IB Maths AA SLTopic 5 — CalculusPaper 1 & 2~9 min read
Calculus for Kinematics
Now the calculus comes in. The three quantities — displacement, velocity, acceleration — are linked by differentiation in one direction and integration in the other. If you’ve got s(t), differentiating gives velocity then acceleration. If you’ve got a(t), integrating gives velocity then displacement (with a constant to find from initial conditions). It’s a chain — and once you see it, every kinematics calculus problem looks the same.
📘 What you need to know
Differentiate to go down the chain: s(t) → v(t) → a(t).
Integrate to go up the chain: a(t) → v(t) → s(t) — and find the +c using initial or boundary conditions.
Acceleration is also the second derivative of displacement: a = d²s/dt².
Definite integral of v = displacement (signed). Definite integral of |v| = distance travelled.
“Initially” → t = 0. Use this to determine the constant of integration.
On Paper 2, your GDC handles the integrals (and the modulus) directly.
The s ⇌ v ⇌ a chain
The whole topic boils down to this picture: differentiation goes one way, integration goes the other.
Differentiation and integration link the three quantities
The three core relationships
v(t) = s‘(t) · a(t) = v‘(t) = s”(t)
✓ in formula booklet
Notice that acceleration is also the second derivative of displacement. So if you start with s(t), you can hop straight to a(t) by differentiating twice.
Going down the chain — differentiation
If you’re given s(t), velocity and acceleration are one and two derivatives away. No constants to worry about — differentiation just gives you a direct answer.
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Common follow-ups
“At what time is the particle at rest?” → solve v(t) = 0. “What’s the initial velocity?” → calculate v(0). “When is the acceleration zero?” → solve a(t) = 0.
Going up the chain — integration
If you’re given a(t), integrate to get v(t), then again to get s(t). But each integration introduces a constant of integration that you must determine.
Integration relationships
v(t) = ∫a(t) dt · s(t) = ∫v(t) dt
The constant comes from an initial or boundary condition the question gives you — like “the particle is initially at rest” (meaning v(0) = 0) or “the particle starts at the origin” (meaning s(0) = 0).
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“Each integration adds a +c to find”
Going from a to v to s means two integrations and two constants to determine. The question always gives you the conditions you need — read carefully for “initially”, “at rest”, “at the origin”, etc.
Definite integrals: displacement vs distance
For a moving particle between time t₁ and t₂:
Displacement vs Distance — same v(t), different integrals
Displacement
s = ∫t₁t₂v(t) dt
signed area — below counts as negative
Distance
d = ∫t₁t₂ |v(t)| dt
absolute area — modulus flips negatives up
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If the velocity never changes sign
If v(t) stays positive (or stays negative) on the whole interval, displacement and distance have the same magnitude. They only differ when the particle reverses direction.
Step-by-step method
Calculus for kinematics recipe
Identify what you have — is it s(t), v(t), or a(t)?
Identify what you need — does the question want a derivative or an antiderivative?
Differentiate or integrate as needed. Each integration gives a + c.
Apply initial/boundary conditions to find each constant. “Initially” means t = 0.
Answer the actual question — “at rest” → solve v = 0; “displacement at t = 5″ → evaluate s(5); “distance from 0 to 5” → integrate |v|.
Worked examples
WE 1
Going down the chain — differentiation
A particle moves along a straight line with displacement (in metres) given by s(t) = t³ − 9t² + 24t − 18, where t is in seconds.
(a) Find v(t) and a(t). (b) Find the times when the particle is at rest. (c) Find the initial velocity.
part (a) — differentiatev(t) = s'(t) = 3t² − 18t + 24a(t) = v'(t) = 6t − 18part (b) — at rest means v = 03t² − 18t + 24 = 03(t² − 6t + 8) = 0 → 3(t − 2)(t − 4) = 0at rest at t = 2 s and t = 4 spart (c) — initial velocityv(0) = 24 m s⁻¹initial velocity = 24 m s⁻¹“initially” always means t = 0 — just plug it in!
WE 2
Going up the chain — integration with conditions
A particle moves with acceleration a(t) = 6t − 4 m s⁻². Initially the particle is at the origin moving with velocity 5 m s⁻¹.
(a) Find v(t). (b) Find s(t). (c) Find the displacement at t = 3 s.
part (a) — integrate a → vv(t) = ∫(6t − 4) dt = 3t² − 4t + cuse v(0) = 5 → c = 5v(t) = 3t² − 4t + 5part (b) — integrate v → ss(t) = ∫(3t² − 4t + 5) dt = t³ − 2t² + 5t + duse s(0) = 0 (at origin) → d = 0s(t) = t³ − 2t² + 5tpart (c) — displacement at t = 3s(3) = 27 − 18 + 15 = 24displacement at t = 3 is 24 mtwo integrations → two constants. Read carefully for two conditions!
WE 3
Definite integrals — displacement vs distance
A particle moves along a straight line with velocity v(t) = t² − 4t + 3 m s⁻¹ for 0 ≤ t ≤ 5.
(a) Find the times when the particle is at rest. (b) Find the displacement during the first 5 seconds. (c) Find the distance travelled in the first 5 seconds.
part (a) — at restv(t) = t² − 4t + 3 = (t − 1)(t − 3)v = 0 at t = 1 and t = 3part (b) — displacement (signed integral)s = ∫(0 to 5) (t² − 4t + 3) dt= [t³/3 − 2t² + 3t] from 0 to 5= (125/3 − 50 + 15) − 0 = 125/3 − 35 = 20/3displacement = 20/3 mpart (c) — distance (use modulus)v > 0 on (0, 1) and (3, 5); v < 0 on (1, 3)∫(0 to 1) v dt = 4/3∫(1 to 3) v dt = −4/3 (take |·| → 4/3)∫(3 to 5) v dt = 20/3distance = 4/3 + 4/3 + 20/3 = 28/3distance travelled = 28/3 mon Paper 2, GDC: ∫(0 to 5) |t² − 4t + 3| dt → 28/3 directly!
💡 Top tips
Identify the chain direction first. Differentiation goes s → v → a; integration reverses it.
Read for initial conditions. “Initially” means t = 0; “at the origin” means s = 0; “at rest” means v = 0.
One integration = one constant. Going from a to s needs two conditions.
Displacement uses signed integral; distance uses |v|. They’re only different when v changes sign.
Sketch the v-t graph to see when the particle reverses — your GDC plots it instantly.
GDC modulus shortcut for distance: ∫|v(t)|dt in one calculation.
⚠ Common mistakes
Forgetting the constant of integration. Without the +c, you can’t find v(t) or s(t) uniquely.
Using only one condition when two integrations were needed.
Treating displacement and distance the same when the particle reverses direction.
Sign errors when integrating velocity that goes negative — split the integral at the roots of v = 0.
Confusing “at rest” with “stationary point” of position — they’re the same thing only if you’re at a turning point of s(t).
Ignoring the GDC on Paper 2 — for messy integrals, it’s faster and a built-in check.
🎉 You’ve finished the entire Kinematics series! You can now move freely between displacement, velocity and acceleration using calculus, find when a particle is at rest or reverses direction, and compute both signed displacement and total distance travelled. Combined with optimisation and the integration toolkit, you’ve got the complete calculus-modelling kit for AA SL.
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