IB Math AA SL Topic 5.9: Kinematics | Notes & Flashcards | Aimnova

IB Math AA SL — Kinematics

Topic 5.9 of IB Mathematics: Analysis and Approaches covers Kinematics, which is part of Unit 5: Calculus. Students explore key concepts including Kinematics. A strong understanding of kinematics is essential for IB Math AA SL exams and builds the foundation for connected topics across the syllabus.

Key concepts in Kinematics

Key Idea: Kinematics applies calculus to motion in a straight line: differentiate down to find velocity and acceleration, integrate up to recover them. It appears on both papers, often as a multi-part 'find when…' question.

🚗 The s ↔ v ↔ a chain

v = \frac{ds}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^{2}s}{dt^{2}}

$s$: displacement (position) at time t
$v$: velocity — the derivative of displacement
$a$: acceleration — the derivative of velocity

Every integration up the chain adds a + C that an initial condition (a value at t = 0) pins down. The key moments: at rest (or changes direction) when v = 0; max/min velocity when a = 0; displacement = ∫v dt (signed) but total distance = ∫|v| dt (split where v = 0, add the magnitudes).

✏️ IB-style worked examples

IB-style question — velocity and acceleration by differentiating

A particle moves with displacement s = t³ − 5t² + 3t metres (t ≥ 0). Find its velocity and acceleration at t = 2.

Step by step:

Differentiate s for velocity, then again for acceleration.
$v = 3t^{2} - 10t + 3, \quad a = 6t - 10$
Substitute t = 2 into each.
$v(2) = -5, \quad a(2) = 2$

Final answer:

At t = 2: velocity −5 m/s, acceleration 2 m/s².

IB-style question — integrate up with an initial condition

A particle has acceleration a = 6t − 4 m/s² and velocity 5 m/s at t = 0. Find an expression for v(t).

Step by step:

Integrate a to get v — don't forget the + C.
$v = \int (6t - 4)\,dt = 3t^{2} - 4t + C$
Use v(0) = 5 to find C.
$C = 5 \Rightarrow v = 3t^{2} - 4t + 5$

Final answer:

v(t) = 3t² − 4t + 5 m/s.

IB-style question — at rest, and maximum velocity

Particle A has velocity v = 2t² − 10t + 12. Find when it is at rest. Particle B has velocity v = 9 + 6t − 3t²; find its maximum velocity.

Step by step:

At rest: set v = 0 and factor.
$2(t-2)(t-3) = 0 \Rightarrow t = 2 \text{ or } t = 3$
Max velocity: set a = 0, i.e. dv/dt = 0, then read off v.
$a = 6 - 6t = 0 \Rightarrow t = 1, \; v(1) = 12$

Final answer:

A is at rest at t = 2 s and t = 3 s; B's maximum velocity is 12 m/s (at t = 1 s).

IB-style question — displacement vs total distance

A particle has velocity v = t² − 4t + 3 m/s. Find the displacement and the total distance travelled from t = 0 to t = 3.

Step by step:

Displacement = ∫v dt over [0, 3].
$\int_{0}^{3} (t^{2}-4t+3)\,dt = \Big[\tfrac{t^{3}}{3} - 2t^{2} + 3t\Big]_{0}^{3} = 0$
Distance: v = 0 at t = 1 and t = 3. Integrate each leg and add magnitudes.
$\Big|\textstyle\int_{0}^{1} v\Big| + \Big|\int_{1}^{3} v\Big| = \big|\tfrac{4}{3}\big| + \big|{-}\tfrac{4}{3}\big| = \tfrac{8}{3}$

Final answer:

Displacement = 0 m; total distance travelled = 8/3 ≈ 2.67 m.

Important: When you integrate up, you must add + C and use an initial condition to find it — a definite-integral shortcut skips this. And for 'how far did it travel', use ∫|v| (distance), not ∫v (displacement) — they differ whenever the particle changes direction.

Tap each card to reveal the answer.

Exam Tips

Differentiate DOWN the chain (s → v → a); integrate UP (a → v → s) and add + C.
Find C from an initial condition — the value of v or s at t = 0.
'At rest' or 'changes direction' → solve v = 0. 'Max/min velocity' → solve a = 0.
Displacement = ∫v dt (signed); total distance = ∫|v| dt (split where v = 0, add magnitudes).
On Paper 2 use fnInt: ∫|v| for distance, ∫v for displacement — read which one is asked.

IB Math AA SL — Kinematics

Key concepts in Kinematics

Key Idea: Kinematics applies calculus to motion in a straight line: differentiate down to find velocity and acceleration, integrate up to recover them. It appears on both papers, often as a multi-part 'find when…' question.

