---
title: Motion Graphs
module: Kinematics
moduleNumber: 1
lessonNumber: 2
order: 102
summary: >
  Draw a motion as a graph and its two most useful facts turn geometric: the slope of
  the position curve is the velocity, and the area under the velocity curve is the
  displacement. We read motion in both directions — differentiating a graph for the
  next rate, integrating it back to recover position — and handle the curved,
  piecewise, and noisy graphs that real measurements produce. Along the way we see why
  a velocity estimated from two positions belongs to the midpoint of their interval,
  not its end.
topics: [Kinematics]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 2 — Motion in One Dimension; §§2-1–2-2, 2-4"
---

## Derivatives and graph interpretation

Position $x(t)$, velocity $v_x(t)$, and acceleration $a_x(t)$ are successive time
derivatives:

$$
v_x=\frac{\d x}{\d t},
\qquad
a_x=\frac{\d v_x}{\d t}=\frac{\d^2x}{\d t^2}.
$$

The slope of a position-time graph is velocity; the slope of a velocity-time graph
is acceleration. Positive acceleration does not necessarily mean increasing speed:
speed decreases whenever velocity and acceleration have opposite signs. A horizontal
position graph has zero instantaneous velocity, while a horizontal velocity graph
has zero acceleration.

Integration reverses differentiation. Between times $t_i$ and $t_f$,

$$
\Delta x=\int_{t_i}^{t_f}v_x\,\d t,
\qquad
\Delta v_x=\int_{t_i}^{t_f}a_x\,\d t.
$$

The areas are signed. A region below the time axis contributes negative displacement
or negative velocity change. Total distance requires integrating $|v_x|$, not
simply taking the signed area.

$$
% caption: The three graph forms describe the same one-dimensional motion. Position
% slope gives velocity, velocity slope gives acceleration, and the shaded signed
% area under the velocity graph gives displacement over the selected time interval.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->] (0,0) -- (3.6,0) node[right] {time};
  \draw[->] (0,0) -- (0,2.5) node[above] {position};
  \draw[thick, domain=0.2:3.2, samples=60]
    plot (\x,{0.35+0.05*(\x-0.2)+0.16*(\x-0.2)*(\x-0.2)});
  \begin{scope}[shift={(4.8,0)}]
    \draw[->] (0,0) -- (3.6,0) node[right] {time};
    \draw[->] (0,0) -- (0,2.5) node[above] {$v$};
    \fill[acc!12] (0.2,0) -- (3.2,1.7) -- (3.2,0) -- cycle;
    \draw[thick] (0.2,0.15) -- (3.2,1.7);
  \end{scope}
  \begin{scope}[shift={(9.6,0)}]
    \draw[->] (0,0) -- (3.6,0) node[right] {time};
    \draw[->] (0,0) -- (0,2.5) node[above] {$a$};
    \draw[thick] (0.2,0.9) -- (3.2,0.9);
  \end{scope}
\end{tikzpicture}
$$

> **Worked example.** A velocity graph rises linearly from $2\ \mathrm{m\,s^{-1}}$ to $8\ \mathrm{m\,s^{-1}}$ over $3\ \mathrm s$.
>
> The constant acceleration is the slope,
> $$
> a_x=\frac{8-2}{3}=2\ \mathrm{m\,s^{-2}}.
> $$
> Displacement is the trapezoid area beneath the line,
> $$
> \Delta x=\frac{2+8}{2}(3)=15\ \mathrm m.
> $$
> The average velocity equals the midpoint velocity here only because the graph is linear; the trapezoid area is the general result.

**Reading the shape of a position graph.**

A position graph records coordinate, not path length. Its height gives the object's
position relative to the origin; its tangent slope gives velocity:

- **Upward slope** — positive velocity.
- **Downward slope** — negative velocity.
- **Steep versus shallow** — larger speed, when both axes share a scale.
- **Leveling off** — velocity approaching zero, even far from the origin.

