\subsection{Motion Graphs, Slopes, and Signed Areas}

This subsection connects the local language of derivatives to the global language of accumulation. In one-dimensional motion, or when motion is analyzed along the $x$-axis, AP problems often ask you to translate among a verbal description, a graph such as $x(t)$, $v_x(t)$, or $a_x(t)$, and equations relating slope and area.

\dfn{Common motion graphs and the meaning of slope}{Let $t$ denote time. Let $x(t)$ denote the position coordinate along the $x$-axis, let $v_x(t)$ denote the $x$-component of velocity, and let $a_x(t)$ denote the $x$-component of acceleration.

An $x(t)$ graph shows position as a function of time. A $v_x(t)$ graph shows velocity as a function of time. An $a_x(t)$ graph shows acceleration as a function of time.

If $t_1$ and $t_2$ are two times with $t_2>t_1$, then the \emph{average slope} of a graph of a quantity $q(t)$ over the interval from $t_1$ to $t_2$ is
\[
\frac{q(t_2)-q(t_1)}{t_2-t_1}.
\]
This is the slope of the secant line through the two points on the graph.

The \emph{slope at a point} is the slope of the tangent line at that time. For motion graphs, the slope at a point gives an instantaneous rate of change, while the average slope over an interval gives an average rate of change over that interval.}

\nt{Do not confuse a graph's \emph{value} with its \emph{slope}. On an $x(t)$ graph, a point high above the axis means large position, not large velocity. On a $v_x(t)$ graph, a point at $v_x=0$ means zero velocity at that instant, while a horizontal tangent means zero acceleration at that instant. Likewise, zero slope is not the same as zero value. Also, signed area under a $v_x(t)$ graph gives displacement, not total distance traveled. If velocity changes sign, distance in one dimension is found from $\int |v_x|\,dt$, so areas below the axis must be counted with positive magnitude when finding distance.}

\mprop{Operational graph rules for one-dimensional motion}{Let $t_1$ and $t_2$ be times with $t_2>t_1$. Let $\Delta x=x(t_2)-x(t_1)$ and let $\Delta v_x=v_x(t_2)-v_x(t_1)$.

For position and velocity graphs,
\[
\text{slope of }x(t)\text{ at time }t=v_x(t),
\]
so the average slope of the position graph over $[t_1,t_2]$ is the average velocity:
\[
\frac{x(t_2)-x(t_1)}{t_2-t_1}=\frac{\Delta x}{t_2-t_1}.
\]

For velocity and acceleration graphs,
\[
\text{slope of }v_x(t)\text{ at time }t=a_x(t),
\]
so the average slope of the velocity graph over $[t_1,t_2]$ is the average acceleration:
\[
\frac{v_x(t_2)-v_x(t_1)}{t_2-t_1}=\frac{\Delta v_x}{t_2-t_1}.
\]

The signed area under the velocity graph from $t_1$ to $t_2$ gives displacement:
\[
\Delta x=\int_{t_1}^{t_2} v_x(t)\,dt.
\]
Area above the time axis contributes positively, and area below the time axis contributes negatively.

The signed area under the acceleration graph from $t_1$ to $t_2$ gives change in velocity:
\[
\Delta v_x=\int_{t_1}^{t_2} a_x(t)\,dt.
\]

In one dimension, the total distance traveled from $t_1$ to $t_2$ is
\[
\text{distance}=\int_{t_1}^{t_2} |v_x(t)|\,dt,
\]
which equals the total unsigned area between the $v_x(t)$ graph and the time axis.}

\qs{Worked example}{A particle moves along the $x$-axis. Its velocity graph $v_x(t)$ is described as follows.

From $t=0$ to $t=2.0\,\mathrm{s}$, the velocity increases linearly from $0$ to $4.0\,\mathrm{m/s}$. From $t=2.0\,\mathrm{s}$ to $t=5.0\,\mathrm{s}$, the velocity is constant at $4.0\,\mathrm{m/s}$. From $t=5.0\,\mathrm{s}$ to $t=7.0\,\mathrm{s}$, the velocity decreases linearly from $4.0\,\mathrm{m/s}$ to $-2.0\,\mathrm{m/s}$.

