Author: Rishabh Kumar (IIT + ISI | Global Math Mentor)
Published: October 2025
Category: Calculus | Applications in Physics
🔹 What Is Kinematics?
Kinematics is the branch of mathematics and physics that describes motion — how objects move, how fast they move, and how their velocity changes over time.
It focuses on what happens (displacement, velocity, acceleration) rather than why (forces).
In calculus, kinematics becomes the perfect playground for understanding rate of change and integration.
🔹 The Core Relationship
In motion along a straight line, the fundamental relationships are: v=dsdtanda=dvdt=d2sdt2v = \frac{ds}{dt} \quad \text{and} \quad a = \frac{dv}{dt} = \frac{d^2s}{dt^2}v=dtdsanda=dtdv=dt2d2s
Where:
- sss = displacement
- vvv = velocity
- aaa = acceleration
- ttt = time
If you integrate these relationships, you can move backward: v=∫a dt,s=∫v dtv = \int a\,dt, \quad s = \int v\,dtv=∫adt,s=∫vdt
In short:
Differentiation connects motion rates; Integration reconstructs motion from rates.
🔹 Step-by-Step Example 1 — Basic Motion
A particle moves so that its displacement (in meters) at time ttt seconds is given by: s=t3−6t2+9ts = t^3 – 6t^2 + 9ts=t3−6t2+9t
1️⃣ Velocity: v=dsdt=3t2−12t+9v = \frac{ds}{dt} = 3t^2 – 12t + 9v=dtds=3t2−12t+9
2️⃣ Acceleration: a=dvdt=6t−12a = \frac{dv}{dt} = 6t – 12a=dtdv=6t−12
3️⃣ At t=2t = 2t=2: v=3(4)−24+9=−3 m/s,a=0v = 3(4) – 24 + 9 = -3 \, \text{m/s}, \quad a = 0v=3(4)−24+9=−3m/s,a=0
✅ The particle is momentarily at rest and changing direction at t=2t = 2t=2.
🔹 Step-by-Step Example 2 — From Acceleration to Displacement
If a particle starts from rest and has constant acceleration a=4 m/s2a = 4 \, \text{m/s}^2a=4m/s2, then: v=∫4 dt=4t+Cv = \int 4\,dt = 4t + Cv=∫4dt=4t+C
Since the particle starts from rest, v=0v = 0v=0 at t=0t = 0t=0, so C=0C = 0C=0. v=4tv = 4tv=4t
Now, s=∫4t dt=2t2+Cs = \int 4t\,dt = 2t^2 + Cs=∫4tdt=2t2+C
If s=0s = 0s=0 when t=0t = 0t=0, then C=0C = 0C=0, giving: s=2t2s = 2t^2s=2t2
✅ Displacement after ttt seconds is s=2t2s = 2t^2s=2t2.
🔹 Average vs Instantaneous Velocity
| Type | Formula | Meaning |
|---|---|---|
| Average Velocity | ΔsΔt\frac{\Delta s}{\Delta t}ΔtΔs | Overall rate of motion over an interval |
| Instantaneous Velocity | dsdt\frac{ds}{dt}dtds | Speed at a specific instant (slope of tangent) |
Example:
If s=t2+3ts = t^2 + 3ts=t2+3t, find average and instantaneous velocity between t=1t = 1t=1 and t=3t = 3t=3. vavg=s(3)−s(1)3−1=(9+9)−(1+3)2=7 m/sv_{avg} = \frac{s(3) – s(1)}{3 – 1} = \frac{(9 + 9) – (1 + 3)}{2} = 7 \, \text{m/s}vavg=3−1s(3)−s(1)=2(9+9)−(1+3)=7m/s vinst=dsdt=2t+3⇒v(3)=9 m/sv_{inst} = \frac{ds}{dt} = 2t + 3 \Rightarrow v(3) = 9 \, \text{m/s}vinst=dtds=2t+3⇒v(3)=9m/s
✅ The instantaneous velocity (9 m/s) is higher than the average (7 m/s) because the particle is speeding up.
🔹 Graphical Interpretation
- The gradient of an s–t graph gives velocity.
- The gradient of a v–t graph gives acceleration.
- The area under a v–t graph gives displacement.
Visualizing motion this way strengthens conceptual intuition — a key skill for high-level problem-solving in IB Math HL Paper 3, AP Calculus FRQs, and STEP/MAT geometry problems.
🔹 Common Student Errors
- ❌ Forgetting to differentiate twice for acceleration.
- ❌ Confusing position vs displacement (sign matters!).
- ❌ Ignoring units — keep everything consistent (e.g., m/s, m/s²).
- ❌ Treating direction as irrelevant — negative velocity indicates opposite motion.
🔹 Advanced Challenge
A particle moves such that v=6t−t2v = 6t – t^2v=6t−t2.
- Find the time when velocity is maximum.
- Find displacement at that time if s=0s = 0s=0 when t=0t = 0t=0.
Solution Sketch: a=dvdt=6−2t=0⇒t=3a = \frac{dv}{dt} = 6 – 2t = 0 \Rightarrow t = 3a=dtdv=6−2t=0⇒t=3 vmax=6(3)−9=9v_{max} = 6(3) – 9 = 9vmax=6(3)−9=9 s=∫v dt=∫(6t−t2) dt=3t2−t33s = \int v\,dt = \int (6t – t^2)\,dt = 3t^2 – \frac{t^3}{3}s=∫vdt=∫(6t−t2)dt=3t2−3t3 s(3)=27−9=18s(3) = 27 – 9 = 18s(3)=27−9=18
✅ At t=3st = 3st=3s:
Velocity is maximum (9 m/s), Displacement = 18 m.
🌟 Why Kinematics Matters
Kinematics links mathematical differentiation with physical intuition — transforming abstract calculus into meaningful motion.
Understanding these ideas is essential for:
- IB HL “Applications of Differentiation”
- A Level Mechanics 1 & 2
- AP Calculus AB/BC motion problems
- STEP/MAT applied reasoning sections
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