Welcome to Straight-Line Motion!

In this chapter, we explore one of the most practical uses of calculus: describing how objects move. Imagine a car driving along a straight highway or a ball being thrown straight up into the air. By using derivatives, we can calculate exactly how fast that object is moving and whether it is picking up speed or slowing down at any specific moment. This topic is a cornerstone of Unit 4: Contextual Applications of Differentiation.

The Hierarchy of Motion

To master straight-line motion, you only need to remember one "ladder." As you move down the ladder, you take the derivative with respect to time \( t \).

  1. Position: \( s(t) \) or \( x(t) \) (Where is it?)
  2. Velocity: \( v(t) = s'(t) \) (How is the position changing?)
  3. Acceleration: \( a(t) = v'(t) = s''(t) \) (How is the velocity changing?)

Quick Review: Remember from Unit 2 that the derivative represents a rate of change. So, velocity is simply the "rate of change of position," and acceleration is the "rate of change of velocity."

1. Position \( s(t) \)

The position function tells us where an object is located on a line relative to an origin (usually \( 0 \)) at a specific time \( t \).
• If \( s(t) > 0 \), the object is to the right of the origin (or above it).
• If \( s(t) < 0 \), the object is to the left of the origin (or below it).
• If \( s(t) = 0 \), the object is at the origin.

2. Velocity \( v(t) \)

Velocity is the first derivative of position: \( v(t) = \frac{ds}{dt} \). Unlike speed, velocity tells us both how fast and in what direction.
• Moving Right/Up: When \( v(t) > 0 \).
• Moving Left/Down: When \( v(t) < 0 \).
• At Rest: When \( v(t) = 0 \). This is a crucial point to check for when an object might be changing direction.

Key Concept: Speed

Speed is the absolute value of velocity: \( \text{Speed} = |v(t)| \). Speed is never negative! Think of the speedometer in a car; it doesn't care if you are driving North or South; it only shows how fast you are going.

3. Acceleration \( a(t) \)

Acceleration is the derivative of velocity: \( a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2} \).
• If \( a(t) > 0 \), the velocity is increasing.
• If \( a(t) < 0 \), the velocity is decreasing.

Key Takeaway: Always include units in your final answer if they are provided (e.g., \( \text{feet/sec}^2 \)). The AP exam requires correct notation and units for full credit in free-response questions!

Analyzing the Motion: Speeding Up vs. Slowing Down

This is one of the most common "trick" questions on the AP Calculus AB exam. Students often think that "negative acceleration" automatically means "slowing down." This is not always true!

To determine if an object is speeding up or slowing down, you must compare the signs of velocity and acceleration at that specific moment:

  • Speeding Up: \( v(t) \) and \( a(t) \) have the same sign (both positive or both negative).
    Analogy: You are moving forward (\( v > 0 \)) and someone pushes you forward (\( a > 0 \)). You go faster! Or, you are moving backward (\( v < 0 \)) and someone pulls you even harder backward (\( a < 0 \)).
  • Slowing Down: \( v(t) \) and \( a(t) \) have different signs (one positive, one negative).
    Analogy: You are moving forward (\( v > 0 \)) but a strong wind is blowing against you (\( a < 0 \)). You slow down.

Did you know? If you are asked when an object "changes direction," you aren't just looking for where \( v(t) = 0 \). You must verify that \( v(t) \) actually changes sign (from positive to negative or vice versa) at that point.

Step-by-Step: Solving a Motion Problem

If you are given a position function \( s(t) = t^3 - 6t^2 + 9t \) and asked to describe the motion at \( t = 2 \):

  1. Find Velocity: Take the derivative. \( v(t) = s'(t) = 3t^2 - 12t + 9 \).
  2. Evaluate Velocity at \( t = 2 \): \( v(2) = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3 \).
    Interpretation: Since \( v(2) < 0 \), the object is moving to the left.
  3. Find Acceleration: Take the derivative of velocity. \( a(t) = v'(t) = 6t - 12 \).
  4. Evaluate Acceleration at \( t = 2 \): \( a(2) = 6(2) - 12 = 0 \).
  5. Conclusion: Since the acceleration is \( 0 \), the speed is momentarily constant at \( t = 2 \). (If \( a(2) \) had been positive, we would say the object is slowing down because the signs of \( v \) and \( a \) would be different).

Common Mistakes to Avoid

1. Confusing "Average" vs. "Instantaneous":
• Instantaneous Velocity is the derivative at a point: \( v(t) \).
• Average Velocity is the change in position divided by change in time: \( \frac{s(b) - s(a)}{b - a} \). (No derivative needed!)

2. Forgetting Units:
If \( s(t) \) is in meters and \( t \) is in seconds:
• Velocity is in \( \text{m/s} \).
• Acceleration is in \( \text{m/s}^2 \).

3. Misinterpreting "Speeding Up":
Don't just look at acceleration. Always look at both \( v(t) \) and \( a(t) \). If the signs match, it's speeding up. If they don't, it's slowing down.

Summary Table for Quick Reference

Object Behavior | Condition
Moving Right | \( v(t) > 0 \)
Moving Left | \( v(t) < 0 \)
Stopped/At Rest | \( v(t) = 0 \)
Changes Direction | \( v(t) \) changes sign
Speeding Up | \( v(t) \) and \( a(t) \) same sign
Slowing Down | \( v(t) \) and \( a(t) \) opposite signs

Note: For further applications of these concepts using integrals (like finding total distance traveled), see Unit 8. For now, focus on mastering the derivatives!