Curvature gives acceleration. A concave-up trace has increasing slope, hence
positive acceleration; a concave-down trace has decreasing slope, hence negative
acceleration. Signed velocity, not apparent steepness, fixes the direction of
motion. A curve can decrease while concave up: its velocity is negative but becoming
less negative, so the object slows while its acceleration is positive.

The tangent is local. A chord drawn between two widely separated position points
gives average velocity over that interval, not the instantaneous velocity at either
endpoint. The two coincide only for a straight position-time graph. A long chord
and a tangent therefore answer different kinematic questions.

$$
% caption: The slope of a position-time trace is signed velocity, while its
% curvature is signed acceleration. The descending concave-up curve has
% negative velocity but positive acceleration; the tangent is negative and becomes
% progressively less steep as the object approaches its turning point.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.2,0) node[right] {time};
  \draw[->, black] (0,0) -- (0,3.8) node[above] {position};
  \draw[acc, very thick, domain=0.4:5.5, samples=120]
    plot (\x,{1.0+2.25*exp(-0.72*(\x-0.4))});
  \draw[black, thick] (0.5,2.90) -- (1.9,1.63);
  \draw[fill=acc!14, draw=acc, thick] (1.2,2.27) circle (1.8pt);
  \node[black, anchor=west] at (2.05,2.55) {steep tangent};
  \draw[black, thick] (2.9,1.34) -- (4.3,1.11);
  \draw[fill=acc!14, draw=acc, thick] (3.6,1.23) circle (1.8pt);
  \node[black, anchor=south west] at (3.5,1.28) {shallow tangent};
\end{tikzpicture}
$$

> **Worked example.** At $t=4.0\ \mathrm s$, a tangent to a position trace passes through the grid points $(3.0\ \mathrm s,\ 7.2\ \mathrm m)$ and $(5.0\ \mathrm s,\ 8.0\ \mathrm m)$.
>
> The instantaneous velocity is the tangent slope,
> $$
> v(4.0\ \mathrm s)=\frac{8.0-7.2}{5.0-3.0}=0.40\ \mathrm{m\,s^{-1}}.
> $$
> A chord through the actual readings at $3.0$ and $5.0\ \mathrm s$ may have a different slope; that slope is the two-second average, not the instantaneous value.

State which line supplied the rise and run before calling a slope instantaneous.

**Velocity graphs: direction, reversal, and distance.**

The sign of a velocity graph has a direct spatial meaning. Above the time axis, the
coordinate increases; below it, the coordinate decreases. A crossing through zero
marks a momentary rest and, when the graph actually changes side, a reversal of
direction. Touching the axis and remaining on the same side does not reverse the
motion: it describes a brief stop followed by travel in the original direction.

Displacement is the signed area between the velocity trace and the time axis.
Distance adds the magnitudes of those regions. Thus a trip with $+12\ \mathrm m$
of signed area followed by $-5\ \mathrm m$ has displacement $+7\ \mathrm m$ and
distance $17\ \mathrm m$. The two quantities agree only when velocity retains one
sign over the interval.

$$
% caption: The positive and negative regions beneath a velocity-time graph have
% opposite effects on displacement. Their absolute areas both contribute to total
% distance; the zero crossing identifies the direction reversal between the outward
% and return segments.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.3,0) node[right] {time};
  \draw[->, black] (0,-2.2) -- (0,2.8) node[above] {$v$};
  \fill[acc!14] (0.45,0) -- (0.45,1.75) -- (2.65,1.75) -- (3.45,0) -- cycle;
  \fill[black] (3.45,0) -- (4.25,-1.25) -- (5.65,-1.25) -- (5.95,0) -- cycle;
  \draw[acc, thick] (0.45,1.75) -- (2.65,1.75) -- (3.45,0) -- (4.25,-1.25) -- (5.65,-1.25) -- (5.95,0);
  \node[anchor=south] at (1.55,1.75) {above axis};
  \node[anchor=north] at (4.90,-1.25) {below axis};
  \draw[fill=acc!12, draw=acc, thick] (3.45,0) circle (1.8pt);
  \node[anchor=north] at (3.45,-0.65) {reversal};
\end{tikzpicture}
$$