Find the acceleration on each time interval, the displacement from $t=0$ to $t=7.0\,\mathrm{s}$, the total distance traveled from $t=0$ to $t=7.0\,\mathrm{s}$, and the average velocity over the full $7.0\,\mathrm{s}$ interval. State when the particle moves in the negative $x$-direction.}

\sol Let $a_x$ denote the slope of the velocity graph.

From $t=0$ to $t=2.0\,\mathrm{s}$,
\[
a_x=\frac{4.0-0}{2.0-0}=2.0\,\mathrm{m/s^2}.
\]

From $t=2.0\,\mathrm{s}$ to $t=5.0\,\mathrm{s}$, the graph is horizontal, so
\[
a_x=0.
\]

From $t=5.0\,\mathrm{s}$ to $t=7.0\,\mathrm{s}$,
\[
a_x=\frac{-2.0-4.0}{7.0-5.0}=-3.0\,\mathrm{m/s^2}.
\]

Now find displacement from the signed area under the $v_x(t)$ graph.

From $t=0$ to $t=2.0\,\mathrm{s}$, the area is a triangle with base $2.0\,\mathrm{s}$ and height $4.0\,\mathrm{m/s}$:
\[
\Delta x_1=\frac{1}{2}(2.0)(4.0)=4.0\,\mathrm{m}.
\]

From $t=2.0\,\mathrm{s}$ to $t=5.0\,\mathrm{s}$, the area is a rectangle:
\[
\Delta x_2=(3.0)(4.0)=12.0\,\mathrm{m}.
\]

From $t=5.0\,\mathrm{s}$ to $t=7.0\,\mathrm{s}$, the signed area is a trapezoid:
\[
\Delta x_3=\frac{4.0+(-2.0)}{2}(2.0)=2.0\,\mathrm{m}.
\]

Therefore the total displacement is
\[
\Delta x=\Delta x_1+\Delta x_2+\Delta x_3=4.0+12.0+2.0=18.0\,\mathrm{m}.
\]

For total distance, any part of the velocity graph below the axis must be counted positively. The velocity becomes zero during the last interval, so first find that time. Starting from $v_x=4.0\,\mathrm{m/s}$ at $t=5.0\,\mathrm{s}$ with slope $-3.0\,\mathrm{m/s^2}$,
\[
0=4.0+(-3.0)(t-5.0).
\]
So,
\[
t-5.0=\frac{4.0}{3.0},
\qquad
t=\frac{19}{3}\,\mathrm{s}.
\]

From $t=5.0\,\mathrm{s}$ to $t=19/3\,\mathrm{s}$, the graph is above the axis, giving a triangle of area
\[
A_+=\frac{1}{2}\left(\frac{4}{3}\right)(4.0)=\frac{8}{3}\,\mathrm{m}.
\]

From $t=19/3\,\mathrm{s}$ to $t=7.0\,\mathrm{s}$, the graph is below the axis, giving a triangle with signed area $-\frac{2}{3}\,\mathrm{m}$ and magnitude
\[
A_-=\frac{1}{2}\left(\frac{2}{3}\right)(2.0)=\frac{2}{3}\,\mathrm{m}.
\]

Thus the total distance traveled is
\[
d=4.0+12.0+\frac{8}{3}+\frac{2}{3}=16.0+\frac{10}{3}=\frac{58}{3}\,\mathrm{m}\approx 19.3\,\mathrm{m}.
\]

The average velocity over the full interval is displacement divided by the total elapsed time of $7.0\,\mathrm{s}$:
\[
v_{x,\mathrm{avg}}=\frac{18.0\,\mathrm{m}}{7.0\,\mathrm{s}}\approx 2.57\,\mathrm{m/s}.
\]

The particle moves in the negative $x$-direction when $v_x<0$, which occurs after the graph crosses the axis. Therefore it moves in the negative $x$-direction for
\[
\frac{19}{3}\,\mathrm{s}<t\le 7.0\,\mathrm{s}.
\]

So the interval accelerations are $2.0\,\mathrm{m/s^2}$, $0$, and $-3.0\,\mathrm{m/s^2}$; the displacement is $18.0\,\mathrm{m}$; the total distance is $58/3\,\mathrm{m}$; and the average velocity is about $2.57\,\mathrm{m/s}$. 
