Three functions, one particle AB & BC

Let $s(t)$ be the position of a particle on a number line at time $t$. Then

$$v(t) = s'(t), \qquad a(t) = v'(t) = s''(t).$$

Position is a location, and it can be negative simply because the particle is left of the origin. Velocity is the rate at which that location changes, and its sign is a direction: positive means moving in the positive direction, negative means moving in the negative direction. Acceleration is the rate at which the velocity changes.

Two consequences get used constantly:

1. At rest means $v(t) = 0$, not $s(t) = 0$. A particle sitting still at position 12 has $v = 0$ and $s = 12$; a particle flying through the origin has $s = 0$ and $v \neq 0$.
2. The particle changes direction only where $v$ changes sign. Touching zero is not enough on its own, the sign has to flip across it.

Going the other way, from $a$ back to $v$ and $s$, requires antiderivatives and an initial condition. That is Unit 6 and Unit 8 work; here everything runs downhill, by differentiating.

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

Velocity carries a direction in its sign. Speed throws that sign away and keeps only the magnitude. A particle with $v = -8$ meters per second has a speed of $8$ meters per second and is moving in the negative direction.

The distinction is not cosmetic. It changes what "increasing" means:

1. Velocity going from $-8$ to $-3$ is increasing (it is moving up the number line) while the speed is decreasing from 8 to 3.
2. Velocity going from $2$ to $7$ has both increasing.

The same split applies to averages. Average velocity over $[t_1, t_2]$ is net displacement over elapsed time, so it can be zero for a particle that has been moving the whole while. Average speed is total distance traveled over elapsed time, and it is zero only if the particle never moved.

Speed is $|v|$, so speed increases exactly when $|v|$ increases, and the test is:

1. $v$ and $a$ have the same sign $\Rightarrow$ the particle is speeding up.
2. $v$ and $a$ have opposite signs $\Rightarrow$ the particle is slowing down.

Both parts of that are worth saying in words. Acceleration in the same direction as the motion pushes the particle along and its speed climbs. Acceleration against the motion pushes back and the speed falls, no matter which way the particle happens to be traveling.

So a negative acceleration does not mean slowing down. If $v = -6$ and $a = -2$, the particle is moving in the negative direction and being pushed further that way: velocity is heading toward $-8$, and the speed is climbing from 6 toward 8. It is speeding up.

Deciding from the sign of $a$ alone is the single most-missed idea in this unit, and it is wrong exactly half the time.

Graphs make the roles easy to mix up, because every graph looks like a graph. The fix is to ask, every time, which function is plotted?

If the graph shown is $v(t)$:

1. Its height is a velocity, never a position.
2. Where it crosses the horizontal axis, the particle is at rest.
3. Its slope is the acceleration.
4. Where it is above the axis the particle moves in the positive direction, and where it is below, the negative direction.

If the graph shown is $s(t)$:

1. Its height is a position.
2. Where its tangent is horizontal, the particle is at rest.
3. Its slope is the velocity, and the concavity carries the sign of the acceleration.
4. Crossing the axis means passing the origin at full speed, not stopping.

Getting acceleration from position takes two differentiations. One is a common miss, and it produces a velocity wearing an acceleration's label.