A change is not a position AB & BC

Differentiating runs $s \to v \to a$. Integrating runs back the other way, and every step gives a change, not a value. This is the net change theorem of Topic 6.1 with units attached:

$$\int_{t_1}^{t_2} v(t)\,dt = s(t_2) - s(t_1), \qquad \int_{t_1}^{t_2} a(t)\,dt = v(t_2) - v(t_1).$$

Read each one left to right and the phrasing writes itself: the integral of velocity over an interval is the displacement across it, and the integral of acceleration is the change in velocity. Neither is a position, and neither is a velocity.

The units confirm it. Velocity in meters per second, integrated against $dt$ in seconds, gives meters: a length, which is what a displacement is. Acceleration in meters per second squared, integrated against seconds, gives meters per second. The units of an integral are always the units of the integrand times the units of the variable, and that product is what tells you which quantity you are holding.

So an integral answers "by how much did it change", and a separate piece of information is needed to answer "where is it now".

Rearranging the first identity gives the form every AP problem actually uses:

$$s(t_2) = s(t_1) + \int_{t_1}^{t_2} v(t)\,dt.$$

The starting position is an addend. It is not a bound, it is not a factor, and it does not cancel. A particle at $s(0) = 5$ whose displacement over $[0, 4]$ is $\frac{4}{3}$ ends at $5 + \frac{4}{3} = \frac{19}{3}$, and reporting $\frac{4}{3}$ answers a question nobody asked.

Three signs that the initial value has gone missing:

1. The answer to "where is the particle" is small. A displacement is often small while a position is not.
2. The given starting value never appeared in the work. If a problem states $s(0)$, it is because $s(0)$ is needed.
3. The same number answers two different questions. "How far did it move" and "where is it" should not have the same answer unless the particle started at the origin.

The same structure governs every applied version in 8.3: a tank holding $30$ gallons into which $45$ more flow holds $75$, not $45$. What changes from problem to problem is the noun, never the shape.

These are different questions and they need different integrals.

$$\text{displacement} = \int_{t_1}^{t_2} v(t)\,dt, \qquad \text{total distance} = \int_{t_1}^{t_2} |v(t)|\,dt.$$

Displacement lets the trips cancel, because that is what "net" means. Total distance counts every meter travelled regardless of direction, so it can never be negative and is never smaller than the size of the displacement.

The absolute value is an instruction to split the interval, not a repair to apply at the end. The procedure:

1. Solve $v(t) = 0$. Every root inside the interval is a place the particle turns around.
2. Integrate over each piece separately. On each one $v$ keeps a single sign.
3. Add the absolute values. One per piece, then sum.

Take $v(t) = (t - 1)(t - 3)$ on $[0, 4]$. It vanishes at $t = 1$ and $t = 3$, and the three pieces contribute $\frac{4}{3}$, $-\frac{4}{3}$ and $\frac{4}{3}$. The displacement is their sum, $\frac{4}{3}$. The total distance is $\frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4$, three times as large.

Two failures are worth naming. Taking $\left|\int v\right|$ instead of $\int |v|$ moves the absolute value outside, where it cannot separate anything: here it returns $\frac{4}{3}$, missing the turn entirely. And splitting at only one of the two roots leaves a piece where $v$ still changes sign, so that piece cancels internally and the answer is too small.

Speed is $|v(t)|$. It has no direction, so it is never negative, and it is what a speedometer reads.

The test for speeding up is about agreement of signs, not about the sign of $a$ alone:

1. $v$ and $a$ share a sign: speed increases. The push is in the direction of travel.
2. $v$ and $a$ have opposite signs: speed decreases. The push opposes the motion.

So a particle with $v = -5$ and $a = 2$ is slowing down even though its acceleration is positive, and its velocity is increasing at the same time. Velocity increasing and speed decreasing are perfectly compatible; they only sound contradictory if speed and velocity have been treated as the same word.

One consequence used constantly on the exam: at a moment when the particle changes direction, $v = 0$, so the speed is momentarily zero. The particle is not at rest for an interval, and $a$ need not be zero there. A velocity of zero at an instant tells you about a turn, not about a stop.