For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.
How this chapter applies
Eclat core: displacement, velocity, acceleration, direction of motion, speed change, turning points, differentiation, and integration for straight-line motion form the main route.
School-sensitive extension: piecewise motion, unfamiliar parameter models, and deeper mechanics interpretation may vary by school.
2027 national comparison: K341 Topic C1.18 explicitly requires applying differentiation and integration to displacement, velocity, and acceleration of a particle moving in a straight line.
Check your school: confirm sign conventions, graph interpretation depth, and whether piecewise motion is assessed.
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.
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)
Velocity or acceleration
Differentiate once for v(t), then again for a(t).
Velocity v(t)
Displacement
Integrate v(t), then use the given initial displacement to find C.
Acceleration a(t)
Velocity or displacement
Integrate once for velocity, use the velocity condition, then integrate again for displacement.
Speed or rest time
When the particle stops or changes direction
Solve v(t)=0, then check whether the question asks for speed, distance, or displacement.
Common trap: distance travelled is not always the same as displacement. If velocity changes sign, split the interval at rest times and add the absolute distances.
1 Relationships
Velocity: v=dtds.
Acceleration: a=dtdv=dt2d2s.
Displacement from velocity: s=∫vdt (plus constant).
Speed-change sign checkpoint
When a question asks whether a particle is speeding up or slowing down, compare the signs of velocity and acceleration. Acceleration sign alone does not tell you what happens to speed.
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.
Worked check: if v(2)=−4 and a(2)=−3, the particle is moving in the negative direction and accelerating in the negative direction, so its speed is increasing at t=2. If v(2)=−4 but a(2)=3, the acceleration opposes the motion, so its speed is decreasing.
Misconception check: negative acceleration does not automatically mean slowing down. It means acceleration acts in the negative direction; speed depends on whether that direction matches the current velocity.
Initial-condition chain checkpoint
When acceleration is given, constants of integration enter in order. Use each condition as soon as its quantity appears.
Starting information
First integration
Condition to use
Second integration
a(t) and v(0)
v(t)=∫a(t)dt+C1
Substitute into v(t) to find C1.
Then integrate v(t) to get s(t).
v(t) and s(0)
Already at velocity.
Integrate once to s(t)=∫v(t)dt+C2
a(t), v(0), and s(0)
Find v(t) first.
Use v(0)
Worked check: if a(t)=6t−4, v(0)=5, and s(0)=0,
v(t)=3t2−4t+C1,5=C1,
so v(t)=3t2−4t+5. Then
s(t)=t3−2t2+5t+C2,0=C2.
Misconception check: do not use s(0) to find the constant in v(t). A velocity condition fixes the velocity constant; a displacement condition fixes the displacement constant.
2 Worked example - From velocity to displacement
Given v(t)=6t2−4t+3 in m⋅s−1, find displacement s(t) if s(0)=2.
Integrate: s(t)=∫(6t2−4t+3),dt=2t3−2t2+3t+C.
Apply initial condition: 2=0−0+0+C⟹C=2.
Hence s(t)=2t3−2t2+3t+2.
3 Worked example - Minimum speed on a given interval
A particle moves with s(t)=t3−6t2+9t. Find when its speed is least on 0≤t≤3.
Velocity: v(t)=3t2−12t+9=3(t−1)(t−3).
Speed is ∣v(t)∣, so the least possible speed occurs when v(t)=0 (the particle is at rest).
Solve v(t)=0: 3(t−1)(t−3)=0⟹t=1
Hence the minimum speed is 0 at t=1 and t=3.
4 Distance versus displacement checkpoint
Before adding up motion over an interval, use the velocity sign to decide whether the particle turns around.
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.
Misconception check: a particle can have zero displacement but non-zero distance travelled if it moves away and returns to the starting point.
5 Worked example - Time to rest and distance travelled
A particle has velocity v(t)=12−6t in m⋅s−1 with displacement s(0)=5m. Find the time when the particle comes to rest and the displacement at that instant.
Set v(t)=0 to locate rest: 12−6t=0⟹t=2 (seconds).
Displacement comes from integrating velocity: s(t)=∫(12−6t)dt=12t−3t2+C.
Use s(0)=5m to obtain C=5. Thus s(t)=12t−3t2+5
Evaluate at t=2: s(2)=24−12+5=17m.
Answer: the particle stops at t=2s with displacement 17m from the origin.
6 Practice Quiz
Check integration-differentiation loops, rest-time checks, and distance computations so motion problems link cleanly to calculus tools.
7 Try this
Given a(t)=12t−8m⋅s−2 with v(0)=5m⋅s−1 and s(0)=0m, find expressions for v(t) and s(t).
