 s, something’s wrong.

Teacher aside: students who write the units at each step catch silly errors before they grow teeth.

⚠️ Common slips (and quick fixes)

Using SUVAT by habit when acceleration varies. If a depends on t, v, or s, switch to calculus.
Forgetting constants of integration. Fix them the moment you integrate. One line later and they’re already awkward.
Sign drift. Choose a positive direction and stick to it. Write “+ to the right” (or up) at the top of the page.
Rounding early. Keep 3 s.f. in working; round once at the end.

Answering the wrong thing. If the prompt says “hence find the time”, they want you to use the previous result — not start a fresh route.

🧰 Quick reference (no-frills formula box)

v = ds/dt
a = dv/dt
a = v dv/ds
If a = a(t): integrate a → v; integrate v → s
If a = a(v): dv/dt = a(v) → ∫ dv/a(v) = t + C
If a = a(s): v dv = a(s) ds → integrate both sides
Graphs:
- v–t area = displacement, gradient = acceleration
- a–t area = change in velocity

Stick this on a half-page card. Glance at it before a practice set. It keeps you honest.

📚 Revision plan that actually works

Ten-minute blocks. Two short variable-acceleration questions most days beat a weekend cram.
Board habits. AQA likes piecewise motion; Edexcel likes resistance models; OCR likes you to write the model before you solve it.
Graph fluency. Five quick v–t questions in a row — same afternoon — locks it in.
Explain aloud. If you can say, “I separated variables and used v(0) = 0 to find C,” you’re already there.
Mistake log. One page. Error, cause, fix. Boring but brilliant.

💬 Mini-FAQ

When can I use SUVAT?
Only when acceleration is constant over the interval. If a varies, use calculus or graphs.

Why does a(v) give exponentials so often?
Because dv/dt = k − c v (or similar) integrates to logs, which rearrange to exponentials. It’s the standard resistance-proportional-to-speed model.

What if my answer comes out negative?
Usually a sign choice. It can also mean the motion is opposite to your assumed positive direction — not the end of the world, but check tension/“speed” statements are still sensible.

Do I have to write the units every line?
No, but checking units on key lines catches disasters before they cost marks.

🏁 Final thoughts (plus your next step)

Variable acceleration isn’t about clever tricks. It’s about choosing the right lens — a(t), a(v), or a(s) — and then keeping your algebra tidy. Fix your constants early, keep the direction consistent, and answer exactly what’s asked. After two or three full examples like the ones above, it starts to feel routine.

Start your revision for A Level Maths today with our 3-day A Level Maths course — variable acceleration, resistance models, and graph skills for AQA, Edexcel, and OCR, all taught step by step with exam-style practice. You do the thinking; we’ll keep the scaffolding clear.

Author Bio

S. Mahandru • Head of Maths, Exam.tips

S. Mahandru is Head of Maths at Exam.tips. With over 15 years of experience, he simplifies complex calculus topics and provides clear worked examples, strategies, and exam-focused guidance.

🧭 Next topic:

“Next, add friction to your mechanics toolkit.”

Master Variable Acceleration (Exam Tips)