Topic 4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration: analyze the motion of a particle along a line using derivatives.

What this topic is asking

The College Board (Topic 4.2) applies derivatives to a particle moving along a straight line. Given a position function $s(t)$, you differentiate to get velocity $v(t) = s'(t)$ and again to get acceleration $a(t) = s''(t)$, and you use the signs of these to describe the motion: which way the particle moves, when it is at rest, and whether it is speeding up or slowing down.

The connections

The speeding-up rule

A reliable procedure

Most motion questions reduce to differentiating, finding where $v$ and $a$ are zero, and testing signs on the resulting intervals.

Distance versus displacement

Two quantities are easy to confuse. Displacement over $[a, b]$ is $s(b) - s(a)$, the net change in position, which can be zero even if the particle moved a lot. Total distance travelled accounts for direction changes: you must split the interval at every time the velocity changes sign, compute the position change on each piece, and add the absolute values. For instance, if a particle goes right then left, the distance is (rightward distance) + (leftward distance), while the displacement subtracts them. AP free-response questions frequently ask for total distance precisely because it forces you to find where $v = 0$ and handle each direction separately. On the calculator section, total distance is often computed as $\int_a^b |v(t)|\,dt$ (a Unit 8 idea), but the conceptual foundation - that direction changes matter - is built here.

Reading the at-rest and turning conditions carefully

A common trap is assuming the particle changes direction every time $v = 0$. It changes direction only where $v$ actually changes sign. If $v(t) = (t - 2)^2$, then $v = 0$ at $t = 2$ but stays non-negative on both sides, so the particle is momentarily at rest without reversing. Always check the sign of $v$ on each side of a zero before concluding a direction change. Similarly, "at rest" is strictly $v = 0$, independent of acceleration; a particle can be instantaneously at rest yet accelerating (about to move off). Keeping the definitions crisp - at rest means $v = 0$, direction change means $v$ changes sign, speeding up means $v$ and $a$ share a sign - is what separates full points from partial credit on these questions.