IB Maths AA HLTopic 5 — CalculusPaper 1 & 2~11 min read
Calculus for Kinematics
Displacement, velocity and acceleration are linked by calculus. v is the derivative of s, a is the derivative of v, and going the other way you integrate — adding a constant when you do. Displacement is ∫v dt; distance is ∫|v| dt. Memorise the four formulae, then everything else is just routine differentiation/integration plus a sign-of-v check.
📘 What you need to know
Differentiate to go DOWN: s → v → a. So v = ds/dt and a = dv/dt = d²s/dt².
Integrate to go UP: a → v → s. So v = ∫a dt and s = ∫v dt.
Constant of integration: every indefinite integral adds + C. Use the initial condition (often s(0) or v(0)) to pin it down.
Times at rest: solve v(t) = 0. These are where the particle stops momentarily — often where it changes direction.
Definite integral of velocity = DISPLACEMENT: ∫t₁t₂v(t) dt = signed change in position.
Definite integral of |velocity| = DISTANCE: ∫t₁t₂ |v(t)| dt = total path length.
If v changes sign on the interval, the two integrals differ. Split at the zero-crossings; flip the sign on the negative pieces.
Use the GDC on Paper 2 — its absolute-value-integral feature does the distance integral directly.
Going right (green) is differentiation; going left (orange) is integration, which introduces a constant C you fix using an initial condition.
Differentiation for kinematics
Given s(t), every other kinematic quantity falls out by differentiation. Velocity is the rate of change of position; acceleration is the rate of change of velocity, which is also the SECOND derivative of position.
Three favourite paper questions from a given s(t) or v(t): (1) find when the particle is at rest — solve v = 0; (2) find maximum/minimum velocity — solve a = 0 (a stationary point of v); (3) decide if the particle is speeding up or slowing down at a given t — check the sign of v(t) · a(t).
Integration for kinematics
The reverse direction needs an initial or boundary condition to pin down the constant of integration. “Initial” always means t = 0; phrases like “initially at the origin” mean s(0) = 0; “initially at rest” means v(0) = 0.
The really powerful tool is the DEFINITE integral. Over a time interval, the signed integral of velocity is the (signed) displacement, and the integral of |velocity| is the total distance travelled — and these are the same thing ONLY if the particle never reverses direction on the interval.
Displacement
∫t₁t₂v(t) dt
signed: forward sections count positive, backward sections negative; tells you NET change in position from t₁ to t₂
Distance
∫t₁t₂ |v(t)| dt
unsigned: every section counts positive; total path length actually travelled — the odometer reading
🧭 Recipe — kinematics problems with calculus
Identify what you’re given — s(t), v(t), or a(t) — and what’s asked. Sketch where to go (differentiate downward, integrate upward).
If integrating, write + C (or + C₁, + C₂ for two stages) and immediately use the given initial/boundary condition to find it.
“At rest” / “max velocity” / “max position” all reduce to setting a derivative to zero: v = 0 for at rest, a = 0 for max/min velocity, v = 0 again for max/min displacement.
For distance vs displacement, find the zeros of v(t) in the interval. If there are none, distance = |displacement|. If there are, split the integral at each zero and flip the sign on the pieces where v < 0.
Always state units: m for s, m s⁻¹ for v, m s⁻² for a, and combine sensibly for integrals.
Worked examples
WE 1
From s(t): find v(t), a(t), and the times the particle is at rest
The displacement (m) of a particle from a fixed origin at time t seconds, for t ≥ 0, is modelled by s(t) = t³ − 6t² + 9t + 4. Find expressions for v(t) and a(t), and find all times at which the particle is at rest.
Step 1 — differentiate s once for vv(t) = s'(t) = 3t² − 12t + 9 = 3(t² − 4t + 3) = 3(t − 1)(t − 3)Step 2 — differentiate again for aa(t) = v'(t) = s”(t) = 6t − 12 = 6(t − 2)Step 3 — particle at rest ⇔ v = 03(t − 1)(t − 3) = 0 → t = 1 or t = 3v(t) = 3(t−1)(t−3); a(t) = 6(t−2); at rest at t = 1 s and t = 3 sfactorising v keeps the at-rest times one short step away. Notice a(1) = −6 (decelerating the positive velocity to zero, then reversing it) and a(3) = 6 (decelerating the negative velocity back to zero).
