Solving Motion Problems Using Parametric and Vector-Valued Functions BC only

A motion problem asks four questions in nearly the same words, and half the confusion is settled by the shape of the answer: position and displacement are vectors, total distance and speed are numbers. Position is $\mathbf{r}(t_0) + \int\mathbf{v}$, displacement is $\int\mathbf{v}$ and uses no initial position, total distance is $\int|\mathbf{v}|$ and is never negative, and speed is $|\mathbf{v}(t_1)|$ with no integration at all. For $\mathbf{v} = \langle t-2,\, t^{2}-4\rangle$ with $\mathbf{r}(0) = \langle 1,1\rangle$ on $[0,3]$ the four answers are $\langle -1.5, -3\rangle$, $\langle -0.5, -2\rangle$, about $8.092$, and $\sqrt{26}$.

Total distance is always at least the length of the displacement, with equality only when the direction never changes, because $\left|\int\mathbf{v}\right|$ lets opposing motion cancel before the magnitude is taken while $\int|\mathbf{v}|$ removes direction first. That is the same structure as 8.2 and 8.6, appearing a third time. A particle returning to its start has displacement $\mathbf{0}$ and a positive distance travelled. Speed is increasing when $\mathbf{v}$ and $\mathbf{a}$ point broadly together, and $|\mathbf{a}|$ is not the rate of change of speed; the particle is at rest only where both components of $\mathbf{v}$ vanish, which for this example is $t = 2$ and is exactly where its path reverses.