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Unit 8.2 – Connecting Position, Velocity, and Acceleration Using Integrals

AP® Calculus AB & BC | Motion Along a Line

Why This Matters: Motion problems are the classic application of calculus! The relationships between position, velocity, and acceleration involve both derivatives (going down) and integrals (going up). This topic connects differentiation and integration in a concrete, physical context. Understanding these connections is essential for physics, engineering, and of course, the AP® exam where motion problems appear frequently!

🎯 The Big Picture: The Motion Connection

The Complete Relationship

DIFFERENTIATION (Going Down) ↓

\[ s(t) \xrightarrow{\text{derivative}} v(t) \xrightarrow{\text{derivative}} a(t) \]
\[ \text{Position} \xrightarrow{s'(t)} \text{Velocity} \xrightarrow{v'(t)} \text{Acceleration} \]

INTEGRATION (Going Up) ↑

\[ a(t) \xrightarrow{\text{integrate}} v(t) \xrightarrow{\text{integrate}} s(t) \]
\[ \text{Acceleration} \xrightarrow{\int a(t)\,dt} \text{Velocity} \xrightarrow{\int v(t)\,dt} \text{Position} \]

📚 Essential Definitions

THE THREE FUNCTIONS

Position Function: \(s(t)\)
  • Location of particle at time \(t\)
  • Units: meters, feet, miles, etc.
  • Can be positive, negative, or zero
Velocity Function: \(v(t) = s'(t)\)
  • Rate of change of position
  • Units: m/s, ft/s, mph, etc.
  • Positive: moving right/up
  • Negative: moving left/down
  • Zero: momentarily stopped
Acceleration Function: \(a(t) = v'(t) = s''(t)\)
  • Rate of change of velocity
  • Units: m/s², ft/s², etc.
  • Positive: velocity increasing
  • Negative: velocity decreasing

📐 The Essential Formulas

Integration Formulas for Motion

1. From Acceleration to Velocity:
\[ v(t) = v_0 + \int_{t_0}^t a(u) \, du \]

where \(v_0 = v(t_0)\) is initial velocity

2. From Velocity to Position:
\[ s(t) = s_0 + \int_{t_0}^t v(u) \, du \]

where \(s_0 = s(t_0)\) is initial position

3. Net Change in Position (Displacement):
\[ \text{Displacement} = s(b) - s(a) = \int_a^b v(t) \, dt \]
4. Net Change in Velocity:
\[ \text{Change in velocity} = v(b) - v(a) = \int_a^b a(t) \, dt \]

⚖️ Displacement vs. Total Distance Traveled

Critical Distinction:

Displacement (Net Change):
\[ \text{Displacement} = \int_a^b v(t) \, dt = s(b) - s(a) \]
  • Can be positive, negative, or zero
  • Final position minus initial position
  • Direction matters!
Total Distance Traveled:
\[ \text{Total Distance} = \int_a^b |v(t)| \, dt \]
  • Always positive (or zero)
  • How far particle actually moved
  • Split integral where \(v(t)\) changes sign

⚠️ Warning: Displacement ≠ Distance!

If a particle moves right 5 meters then left 3 meters:

  • Displacement = 2 meters (net change)
  • Total distance = 8 meters (5 + 3)

🏃 Speed vs. Velocity

Important Distinction:

Velocity (vector):
\[ v(t) = s'(t) \]

Has direction (sign matters)

Speed (scalar):
\[ \text{Speed} = |v(t)| = |s'(t)| \]

Always non-negative (just magnitude)

📖 Comprehensive Worked Examples

Example 1: Finding Position from Velocity

Problem: A particle moves along a line with velocity \(v(t) = 3t^2 - 12t + 9\) m/s. At \(t = 0\), the position is \(s(0) = 2\) meters. Find the position function \(s(t)\).

Solution:

Step 1: Use the formula

\[ s(t) = s_0 + \int_0^t v(u) \, du \]

Step 2: Integrate

\[ s(t) = 2 + \int_0^t (3u^2 - 12u + 9) \, du \]
\[ = 2 + [u^3 - 6u^2 + 9u]_0^t \]
\[ = 2 + (t^3 - 6t^2 + 9t) - 0 \]

Answer:

\[ s(t) = t^3 - 6t^2 + 9t + 2 \text{ meters} \]

Example 2: Displacement vs. Distance

Problem: A particle moves with velocity \(v(t) = t^2 - 5t + 4\) m/s for \(0 \leq t \leq 5\). Find:
(a) Displacement
(b) Total distance traveled

Solution:

(a) Displacement:

\[ \text{Displacement} = \int_0^5 (t^2 - 5t + 4) \, dt \]
\[ = \left[\frac{t^3}{3} - \frac{5t^2}{2} + 4t\right]_0^5 \]
\[ = \frac{125}{3} - \frac{125}{2} + 20 = -\frac{5}{6} \text{ meters} \]

(b) Total distance: Find when \(v(t) = 0\)

\[ t^2 - 5t + 4 = 0 \]
\[ (t-1)(t-4) = 0 \quad \Rightarrow \quad t = 1, 4 \]

Split integral at \(t = 1\) and \(t = 4\):

\[ \text{Distance} = \left|\int_0^1 v(t)\,dt\right| + \left|\int_1^4 v(t)\,dt\right| + \left|\int_4^5 v(t)\,dt\right| \]

After calculation:

\[ = \frac{11}{6} + \frac{9}{2} + \frac{7}{6} = \frac{37}{6} \text{ meters} \]

Example 3: Finding Velocity from Acceleration

Problem: A particle has acceleration \(a(t) = 6t - 2\) m/s². At \(t = 0\), velocity is \(v(0) = 5\) m/s. Find \(v(t)\) and \(v(3)\).

