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Variable Acceleration Unveiled: A-Level Maths Simplified

Variable Acceleration Unveiled

⚡ Variable Acceleration Unveiled: A-Level Maths Simplified

Constant acceleration (the friendly SUVAT world) is tidy. Real motion? Not so tidy. Cars speed up then ease off, cyclists coast, a falling object picks up speed but air resistance kicks in. That’s variable acceleration: acceleration that changes with time, speed, or position.

Examiners on AQA, Edexcel, OCR like it because it checks whether you can choose the right model and carry out the calculus cleanly. If you can link displacement, velocity, and acceleration with derivatives and integrals — and keep your signs straight — you’ll turn a “scary” 8-mark part into steady method marks. (Small aside: it’s fine if your algebra looks a bit scruffy. Clear structure beats perfect handwriting.)

🔙 Previous topic:

“Review friction before tackling variable acceleration.”

🧠 The idea in one line

Acceleration is the rate of change of velocity. If that rate isn’t constant, we call it variable acceleration. It might depend on time (a = a(t)), on velocity (a = a(v)), or on position (a = a(s)). Your job is to pick the description that fits the question and then move between s, v, and t with calculus.

🔩 The three glue rules (write these at the top of your page)

  1. v = ds/dt

  2. a = dv/dt

  3. a = v dv/ds (this is just the chain rule dressed up for mechanics)

That’s your ladder. Different questions start on different rungs. You either integrate up (a → v → s) or differentiate down (s → v → a). Decide once. Stick to it.

Teacher aside: draw a tiny arrow on your page: “+ right, + up”. That one line saves half the sign errors.

🧰 Toolbox you’ll actually use (plain text, copy as-is)

  • If a = a(t):
     v(t) = ∫ a(t) dt + C₁
     s(t) = ∫ v(t) dt + C₂

  • If a = a(v):
     dv/dt = a(v) ⇒ separate variables: ∫ dv / a(v) = ∫ dt

  • If a = a(s):
     a = v dv/ds ⇒ v dv = a(s) ds ⇒ integrate both sides to link v and s

  • Graphs:

    • area under a–t = change in velocity

    • area under v–t = displacement

    • gradient of v–t = acceleration

Constants of integration: use the initial condition immediately (for example “t = 0, v = …”). Don’t leave constants floating around; they cause chaos later.

🔎 When each model shows up (board flavour)

  • AQA: piecewise behaviour or graph blends — accelerate, cruise, brake; or a(v) with quick separation of variables.

     

  • Edexcel: air resistance proportional to speed (a = g − k v), or a depending on position (a = k s).

     

  • OCR: light modelling prompts — “assume resistance is proportional to speed” — set up the differential equation yourself, then solve.

     

Different wrappers, same techniques: pick the form, separate, integrate, apply the condition, answer the exact wording.

🧪 Quick pictures (so it feels real)

  • Parachutist: dv/dt = g − k v. Starts near g, then acceleration shrinks to zero; speed tends to a limit.

  • Spring-like pull from a point O: a = k s (towards O). As you move away, pull increases; chain-rule form a = v dv/ds is perfect here.

  • Car in traffic: a = a(t). Short bursts, then flatter acceleration — very typical v–t graph reading.

Right — let’s do the work..

🧮 Worked Example 1 — acceleration depends on velocity

A particle moves on a straight line. dv/dt = 6 − 0.5 v. At t = 0, v = 0. Find v(t), then the time 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 v = 10:
 10 = 12 (1 − e^(−t/2)) ⇒ e^(−t/2) = 1/6 ⇒ t = 2 ln 6 (about 3.58 s)

Sanity check: limiting speed 12 m/s as t → ∞. 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 and the speed when 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
 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, and exactly what the examiner wants.

📈 Graph skills (easy marks you shouldn’t drop)

  • v–t graph: area is displacement (watch for negative area below the axis), gradient is acceleration.

  • a–t graph: area is change in velocity (units: m/s).

  • Piecewise motion: treat each segment separately, then add. State what you’re doing in one line: “From 0–4 s, a is constant and positive; from 4–6 s, a = 0.”

Label the graph. It signals control and earns method marks even before numbers appear.

