IB Mathematics AA – Topic 5: Calculus
Comprehensive Guide to Kinematics
Introduction to Kinematics
Kinematics is the study of motion without considering the forces that cause it. It connects calculus to the physical world, describing how objects move through space and time using displacement, velocity, and acceleration. From analyzing the motion of vehicles to understanding projectile trajectories and modeling population growth rates, kinematics provides the mathematical framework for describing change.
Key concepts: Displacement measures change in position, velocity describes the rate of change of displacement, and acceleration quantifies how velocity changes over time. These three quantities are intimately connected through differentiation and integration—velocity is the derivative of displacement, acceleration is the derivative of velocity, and the reverse relationships hold through integration. Understanding the distinction between total distance traveled and displacement is crucial for solving real-world problems.
Why kinematics matters: Kinematics transforms abstract calculus into tangible applications. Engineers use it to design braking systems, physicists to analyze particle motion, economists to model rates of change in markets, and biologists to study population dynamics. The graphical representation of motion—position-time and velocity-time graphs—provides visual insight into how objects behave, while calculus gives us the tools to extract precise numerical information from these graphs.
In this guide: We'll master the fundamental relationships between displacement, velocity, and acceleration using differentiation and integration, distinguish between displacement and total distance traveled, analyze motion using calculus techniques, interpret and create position-time and velocity-time graphs, use the GDC efficiently for kinematic calculations, and solve comprehensive IB-style problems combining all these concepts—essential skills for exam success.
1. Displacement, Velocity, and Acceleration
Fundamental Definitions
Three Key Quantities:
Displacement \(s(t)\)
Position of an object at time \(t\) (can be positive or negative)
Units: meters (m), kilometers (km), etc.
Vector quantity—includes direction
Velocity \(v(t)\)
Rate of change of displacement with respect to time
Units: m/s, km/h, etc.
Positive = moving in positive direction; Negative = moving in negative direction
Acceleration \(a(t)\)
Rate of change of velocity with respect to time
Units: m/s², km/h², etc.
Positive = speeding up in positive direction or slowing down in negative direction
Calculus Relationships
Differentiation and Integration Formulas
Velocity from Displacement (Differentiation)
\(v(t) = \frac{ds}{dt} = s'(t)\)
Velocity is the first derivative of displacement
Acceleration from Velocity (Differentiation)
\(a(t) = \frac{dv}{dt} = v'(t) = \frac{d^2s}{dt^2} = s''(t)\)
Acceleration is the derivative of velocity or second derivative of displacement
Displacement from Velocity (Integration)
\(s(t) = \int v(t) \, dt + C\)
Use initial conditions to find constant \(C\)
Velocity from Acceleration (Integration)
\(v(t) = \int a(t) \, dt + C\)
Use initial velocity to find constant \(C\)
⚠ Common Pitfalls:
- Signs matter: Negative velocity means moving backward, not slowing down!
- Acceleration sign: Negative acceleration doesn't always mean slowing down
- Don't forget initial conditions: When integrating, always use given initial values
- Units consistency: Keep track of units throughout calculations
2. Total Distance vs Displacement
Key Distinction
Displacement vs Distance
Displacement (Net Change in Position)
\(\text{Displacement} = \int_{t_1}^{t_2} v(t) \, dt = s(t_2) - s(t_1)\)
Can be positive, negative, or zero
Example: Walk 5m forward, then 3m backward → displacement = 2m
Total Distance Traveled
\(\text{Distance} = \int_{t_1}^{t_2} |v(t)| \, dt\)
Always positive or zero—sum of all movement
Example: Walk 5m forward, then 3m backward → distance = 8m
Calculating Total Distance
Step-by-Step Method:
- Find when velocity changes sign (when \(v(t) = 0\))
- These points split the time interval into sections
- Calculate displacement in each section separately
- Take absolute value of each section
- Add all absolute values together
Formula:
\(\text{Distance} = \left|\int_{t_1}^{t_a} v(t) \, dt\right| + \left|\int_{t_a}^{t_b} v(t) \, dt\right| + \left|\int_{t_b}^{t_2} v(t) \, dt\right|\)
where \(t_a, t_b\) are times when velocity equals zero
💡 Distance Tips:
- Always check if velocity changes sign in the interval
- Use GDC to find zeros of velocity function
- Distance ≥ |displacement| always
- If velocity doesn't change sign, distance = |displacement|
Example 1: Displacement vs Distance (IB-Style)
Problem: A particle moves along a straight line with velocity \(v(t) = t^2 - 4t + 3\) m/s for \(0 \leq t \leq 4\)
(a) Find when the particle is at rest
(b) Find the displacement from \(t = 0\) to \(t = 4\)
(c) Find the total distance traveled
Solution:
(a) When particle is at rest:
Particle at rest when \(v(t) = 0\)
\(t^2 - 4t + 3 = 0\)
\((t-1)(t-3) = 0\)
At rest at \(t = 1\) s and \(t = 3\) s