🚗 The s ↔ v ↔ a chain

v = \frac{ds}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^{2}s}{dt^{2}}

$s$: displacement (position) at time t
$v$: velocity — the derivative of displacement
$a$: acceleration — the derivative of velocity

Every integration up the chain adds a + C that an initial condition (a value at t = 0) pins down. The key moments: at rest (or changes direction) when v = 0; max/min velocity when a = 0; displacement = ∫v dt (signed) but total distance = ∫|v| dt (split where v = 0, add the magnitudes).

✏️ IB-style worked examples

IB-style question — velocity and acceleration by differentiating

A particle moves with displacement s = t³ − 5t² + 3t metres (t ≥ 0). Find its velocity and acceleration at t = 2.

Step by step:

Differentiate s for velocity, then again for acceleration.
$v = 3t^{2} - 10t + 3, \quad a = 6t - 10$
Substitute t = 2 into each.
$v(2) = -5, \quad a(2) = 2$

Final answer:

At t = 2: velocity −5 m/s, acceleration 2 m/s².

IB-style question — integrate up with an initial condition

A particle has acceleration a = 6t − 4 m/s² and velocity 5 m/s at t = 0. Find an expression for v(t).

Step by step:

Integrate a to get v — don't forget the + C.
$v = \int (6t - 4)\,dt = 3t^{2} - 4t + C$
Use v(0) = 5 to find C.
$C = 5 \Rightarrow v = 3t^{2} - 4t + 5$

Final answer:

v(t) = 3t² − 4t + 5 m/s.

IB-style question — at rest, and maximum velocity

Particle A has velocity v = 2t² − 10t + 12. Find when it is at rest. Particle B has velocity v = 9 + 6t − 3t²; find its maximum velocity.

Step by step:

At rest: set v = 0 and factor.
$2(t-2)(t-3) = 0 \Rightarrow t = 2 \text{ or } t = 3$
Max velocity: set a = 0, i.e. dv/dt = 0, then read off v.
$a = 6 - 6t = 0 \Rightarrow t = 1, \; v(1) = 12$

Final answer:

A is at rest at t = 2 s and t = 3 s; B's maximum velocity is 12 m/s (at t = 1 s).

IB-style question — displacement vs total distance

A particle has velocity v = t² − 4t + 3 m/s. Find the displacement and the total distance travelled from t = 0 to t = 3.

Step by step:

Displacement = ∫v dt over [0, 3].
$\int_{0}^{3} (t^{2}-4t+3)\,dt = \Big[\tfrac{t^{3}}{3} - 2t^{2} + 3t\Big]_{0}^{3} = 0$
Distance: v = 0 at t = 1 and t = 3. Integrate each leg and add magnitudes.
$\Big|\textstyle\int_{0}^{1} v\Big| + \Big|\int_{1}^{3} v\Big| = \big|\tfrac{4}{3}\big| + \big|{-}\tfrac{4}{3}\big| = \tfrac{8}{3}$

Final answer:

Displacement = 0 m; total distance travelled = 8/3 ≈ 2.67 m.

Important: When you integrate up, you must add + C and use an initial condition to find it — a definite-integral shortcut skips this. And for 'how far did it travel', use ∫|v| (distance), not ∫v (displacement) — they differ whenever the particle changes direction.

Tap each card to reveal the answer.

Exam Tips

Differentiate DOWN the chain (s → v → a); integrate UP (a → v → s) and add + C.
Find C from an initial condition — the value of v or s at t = 0.
'At rest' or 'changes direction' → solve v = 0. 'Max/min velocity' → solve a = 0.
Displacement = ∫v dt (signed); total distance = ∫|v| dt (split where v = 0, add magnitudes).
On Paper 2 use fnInt: ∫|v| for distance, ∫v for displacement — read which one is asked.

IB Math AA SL — Kinematics

Key concepts in Kinematics

🚗 The s ↔ v ↔ a chain

✏️ IB-style worked examples

IB-style question — velocity and acceleration by differentiating

IB-style question — integrate up with an initial condition

IB-style question — at rest, and maximum velocity

IB-style question — displacement vs total distance

Exam Tips

What you'll learn in Topic 5.9

Study resources — 5.9 Kinematics

Kinematics

Ready to study Kinematics?

Ready to practice?

IB Math AA SL — Kinematics

Key concepts in Kinematics

🚗 The s ↔ v ↔ a chain

✏️ IB-style worked examples

IB-style question — velocity and acceleration by differentiating

IB-style question — integrate up with an initial condition

IB-style question — at rest, and maximum velocity

IB-style question — displacement vs total distance

Exam Tips

What you'll learn in Topic 5.9

Study resources — 5.9 Kinematics

Kinematics

Ready to study Kinematics?

Ready to practice?