Four questions that sound alike BC only

Given a velocity $\mathbf{v}(t)$ and an initial position $\mathbf{r}(t_0)$:

1. Where is it at $t_1$? $\mathbf{r}(t_1) = \mathbf{r}(t_0) + \int_{t_0}^{t_1}\mathbf{v}\,dt$. A vector.
2. How far did it move, net? $\int_{t_0}^{t_1}\mathbf{v}\,dt$, the displacement. A vector, and it does not use the initial position at all.
3. How far did it travel? $\int_{t_0}^{t_1}|\mathbf{v}|\,dt$, the total distance. A number, never negative.
4. How fast is it going at $t_1$? $|\mathbf{v}(t_1)|$, the speed. A number, and no integration is involved.

Two of these are vectors and two are scalars, so half the errors are visible from the type of the answer alone. An ordered pair offered as a distance, or a single number offered as a position, is wrong before its value is examined.

The remaining confusions are within each type. Between the two vectors, the difference is whether the initial position was added. Between the two numbers, the difference is whether anything was integrated.

Take $\mathbf{v}(t) = \langle t - 2,\; t^{2} - 4\rangle$ with $\mathbf{r}(0) = \langle 1,\, 1\rangle$, over $[0, 3]$.

1. Displacement: $\int_0^3 \langle t-2,\, t^{2}-4\rangle\,dt = \left\langle \tfrac{9}{2} - 6,\; 9 - 12 \right\rangle = \langle -1.5,\, -3\rangle$.
2. Position at $t = 3$: $\langle 1, 1\rangle + \langle -1.5, -3\rangle = \langle -0.5,\, -2\rangle$.
3. Total distance: $\int_0^3 \sqrt{(t-2)^{2} + (t^{2}-4)^{2}}\;dt \approx 8.092$.
4. Speed at $t = 3$: $|\langle 1,\, 5\rangle| = \sqrt{26} \approx 5.099$.

Four different answers, and the gap between the second and third is the point. The particle travels about $8.09$ units of path while ending only $|\langle -1.5, -3\rangle| = \sqrt{11.25} \approx 3.354$ from where it started, because both components change sign inside the interval and the motion doubles back.

That inequality is general and worth carrying as a check: total distance $\geq$ the length of the displacement, with equality only when the direction of travel never changes. A computed distance smaller than the displacement's length is impossible.

Compare the two:

$$\left|\int_{t_0}^{t_1}\mathbf{v}\,dt\right| \qquad\text{against}\qquad \int_{t_0}^{t_1}\left|\mathbf{v}\right|dt.$$

The first integrates with direction and takes the magnitude at the very end, so opposing motion cancels before anything is measured. The second discards direction inside the integral, so every instant contributes positively and nothing can cancel.

This is exactly the structure of 8.6's $\left|\int (f-g)\right|$ against $\int|f-g|$, and of 8.2's displacement against total distance. Third appearance, same shape: taking an absolute value before adding is a different operation from taking it after.

The clean extreme case: a particle that returns to its starting point has displacement $\mathbf{0}$, so the first expression is $0$, while the second is the whole length of the loop it travelled. No amount of care with the first expression recovers the second.

One practical note. The distance integral almost never has an elementary antiderivative, because a square root of a sum of squares rarely simplifies. On the exam these are calculator-active by design, and setting the integral up correctly is what is being scored.

Speed is $|\mathbf{v}(t)|$, and whether it is rising is a question about the derivative of that magnitude, not about $|\mathbf{a}|$.

The reliable route is to write the speed as a function of $t$ and differentiate it, or, on a calculator-active problem, to evaluate the speed at nearby times. The one-dimensional shortcut from 8.2 (speed rises when $v$ and $a$ share a sign) generalises to the speed rises when $\mathbf{v}$ and $\mathbf{a}$ point in broadly the same direction, and its precise form uses the dot product, which is outside the AP scope.

Two facts worth holding, both already met in 9.4:

1. $|\mathbf{a}|$ is not the rate of change of speed. Uniform circular motion has $|\mathbf{a}| = 1$ and constant speed.
2. A speed of zero is where the particle is at rest, and that requires both components of $\mathbf{v}$ to vanish at once. In the running example $t - 2$ and $t^{2} - 4$ both vanish at $t = 2$, so the particle genuinely stops there, which is exactly what lets the path double back.

That second observation ties the whole topic together: the instant the particle is momentarily at rest is the instant its path can reverse, and a reversal is what makes distance exceed displacement.