\subsection{Velocity and Acceleration as Derivatives}

This subsection describes motion locally: starting from the position vector $\vec{r}(t)$, velocity and acceleration are defined by derivatives at an instant. Later sections will reverse these local definitions with definite integrals to recover displacement and changes in velocity over a time interval.

\dfn{Average and instantaneous velocity and acceleration}{Let $t$ denote time, let $\Delta t$ denote a nonzero time interval, and let $\vec{r}(t)$ denote the position vector of a particle.

The \emph{average velocity} from time $t$ to time $t+\Delta t$ is
\[
\vec{v}_{\mathrm{avg}}=\frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t}.
\]
If the limit exists, the \emph{instantaneous velocity} at time $t$ is the vector
\[
\vec{v}(t)=\lim_{\Delta t\to 0}\frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t}=\frac{d\vec{r}}{dt}.
\]

Let $\vec{v}(t)$ denote the instantaneous velocity. Then the \emph{average acceleration} from time $t$ to time $t+\Delta t$ is
\[
\vec{a}_{\mathrm{avg}}=\frac{\vec{v}(t+\Delta t)-\vec{v}(t)}{\Delta t}.
\]
If the limit exists, the \emph{instantaneous acceleration} at time $t$ is the vector
\[
\vec{a}(t)=\lim_{\Delta t\to 0}\frac{\vec{v}(t+\Delta t)-\vec{v}(t)}{\Delta t}=\frac{d\vec{v}}{dt}.
\]
The \emph{speed} at time $t$ is the scalar magnitude $|\vec{v}(t)|$.}

\thm{Derivative relations in vector and component form}{Let
\[
\vec{r}(t)=x(t)\,\hat{\imath}+y(t)\,\hat{\jmath}+z(t)\,\hat{k},
\]
where $x(t)$, $y(t)$, and $z(t)$ are coordinate functions.

Then the velocity vector is
\[
\vec{v}(t)=\frac{d\vec{r}}{dt}=\frac{dx}{dt}\,\hat{\imath}+\frac{dy}{dt}\,\hat{\jmath}+\frac{dz}{dt}\,\hat{k}.
\]
If $v_x(t)$, $v_y(t)$, and $v_z(t)$ denote the components of $\vec{v}(t)$, then
\[
v_x=\frac{dx}{dt},\qquad v_y=\frac{dy}{dt},\qquad v_z=\frac{dz}{dt}.
\]
The acceleration vector is
\[
\vec{a}(t)=\frac{d\vec{v}}{dt}=\frac{d^2\vec{r}}{dt^2}=\frac{dv_x}{dt}\,\hat{\imath}+\frac{dv_y}{dt}\,\hat{\jmath}+\frac{dv_z}{dt}\,\hat{k}.
\]
If $a_x(t)$, $a_y(t)$, and $a_z(t)$ denote the components of $\vec{a}(t)$, then
\[
a_x=\frac{dv_x}{dt}=\frac{d^2x}{dt^2},\qquad a_y=\frac{dv_y}{dt}=\frac{d^2y}{dt^2},\qquad a_z=\frac{dv_z}{dt}=\frac{d^2z}{dt^2}.
\]
In two-dimensional motion, the same formulas hold with the $z$-terms omitted.}

\pf{Why these formulas are true}{By definition,
\[
\vec{v}(t)=\lim_{\Delta t\to 0}\frac{\vec{r}(t+\Delta t)-\vec{r}(t)}{\Delta t}=\frac{d\vec{r}}{dt}.
\]
If
\[
\vec{r}(t)=x(t)\,\hat{\imath}+y(t)\,\hat{\jmath}+z(t)\,\hat{k},
\]
then differentiating component-by-component gives
\[
\frac{d\vec{r}}{dt}=\frac{dx}{dt}\,\hat{\imath}+\frac{dy}{dt}\,\hat{\jmath}+\frac{dz}{dt}\,\hat{k}.
\]
Likewise,
\[
\vec{a}(t)=\lim_{\Delta t\to 0}\frac{\vec{v}(t+\Delta t)-\vec{v}(t)}{\Delta t}=\frac{d\vec{v}}{dt}.
\]
Differentiating the velocity components gives
\[
a_x=\frac{dv_x}{dt},\qquad a_y=\frac{dv_y}{dt},\qquad a_z=\frac{dv_z}{dt},
\]
and substituting $v_x=dx/dt$, $v_y=dy/dt$, and $v_z=dz/dt$ gives
\[
a_x=\frac{d^2x}{dt^2},\qquad a_y=\frac{d^2y}{dt^2},\qquad a_z=\frac{d^2z}{dt^2}.
\]}

