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Calculus BC · Unit 4: Contextual Applications of Differentiation · 14 min read · Updated 2026-05-11

Straight-line motion: position, velocity, acceleration — AP Calculus BC

AP Calculus BC · Unit 4: Contextual Applications of Differentiation · 14 min read

1. Position and Instantaneous Velocity ★★☆☆☆ ⏱ 3 min

Straight-line motion models a particle constrained to move along a single axis, with all motion quantities expressed as single-valued functions of time $t \geq 0$. This is a foundational application of derivatives to real-world rates, appearing on both multiple-choice and free-response sections of the AP Calculus BC exam.

v(t) = s'(t) = \lim_{\Delta t \to 0} \frac{s(t+\Delta t) - s(t)}{\Delta t}

$v(t) > 0$ means the particle moves right/forward, $v(t) < 0$ means it moves left/backward, and $v(t) = 0$ means the particle is momentarily at rest.

2. Acceleration and Speed Change ★★★☆☆ ⏱ 4 min

Acceleration is the instantaneous rate of change of velocity with respect to time, so it connects directly to both position and velocity via differentiation.

a(t) = v'(t) = s''(t)

A common misconception is that positive acceleration means speeding up and negative acceleration means slowing down. This is incorrect. Whether a particle speeds up or slows down depends on the relative signs of $v(t)$ and $a(t)$:

  • If $v(t)$ and $a(t)$ have the **same sign**: the particle is speeding up
  • If $v(t)$ and $a(t)$ have **opposite signs**: the particle is slowing down

3. Speed and Direction Change ★★★☆☆ ⏱ 4 min

\text{speed}(t) = |v(t)|

A common AP exam question asks for all times when a particle changes direction. For continuous velocity (always the case in AP problems), a particle can only change direction if velocity changes sign, which can only happen at points where $v(t) = 0$. Follow this process to confirm direction change:

  1. Find all $t>0$ where $v(t) = 0$ (candidate times)
  2. Test the sign of $v(t)$ on intervals on either side of each candidate
  3. Confirm a sign change occurs: only then does the particle change direction

4. Concept Check ★★★☆☆ ⏱ 3 min

Common Pitfalls

Why: Students confuse the sign of acceleration with change in speed, relying on an incorrect intuitive rule that only holds when velocity is already positive.

Why: Students mix up conditions for direction change, confusing zeros of velocity with zeros of acceleration.

Why: Students forget that speed is a magnitude, so they directly write $v(t)$ instead of $|v(t)|$.

Why: Many problems have $v(t) = 0$ at a point where velocity touches zero but does not cross it, so no direction change occurs.

Why: Students confuse position and velocity, thinking negative position means no movement.

Quick Reference Cheatsheet

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