🧭 A reliable method (use it every time)

  1. Underline givens (units, initial conditions, what’s asked).

  2. Choose the model (a(t), a(v), a(s), or graph first).

  3. Write the governing equation (dv/dt = … or v dv/ds = …).

  4. Separate variables and integrate (or differentiate, if starting from s(t)).

  5. Apply initial condition immediately to fix constants.

  6. Finish the ask (time, speed, displacement — don’t stop one step early).

  7. Sense-check (signs, limits, units). If a car “travels 3 km in 2 s,” something slipped.

Teacher aside: put the units on your intermediate lines when something looks large or tiny. It catches disasters before they cost marks.

⚠️ Common mistakes (and quick fixes)

  • Using SUVAT by habit. If acceleration varies with t, v, or s, SUVAT doesn’t apply. Use calculus or graphs.

  • Forgetting constants of integration. Fix them straight away using the given condition.

  • Sign drift. Choose a positive direction and stick with it. Write “+ to the right (and up)” at the top.

  • Early rounding. Keep 3 s.f. in working; round once per part at the end.

Answering the wrong thing. “Hence find the time” means use the previous result — not a brand-new route.

🧰 Mini “no-frills” formula card (pin to the front of your notes)

  • v = ds/dt

  • a = dv/dt

  • a = v dv/ds

  • If a = a(t): integrate to get v(t), then s(t)

  • If a = a(v): ∫ dv/a(v) = t + C

  • If a = a(s): v dv = a(s) ds

  • Graphs: area under v–t = displacement; area under a–t = change in velocity

That half-page covers 90% of variable-acceleration questions you’ll see at A Level.

🧪 Practice prompts (quick to set, high yield)

  1. dv/dt = 8 − 0.4 v, v(0) = 0. Find v(t). Time to reach v = 12 m/s? Limiting speed?

  2. a = 0.6 s, v = 1.5 m/s at s = 0. Find v when s = 8 m.

  3. A v–t graph is a triangle from (0,0) to (4,12) to (8,0). Find total displacement and maximum acceleration (use gradients).

  4. Resistance model: dv/dt = 9.8 − 1.2 v with v(0) = 0. Find v(t) and the time when v first exceeds 6 m/s.

Try to complete each inside 6–8 minutes. Talk through the method aloud — if you can explain it, you own it.

📚 Revision plan that actually sticks

  • Ten-minute blocks. Two calculus-motion questions most days beat a weekend cram.

  • Board habits. AQA: mixed graphs; Edexcel: resistance and a(s) chain-rule; OCR: set up the DE first.

  • Graph fluency. Five a–t and v–t reads in a row — low effort, reliable marks.

  • Explain to a friend. “I separated variables and used v(0) = 0 to find C” — that sentence is more valuable than another page of algebra.

Mistake log. One line per error: issue → cause → fix. Boring, but it reduces repeat errors fast.

💬 Quick FAQ

When can I still use SUVAT?
 Only when acceleration is constant on the interval. If a varies, move to calculus or graph methods.

Why do exponentials appear with resistance?
 Because dv/dt = k − c v integrates to logs; rearranging gives exponentials. It’s the standard linear first-order DE.

What if my answer is negative?
 Often a sign choice. It can also indicate motion opposite your positive direction. Check units and your initial condition.

Do I need to show every integration step?
 Show separation and clearly state the integral result. The constant must be fixed with the given condition — that’s where many marks sit.

🏁 Final thoughts

Variable acceleration isn’t about clever tricks. It’s about choosing the right lens — a(t), a(v), or a(s) — and then keeping the algebra tidy. Separate, integrate, use the initial condition, finish exactly what’s asked, and sanity-check. After a couple of full worked examples, the routine becomes second nature.

Start your revision for A Level Maths today with our A Level Maths Revision Course — step-by-step teaching for variable acceleration, resistance models, and graph skills across AQA, Edexcel, and OCR, with exam-style practice and clear worked solutions.

About the Author

S. Mahandru is Head of Maths at Exam.tips and has more than 15 years of experience in simplifying difficult subjects such as pure maths, mechanics and statistics. He gives worked examples, clear explanations and strategies to make students succeed.

🧭 Next topic:

“Next, look at moments — forces that make things turn.”

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