(b) Displacement from \(t = 0\) to \(t = 4\):
\(\text{Displacement} = \int_0^4 (t^2 - 4t + 3) \, dt\)
\(= \left[\frac{t^3}{3} - 2t^2 + 3t\right]_0^4\)
\(= \left(\frac{64}{3} - 32 + 12\right) - 0\)
\(= \frac{64 - 96 + 36}{3} = \frac{4}{3}\)
Displacement = \(\frac{4}{3}\) m ≈ 1.33 m
(c) Total distance traveled:
Velocity changes sign at \(t = 1\) and \(t = 3\)
Split into three intervals: [0,1], [1,3], [3,4]
Interval [0, 1]:
\(s_1 = \int_0^1 (t^2 - 4t + 3) \, dt = \left[\frac{t^3}{3} - 2t^2 + 3t\right]_0^1\)
\(= \frac{1}{3} - 2 + 3 = \frac{4}{3}\) m
Interval [1, 3]:
\(s_2 = \int_1^3 (t^2 - 4t + 3) \, dt = \left[\frac{t^3}{3} - 2t^2 + 3t\right]_1^3\)
\(= (9 - 18 + 9) - (\frac{1}{3} - 2 + 3) = 0 - \frac{4}{3} = -\frac{4}{3}\) m
Interval [3, 4]:
\(s_3 = \int_3^4 (t^2 - 4t + 3) \, dt = \left[\frac{t^3}{3} - 2t^2 + 3t\right]_3^4\)
\(= (\frac{64}{3} - 32 + 12) - (9 - 18 + 9) = \frac{4}{3} - 0 = \frac{4}{3}\) m
Total distance:
\(\text{Distance} = |s_1| + |s_2| + |s_3| = \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4\) m
Total distance = 4 m
3. Motion Graphs
Position-Time Graphs
Reading Position-Time Graphs:
- Slope = velocity: Steeper slope = greater speed
- Positive slope: Moving in positive direction
- Negative slope: Moving in negative direction
- Zero slope (horizontal): At rest
- Curved graph: Acceleration present (velocity changing)
Velocity-Time Graphs
Reading Velocity-Time Graphs:
- Slope = acceleration: Steeper slope = greater acceleration
- Area under curve = displacement: Use definite integral
- Area above x-axis: Positive displacement
- Area below x-axis: Negative displacement
- Total area (absolute): Total distance traveled
- Horizontal line: Constant velocity (zero acceleration)
Key Relationships
Graph Connections:
- Slope of s-t graph = velocity
- Slope of v-t graph = acceleration
- Area under v-t graph = displacement
- Area under a-t graph = change in velocity
4. Using GDC for Kinematics
Essential GDC Functions
Key Calculator Operations:
- Finding zeros: Solve for when \(v(t) = 0\) to find rest times
- Numerical differentiation: Find \(v(t)\) from \(s(t)\) or \(a(t)\) from \(v(t)\)
- Numerical integration: Calculate displacement or distance
- Graphing: Visualize motion to understand behavior
- Finding max/min: Determine extreme positions or velocities
Example 2: Complete Kinematics Problem
Problem: A particle moves with acceleration \(a(t) = 6t - 4\) m/s². At \(t = 0\), velocity is 2 m/s and displacement is 0 m.
(a) Find the velocity function \(v(t)\)
(b) Find the displacement function \(s(t)\)
(c) Find when the particle is at rest
(d) Find the displacement at \(t = 2\)
Solution:
(a) Find velocity function:
Integrate acceleration: \(v(t) = \int a(t) \, dt = \int (6t - 4) \, dt\)
\(v(t) = 3t^2 - 4t + C\)
Use initial condition: \(v(0) = 2\)
\(2 = 3(0)^2 - 4(0) + C\)
\(C = 2\)
\(v(t) = 3t^2 - 4t + 2\) m/s
(b) Find displacement function:
Integrate velocity: \(s(t) = \int v(t) \, dt = \int (3t^2 - 4t + 2) \, dt\)
\(s(t) = t^3 - 2t^2 + 2t + C\)
Use initial condition: \(s(0) = 0\)
\(0 = 0 - 0 + 0 + C\)
\(C = 0\)
\(s(t) = t^3 - 2t^2 + 2t\) m
(c) When particle is at rest:
Set \(v(t) = 0\): \(3t^2 - 4t + 2 = 0\)
Using quadratic formula or GDC:
\(t = \frac{4 \pm \sqrt{16 - 24}}{6} = \frac{4 \pm \sqrt{-8}}{6}\)
No real solutions—particle is never at rest
(d) Displacement at \(t = 2\):
\(s(2) = (2)^3 - 2(2)^2 + 2(2)\)
\(= 8 - 8 + 4 = 4\)
Displacement = 4 m
📋 Kinematics Quick Reference
🎯 IB Exam Strategy
Common Question Types:
- "Given v(t), find s(t)": Integrate velocity, use initial position for constant
- "When is particle at rest?": Solve \(v(t) = 0\)
- "Find total distance": Find when \(v = 0\), split integral, add absolute values
- "Maximum velocity": Set \(a(t) = 0\) and test, or use GDC
- "Interpret graph": Use slope and area relationships
- "Given initial conditions": Always use them to find integration constants
Key Reminders:
- Always use initial conditions to find constants after integration
- Distance ≠ displacement—check if velocity changes sign
- Negative velocity means backward motion, not slowing down
- Use GDC to verify calculations and find zeros
- Sketch graphs to visualize motion behavior
- Check units throughout your solution
🎉 Master Kinematics!
Kinematics brings calculus to life by describing how objects move through space and time. From analyzing vehicle motion to understanding rates of change in any context, kinematics provides the essential connection between mathematics and the physical world. Master differentiation and integration relationships, understand displacement versus distance, and excel at interpreting motion graphs to succeed in IB exams and beyond!
Key Success Factors:
- ✓ Velocity: \(v = \frac{ds}{dt}\) (differentiate displacement)
- ✓ Acceleration: \(a = \frac{dv}{dt}\) (differentiate velocity)
- ✓ Displacement: \(s = \int v \, dt\) (integrate velocity, use initial conditions)
- ✓ Distance: \(\int |v| \, dt\) (split at zeros, add absolute values)
- ✓ At rest: solve \(v(t) = 0\)
- ✓ Always check velocity sign changes for distance calculations
Use Initial Conditions • Check Signs • Visualize Motion
Master kinematics and excel in IB Mathematics! 🚀