One Particle Carries Three Functions, and Each Answers a Different Question

You'll learnto read position, velocity and acceleration off one motion without swapping their jobs, and to decide speeding up or slowing down by comparing two signs instead of reading one.

A particle on a line has one job: be somewhere. Position s(t) says where it is. Differentiate once and you get velocity — how fast and which way, because the sign is a direction. Differentiate again and you get acceleration — how that velocity is changing. Three functions, one particle, and almost every error in this topic is a question asked of the wrong one. At rest means v = 0, not s = 0. Speed is |v|, so it is never negative. And whether the particle is speeding up is settled by comparing the sign of v with the sign of a — never by the sign of a alone, which on this trip is wrong for exactly half of it.

8 STEPS · 6 QUICK CHECKS · AT REST IS v = 0, NOT s = 0 · SPEED IS |v| · SPEEDING UP COMPARES TWO SIGNS · v1

s(t) = t³ − 6t² + 9t − 2 ⇒ v(t) = 3t² − 12t + 9 ⇒ a(t) = 6t − 12

Before you start

What you're looking at

A particle on a number line, moving under s(t) = t³ − 6t² + 9t − 2 for t from 0 to 4 seconds. The arrow above it is its velocity; the arrow below it is its acceleration. Both are pinned to the particle and ride with it.

The question

At any instant, is the particle speeding up or slowing down — and which of the three functions actually answers that?

Watch for

The two instants where the particle stops sitting nowhere near the three where it passes the origin, and the sign strips where acceleration changes sign once while the verdict changes three times.

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