WE 2
From v(t): find a(t), times at rest, and speeding up vs slowing down at t = 3
A particle moves along a straight line with velocity v(t) = t² − 5t + 4 (m s⁻¹) for t ≥ 0. (a) Find a(t). (b) Find the times at which the particle is at rest. (c) Determine, with reason, whether the particle is speeding up or slowing down at t = 3.
(a) differentiate va(t) = v'(t) = 2t − 5(b) at rest: v = 0t² − 5t + 4 = 0 → (t − 1)(t − 4) = 0→ t = 1 s or t = 4 s(c) evaluate v and a at t = 3v(3) = 9 − 15 + 4 = −2a(3) = 2(3) − 5 = 1v · a = (−2)(1) = −2 < 0 → OPPOSITE signsa(t) = 2t − 5; at rest at t = 1 and t = 4; at t = 3 the particle is SLOWING DOWNbetween t = 1 and t = 4 the particle moves in the negative direction (v < 0), but a > 0 acts against it, gradually braking it back toward rest at t = 4.
WE 3
Trigonometric velocity — max speed and at-rest times
A particle moves along a straight line with velocity v(t) = 4 cos(t) m s⁻¹ for 0 ≤ t ≤ 2π. Find (a) an expression for the acceleration a(t), (b) the times in the interval at which the particle is at rest, and (c) the maximum speed of the particle.
(a) differentiate va(t) = v'(t) = −4 sin(t)(b) at rest: v = 04 cos(t) = 0 → cos(t) = 0in [0, 2π]: t = π/2 and t = 3π/2(c) speed = |v(t)| = |4 cos(t)| = 4 |cos(t)||cos(t)| has max value 1, attained at t = 0, π, 2π→ max speed = 4 · 1 = 4 m s⁻¹a(t) = −4 sin(t); at rest at t = π/2 and t = 3π/2; max speed = 4 m s⁻¹“max VELOCITY” would be +4 (when cos t = +1); “max SPEED” is 4 m/s achieved at t = 0, π, 2π — including t = π where the velocity is −4 (full speed backwards).
WE 4
Integrate v(t) with initial condition to find s(t)
A particle moves with velocity v(t) = 6t² − 18t + 12 (m s⁻¹) for t ≥ 0. Its position relative to a fixed origin at time t = 0 is s(0) = 5 m. (a) Find an expression for s(t). (b) Find the displacement of the particle from its starting position at t = 4 s.
(a) integrate v with respect to ts(t) = ∫(6t² − 18t + 12) dt = 2t³ − 9t² + 12t + Cuse the initial condition s(0) = 5 to find Cs(0) = 0 − 0 + 0 + C = 5 → C = 5→ s(t) = 2t³ − 9t² + 12t + 5(b) displacement from starting position = s(4) − s(0)s(4) = 2(64) − 9(16) + 12(4) + 5 = 128 − 144 + 48 + 5 = 37displacement = 37 − 5 = 32 ms(t) = 2t³ − 9t² + 12t + 5; displacement from start at t = 4 is 32 m“position” and “displacement from start” differ by s(0). At t = 4 the particle is 37 m from the origin, but only 32 m from where it began at t = 0.
WE 5
Displacement vs distance when velocity changes sign
A particle moves along a straight line with velocity v(t) = 3t² − 12t (m s⁻¹) for t ≥ 0. Find (a) the displacement of the particle between t = 1 and t = 5, and (b) the total distance travelled by the particle between t = 1 and t = 5.
