IP AMaths Notes (Upper Sec, Year 3-4): 19) Kinematics

Study guide08 Nov 2025, 00:00 ZUpdated 30 Nov 2025

Displacement, velocity, and acceleration relationships with calculus for IP AMaths motion problems.

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Q: What does IP AMaths Notes (Upper Sec, Year 3-4): 19) Kinematics cover?
A: Displacement, velocity, and acceleration relationships with calculus for IP AMaths motion problems.

Kinematics connects calculus with motion. Derivatives give velocity and acceleration; integration recovers displacement.

Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.

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These kinematics relationships align with the SEAB GCE O-Level Additional Mathematics (4049) syllabus: displacement, velocity, and acceleration links via differentiation/integration, rest-time checks, and speed extrema within a stated interval.

Status: SEAB O-Level Additional Mathematics 4049 syllabus (exams from 2025) checked 2025-11-30 - scope unchanged; remains the reference for this note.

The core idea is simple: Kinematics links displacement, velocity, and acceleration through calculus.

Use it as a working check: Differentiate displacement to get velocity and acceleration. Integrate velocity to recover displacement, then use the initial condition.

Then go one layer deeper: Example: if v(t) = 12 - 6t, set v = 0 to find the rest time, then integrate v(t) to find displacement at that time.

Choosing the kinematics route

Start from the quantity the question gives you, then move one step at a time through the calculus chain.

Given quantity	To find	First move
Displacement $s(t)$

Reviewed by

Marcus Pang·Managing Director (Maths)

Sources

SEAB GCE O-Level Additional Mathematics (4049) syllabus (examinations from 2025)

Velocity sign	Acceleration sign	What happens to speed	Reason
$v > 0$	$a > 0$	Speed increases.	Motion is in the positive direction and acceleration is also positive.
$v > 0$	$a < 0$	Speed decreases.	Acceleration opposes the direction of motion.
$v < 0$	$a < 0$	Speed increases.	Motion is in the negative direction and acceleration is also negative.
$v < 0$	$a > 0$	Speed decreases.	Acceleration opposes the negative-direction motion.
$v = 0$	any non-zero $a$	The particle is instantaneously at rest, then starts moving in the acceleration direction.	Do not call this "speeding up" over an interval without checking nearby times.

Starting information	First integration	Condition to use	Second integration
$a(t)$ and $v(0)$	$v(t)=\int a(t)\space dt + C_1$	Substitute into $v(t)$ to find $C_1$ .	Then integrate $v(t)$ to get $s(t)$ .
$v(t)$ and $s(0)$	Already at velocity.	Integrate once to $s(t)=\int v(t)\space dt + C_2$
$a(t)$ , $v(0)$ , and $s(0)$	Find $v(t)$ first.	Use $v(0)$

Question asks for	First move	How to finish
Displacement from $t=a$ to $t=b$	Integrate $v(t)$ over the whole interval.	Keep signed areas, so movement in the negative direction subtracts.
Total distance travelled	Find rest times where $v(t)=0$ inside the interval.	Split the interval at those times and add absolute changes in displacement.
Speed at a time	Find $v(t)$ , then take its magnitude.	Do not report a negative speed.

IP AMaths Notes (Upper Sec, Year 3-4): 19) Kinematics

Choosing the kinematics route

Sources

1 Relationships

Speed-change sign checkpoint

Initial-condition chain checkpoint

2 Worked example - From velocity to displacement

3 Worked example - Minimum speed on a given interval

4 Distance versus displacement checkpoint

5 Worked example - Time to rest and distance travelled

6 Practice Quiz

7 Try this

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