IB Mathematics AI – Topic 5

Key Variables:

• s(t): Displacement (position) at time t (meters, m)
• v(t): Velocity at time t (meters per second, m/s)
• a(t): Acceleration at time t (meters per second squared, m/s²)
• t: Time (seconds, s)

Calculus Relationships:

1. Velocity is derivative of displacement:

\[ v(t) = \frac{ds}{dt} = s'(t) \]

2. Acceleration is derivative of velocity:

\[ a(t) = \frac{dv}{dt} = v'(t) \]

3. Acceleration is second derivative of displacement:

\[ a(t) = \frac{d^2s}{dt^2} = s''(t) \]

Integration Relationships (going backwards):

1. Displacement from velocity:

\[ s(t) = \int v(t)\,dt \]

2. Velocity from acceleration:

\[ v(t) = \int a(t)\,dt \]

Summary Diagram:

Displacement s(t)
↓ differentiate / ↑ integrate
Velocity v(t)
↓ differentiate / ↑ integrate
Acceleration a(t)

⚠️ Common Pitfalls & Tips:

• Direction matters: positive/negative indicates direction
• Don't confuse speed (scalar) with velocity (vector)
• Units: s in m, v in m/s, a in m/s²
• When integrating, don't forget +C (unless definite integral)

Displacement (from t₁ to t₂):

Change in position (vector - includes direction)

\[ \text{Displacement} = \int_{t_1}^{t_2} v(t)\,dt = s(t_2) - s(t_1) \]

Can be positive, negative, or zero

Example: Walk 5m forward, then 5m back → displacement = 0

Total Distance Travelled (from t₁ to t₂):

Total length of path (scalar - always positive)

\[ \text{Distance} = \int_{t_1}^{t_2} |v(t)|\,dt \]

Always positive or zero

Example: Walk 5m forward, then 5m back → distance = 10m

Key Difference:

Finding Distance When Velocity Changes Sign:

1. Find when v(t) = 0 (particle changes direction)
2. Split integral at these points
3. Take absolute value of each part
4. Add all parts together

⚠️ Common Pitfalls & Tips:

• CRITICAL: Displacement can be negative, distance cannot
• If particle changes direction, must split the integral
• Use absolute value |v(t)| for distance
• Check when v(t) = 0 to find direction changes

General Strategy:

1. Identify what you have: s(t), v(t), or a(t)?
2. Identify what you need: s, v, or a?
3. Decide operation: Differentiate (to go down) or Integrate (to go up)
4. Apply calculus: Use GDC for calculations
5. Check units: Ensure answer has correct units

⚠️ Common Pitfalls & Tips:

• Always state initial conditions when integrating
• Check if particle is at rest (v = 0) or changes direction
• Negative velocity means moving in negative direction
• Use GDC to verify all calculations

Solution:

(a) When is particle at rest?

Particle at rest when v(t) = 0

\[ 3t^2 - 12t + 9 = 0 \]

\[ t^2 - 4t + 3 = 0 \]

\[ (t-1)(t-3) = 0 \]

Answer: t = 1s and t = 3s

(b) Find acceleration at t = 2

Acceleration is derivative of velocity:

\[ a(t) = \frac{dv}{dt} = 6t - 12 \]

At t = 2:

\[ a(2) = 6(2) - 12 = 0 \text{ m/s}^2 \]

Answer: 0 m/s²

(c) Find displacement function s(t)

Displacement is integral of velocity:

\[ s(t) = \int (3t^2 - 12t + 9)\,dt \]

\[ s(t) = t^3 - 6t^2 + 9t + C \]

Use initial condition: s(0) = 2

\[ 2 = 0 - 0 + 0 + C \]

\[ C = 2 \]

Answer: s(t) = t³ - 6t² + 9t + 2 meters

Solution:

(a) Total displacement

\[ \text{Displacement} = \int_0^4 (t^2 - 4t + 3)\,dt \]

\[ = \left[\frac{t^3}{3} - 2t^2 + 3t\right]_0^4 \]

At t = 4: \(\frac{64}{3} - 32 + 12 = \frac{64 - 60}{3} = \frac{4}{3}\)

At t = 0: 0

Answer: Displacement = \(\frac{4}{3}\) m or 1.33m

(b) Total distance travelled

Step 1: Find when v(t) = 0 (direction changes)

\[ t^2 - 4t + 3 = 0 \]

\[ (t-1)(t-3) = 0 \]

Particle changes direction at t = 1 and t = 3

Step 2: Split into three intervals

Check signs: v(0.5) > 0, v(2) < 0, v(3.5) > 0

Step 3: Calculate each part

\[ \text{Distance} = \int_0^1 v\,dt + \left|\int_1^3 v\,dt\right| + \int_3^4 v\,dt \]

From 0 to 1: \(\frac{4}{3}\)

From 1 to 3: \(-\frac{4}{3}\), so absolute value is \(\frac{4}{3}\)

From 3 to 4: \(\frac{4}{3}\)

\[ \text{Total} = \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4 \text{ m} \]

Answer: Total distance = 4 m

Key Relationships Between Graphs:

1. Displacement-Time Graph (s vs t):

• Gradient: gives velocity
• Horizontal line: particle at rest (v = 0)
• Positive slope: moving in positive direction
• Negative slope: moving in negative direction
• Turning point: particle changes direction

2. Velocity-Time Graph (v vs t):

• Gradient: gives acceleration
• Area under curve: gives displacement
• Area with absolute value: gives distance
• Crosses x-axis: particle changes direction
• Above x-axis: positive velocity
• Below x-axis: negative velocity

3. Acceleration-Time Graph (a vs t):

• Area under curve: gives change in velocity
• Positive: velocity increasing
• Negative: velocity decreasing (deceleration)

Summary Table:

⚠️ Common Pitfalls & Tips:

• Gradient = differentiation, Area = integration
• Velocity graph crossing x-axis = direction change
• Area below x-axis on v-t graph is negative displacement
• Use GDC to sketch and analyze graphs

Calculus Chain

• s → v → a (differentiate)
• a → v → s (integrate)
• v = ds/dt, a = dv/dt

Distance vs Displacement

• Displacement: ∫v dt
• Distance: ∫|v| dt
• Split when v = 0

Key Indicators

• At rest: v = 0
• Changes direction: v = 0
• Constant velocity: a = 0

Graphs

• Gradient → differentiate
• Area → integrate
• v-t: area = displacement