Step 1 — locate sign-change times of v in [1, 5]v = 3t² − 12t = 3t(t − 4) = 0→ t = 0 or t = 4; in [1, 5] only t = 4 matterscheck signs: at t = 2, v = 12 − 24 = −12 < 0 (so v < 0 on [1, 4]) at t = 5, v = 75 − 60 = 15 > 0 (so v > 0 on [4, 5])(a) displacement = ∫₁⁵ v dt (signed, single integral)∫₁⁵ (3t² − 12t) dt = [t³ − 6t²]₁⁵ = (125 − 150) − (1 − 6) = −25 − (−5) = −20(b) distance = split at t = 4, take |area| on each pieceon [1, 4], v < 0: ∫₁⁴ −v dt = ∫₁⁴ (12t − 3t²) dt = [6t² − t³]₁⁴ = (96 − 64) − (6 − 1) = 27on [4, 5], v > 0: ∫₄⁵ v dt = [t³ − 6t²]₄⁵ = (125 − 150) − (64 − 96) = −25 + 32 = 7total distance = 27 + 7 = 34Displacement = −20 m; Distance = 34 mconsistency check: 27 m backward, then 7 m forward → net 7 − 27 = −20 m ✓. The minus sign means the particle ends up 20 m to the negative side of where it was at t = 1.
A ball is thrown vertically upward from ground level with initial velocity 30 m s⁻¹. Taking upward as the positive direction, the only acceleration on the ball is gravity, modelled as a(t) = −10 m s⁻². Find (a) an expression for v(t), (b) an expression for s(t) — the height above ground, (c) the maximum height reached, and (d) the total time the ball is in the air before returning to the ground.
(a) integrate a with v(0) = 30v(t) = ∫(−10) dt = −10t + C₁v(0) = 30 → C₁ = 30→ v(t) = 30 − 10t(b) integrate v with s(0) = 0s(t) = ∫(30 − 10t) dt = 30t − 5t² + C₂s(0) = 0 → C₂ = 0→ s(t) = 30t − 5t²(c) max height when v = 0 (instantaneously at rest at the top)30 − 10t = 0 → t = 3s(3) = 30(3) − 5(9) = 90 − 45 = 45(d) returns to ground when s = 0 (and t > 0)30t − 5t² = 0 → 5t(6 − t) = 0→ t = 0 (start) or t = 6 (return)v(t) = 30 − 10t; s(t) = 30t − 5t²; max height = 45 m; flight time = 6 stwo stages of integration, two constants. Each constant pinned down by an initial condition. By symmetry the total distance is 90 m (up 45, down 45) while the displacement is 0 (ball returns to where it started).
💡 Top tips
Always write + C immediately when you integrate — and pin it down with the initial condition BEFORE moving on. Forgetting + C is one of the easiest marks to lose.
Factor v(t) when you can — it makes the at-rest times leap out and tells you the sign of v on each interval (which you need for distance).
For distance, find the sign changes FIRST, then integrate piece by piece. Or use the GDC’s ∫|v| feature directly on Paper 2.
Sketching v(t) helps — even a rough GDC plot tells you where it crosses zero, whether it’s a parabola opening up/down, and visualises the displacement-vs-distance situation.
“Max velocity” and “max speed” are different — max velocity solves a = 0 (a stationary point of v); max speed is the max of |v(t)| and may occur at an endpoint or at a sign-change.
⚠ Common mistakes
Computing ∫v dt and calling it distance — that’s displacement. For distance you need ∫|v| dt, which requires splitting at the sign changes.
Forgetting + C when integrating, or finding it but then dropping it from the final expression. Always include it explicitly.
Confusing “initially at the origin” with “initially at rest” — the former is s(0) = 0; the latter is v(0) = 0. They give different constants.
Solving a = 0 thinking it gives at-rest times — a = 0 gives where the velocity is stationary (max/min velocity). At-rest times come from v = 0.
Differentiating with respect to the wrong variable — if s is a function of t, you compute ds/dt, NOT ds/dx.
Up next: Basic Limits & Continuity. We’ve been freely differentiating and integrating functions of t all chapter, taking it for granted that “the derivative exists”. The next chapter zooms in on what derivatives REALLY mean as limits, where they fail to exist (corners, vertical tangents, discontinuities), and gives you the formal limit notation and continuity definitions that the rest of HL calculus builds on.
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