Solution:

\[ v(t) = v_0 + \int_0^t a(u) \, du \]
\[ v(t) = 5 + \int_0^t (6u - 2) \, du \]
\[ = 5 + [3u^2 - 2u]_0^t \]
\[ = 5 + 3t^2 - 2t \]

At \(t = 3\):

\[ v(3) = 5 + 3(9) - 2(3) = 26 \text{ m/s} \]

🔑 Key Concepts Summary

Motion Relationships
To Find From Formula
Velocity Position \(v(t) = s'(t)\)
Acceleration Velocity \(a(t) = v'(t) = s''(t)\)
Position Velocity \(s(t) = s_0 + \int_{t_0}^t v(u)\,du\)
Velocity Acceleration \(v(t) = v_0 + \int_{t_0}^t a(u)\,du\)
Displacement Velocity \(\int_a^b v(t)\,dt\)
Distance Velocity \(\int_a^b |v(t)|\,dt\)

💡 Essential Tips & Strategies

✅ Success Strategies:

  • Initial conditions are crucial: Always include \(s_0\) or \(v_0\)
  • Check units: Position (m), velocity (m/s), acceleration (m/s²)
  • Direction matters: Positive vs negative values
  • For distance: Find where \(v(t) = 0\) and split integral
  • Speed = |velocity|: Always positive
  • When stopped: \(v(t) = 0\), not \(a(t) = 0\)
  • Speeding up/slowing down: Look at signs of \(v\) and \(a\)

🔥 Speeding Up vs. Slowing Down:

  • Speeding up: \(v\) and \(a\) have same sign
  • Slowing down: \(v\) and \(a\) have opposite signs
  • Moving right: \(v > 0\)
  • Moving left: \(v < 0\)

❌ Common Mistakes to Avoid

  • Mistake 1: Confusing displacement with distance
  • Mistake 2: Forgetting initial conditions (\(s_0\) or \(v_0\))
  • Mistake 3: Not finding where \(v(t) = 0\) for distance problems
  • Mistake 4: Wrong limits of integration
  • Mistake 5: Confusing when particle is stopped (\(v=0\)) vs at origin (\(s=0\))
  • Mistake 6: Sign errors with negative velocities
  • Mistake 7: Not using absolute value for distance
  • Mistake 8: Mixing up velocity and speed
  • Mistake 9: Integration errors (especially with bounds)
  • Mistake 10: Not including units in final answer

📝 Practice Problems

Try these problems:

  1. If \(v(t) = 2t - 3\) m/s and \(s(0) = 5\) m, find \(s(t)\).
  2. If \(a(t) = 12t\) m/s² and \(v(0) = -4\) m/s, find \(v(2)\).
  3. For \(v(t) = t^2 - 4\) on \([0, 3]\), find displacement and distance.
  4. When is a particle at rest if \(v(t) = t^2 - 9\)?

Answers:

  1. \(s(t) = t^2 - 3t + 5\) m
  2. \(v(2) = 20\) m/s
  3. Displacement: \(-3\) m, Distance: \(\frac{35}{3}\) m
  4. \(t = 3\) (only positive value)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

  • Show integral setup: Write \(\int_a^b v(t)\,dt\) before evaluating
  • Include initial conditions: Show \(s_0\) or \(v_0\)
  • For distance: Show finding zeros of \(v(t)\)
  • Show splitting integrals: When velocity changes sign
  • Include units: Throughout your work
  • Label clearly: Displacement vs. distance
  • Show antiderivative: Before applying bounds
  • Explain reasoning: For speeding up/slowing down questions

💯 Exam Strategy:

  1. Read carefully: displacement or distance?
  2. Identify given: position, velocity, or acceleration?
  3. Note initial conditions (always given!)
  4. Set up integral with correct bounds
  5. For distance: find where \(v(t) = 0\) first
  6. Show all work (even calculator allowed problems)
  7. Include units in final answer
  8. Check reasonableness of answer

⚡ Quick Reference Guide

MOTION FORMULAS ESSENTIALS

The Relationship:

\[ a(t) = v'(t) = s''(t) \]
\[ v(t) = s'(t) = \int a(t)\,dt + C \]
\[ s(t) = \int v(t)\,dt + C \]

Key Formulas:

  • Displacement: \(\int_a^b v(t)\,dt = s(b) - s(a)\)
  • Distance: \(\int_a^b |v(t)|\,dt\)
  • Speed: \(|v(t)|\)

Remember:

  • Always include initial conditions
  • Displacement can be negative
  • Distance is always non-negative
  • Find zeros of \(v(t)\) for distance problems

Master Motion with Integration! The fundamental relationships: velocity is the derivative of position (\(v = s'\)) and acceleration is the derivative of velocity (\(a = v' = s''\)). Going the other direction: integrate acceleration to get velocity \(v(t) = v_0 + \int a(t)\,dt\) and integrate velocity to get position \(s(t) = s_0 + \int v(t)\,dt\). Always include initial conditions! Displacement (net change) = \(\int_a^b v(t)\,dt\) can be negative. Total distance = \(\int_a^b |v(t)|\,dt\) is always non-negative—find where \(v(t) = 0\) and split the integral. Speed = \(|v(t)|\) (magnitude only). Particle is stopped when \(v = 0\). Speeding up when \(v\) and \(a\) have same sign; slowing down when opposite signs. Units: position (m), velocity (m/s), acceleration (m/s²). This is THE most common application of integration on AP® exams—practice until automatic! 🎯✨