\cor{One-dimensional motion and a speed caution}{If motion is confined to the $x$-axis, so that
\[
\vec{r}(t)=x(t)\,\hat{\imath},
\]
then
\[
\vec{v}(t)=v_x(t)\,\hat{\imath}\qquad\text{and}\qquad \vec{a}(t)=a_x(t)\,\hat{\imath},
\]
with
\[
v_x=\frac{dx}{dt},\qquad a_x=\frac{dv_x}{dt}=\frac{d^2x}{dt^2}.
\]
However, in two or three dimensions, constant speed $|\vec{v}|$ does not by itself imply zero acceleration, because the direction of $\vec{v}$ can change even when its magnitude stays the same.}

\qs{Worked example}{Let $t$ denote time in seconds. A particle moves in the $xy$-plane with position vector
\[
\vec{r}(t)=\bigl(t^2-4t\bigr)\hat{\imath}+\bigl(3t-t^2\bigr)\hat{\jmath}\;\mathrm{m}.
\]
Let $x(t)=t^2-4t$ and let $y(t)=3t-t^2$, where $x(t)$ and $y(t)$ are measured in meters.

Find $\vec{v}(t)$ and $\vec{a}(t)$. Then find $\vec{v}(1.0\,\mathrm{s})$, $\vec{a}(1.0\,\mathrm{s})$, and the speed at $t=1.0\,\mathrm{s}$. Interpret the signs of the velocity components at $t=1.0\,\mathrm{s}$.}

\sol From
\[
x(t)=t^2-4t\qquad\text{and}\qquad y(t)=3t-t^2,
\]
the component formulas for velocity give
\[
v_x=\frac{dx}{dt}=2t-4\qquad\text{and}\qquad v_y=\frac{dy}{dt}=3-2t.
\]
Therefore,
\[
\vec{v}(t)=\bigl(2t-4\bigr)\hat{\imath}+\bigl(3-2t\bigr)\hat{\jmath}\;\mathrm{m/s}.
\]

Differentiate again to find the acceleration components:
\[
a_x=\frac{dv_x}{dt}=2\qquad\text{and}\qquad a_y=\frac{dv_y}{dt}=-2.
\]
So,
\[
\vec{a}(t)=2\hat{\imath}-2\hat{\jmath}\;\mathrm{m/s^2}.
\]

At $t=1.0\,\mathrm{s}$,
\[
\vec{v}(1.0\,\mathrm{s})=\bigl(2(1.0)-4\bigr)\hat{\imath}+\bigl(3-2(1.0)\bigr)\hat{\jmath}=-2\hat{\imath}+\hat{\jmath}\;\mathrm{m/s},
\]
and
\[
\vec{a}(1.0\,\mathrm{s})=2\hat{\imath}-2\hat{\jmath}\;\mathrm{m/s^2}.
\]

The speed at $t=1.0\,\mathrm{s}$ is the magnitude of the velocity vector:
\[
|\vec{v}(1.0\,\mathrm{s})|=\sqrt{(-2)^2+(1)^2}\;\mathrm{m/s}=\sqrt{5}\;\mathrm{m/s}.
\]

Because $v_x(1.0\,\mathrm{s})=-2\,\mathrm{m/s}$, the particle is moving in the negative $x$-direction at that instant. Because $v_y(1.0\,\mathrm{s})=1\,\mathrm{m/s}$, the particle is moving in the positive $y$-direction at that instant. So at $t=1.0\,\mathrm{s}$ the particle is moving left and upward, with speed $\sqrt{5}\,\mathrm{m/s}$.