> **Worked example.** A walker moves at $3.0\ \mathrm{m\,s^{-1}}$ for $4.0\ \mathrm s$, then at $-1.5\ \mathrm{m\,s^{-1}}$ for $2.0\ \mathrm s$.
>
> The two rectangular signed areas give displacement,
> $$
> \Delta x=(3.0)(4.0)+(-1.5)(2.0)=12-3.0=9.0\ \mathrm m,
> $$
> while distance adds their magnitudes,
> $$
> s=|12|+|-3.0|=15\ \mathrm m.
> $$
> Reporting $9.0\ \mathrm m$ as a distance confuses displacement with path length; the return segment adds $3.0\ \mathrm m$ to the distance.

**Acceleration graphs and changes in velocity.**

Acceleration is read from a velocity graph as a slope, but its own graph is read
through signed area. A constant positive acceleration appears as a horizontal line
above zero. Over any selected interval, the rectangular area under that line is the
positive change in velocity. A constant negative acceleration appears below zero
and produces a negative velocity change. Neither sign alone says whether the
object speeds up: that conclusion also requires the sign of the current velocity.

For example, a car moving in the positive direction with $a<0$ loses speed. A cart
moving in the negative direction with the same $a<0$ gains speed, since its velocity
becomes more negative. The velocity-time graph separates these cases plainly. Its
slope is negative in both, while the graph lies above the axis in the first case and
below it in the second.

Units follow directly from the area operation. On an acceleration-time graph, vertical
units are $\mathrm{m\,s^{-2}}$ and horizontal units are seconds, so area has units
$\mathrm{m\,s^{-1}}$, the required units for a velocity change. A graph
whose vertical scale is labelled merely “acceleration” without units cannot support
a numerical calculation until the scale is supplied.

> **Worked example.** An acceleration rises uniformly from $0$ to $4.0\ \mathrm{m\,s^{-2}}$ over $3.0\ \mathrm s$, then holds at $4.0\ \mathrm{m\,s^{-2}}$ for another $2.0\ \mathrm s$, starting from $v_0=-5.0\ \mathrm{m\,s^{-1}}$.
>
> The acceleration-time area is a triangle plus a rectangle,
> $$
> \Delta v=\frac12(3.0)(4.0)+(2.0)(4.0)=14\ \mathrm{m\,s^{-1}}.
> $$
> The velocity ends at $-5.0+14=+9.0\ \mathrm{m\,s^{-1}}$. During the rising stage $a=(4.0/3.0)t$, so the accumulated change is $\Delta v=(2.0/3.0)t^2$; setting it to $5.0\ \mathrm{m\,s^{-1}}$ puts the direction reversal at $2.74\ \mathrm s$.

**Piecewise graphs and boundary times.**

Motion graphs can change rule when a motor starts, a brake engages, or a measured
force switches level. The graph should preserve the physical
quantities that remain continuous. Position is continuous for ordinary motion;
an object cannot jump from one coordinate to another without passing through the
intermediate positions. Velocity is also continuous unless an idealized impulse is
being used. Acceleration can jump abruptly when the net force changes, and that
jump appears as a corner in the velocity graph or as a step in the acceleration
graph.

Piecewise calculations proceed interval by interval. The final position and
velocity from one interval become the initial values for the next. Reusing the
original initial velocity in the second interval erases the first stage. Each new
signed acceleration area changes the velocity already reached, so the prior endpoint
becomes the initial value of the next interval.

$$
% caption: A step in acceleration creates a corner, not a break, in the velocity
% graph. The first and second slopes equal the two acceleration values; continuity
% at the switch time preserves the accumulated velocity from the first stage into
% the second.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.2,0) node[right] {time};
  \draw[->, black] (0,0) -- (0,3.8) node[above] {$v$};
  \draw[acc, thick] (0.45,0.42) -- (2.90,2.62) -- (5.75,2.62);
  \draw[dashed, black] (2.90,0) -- (2.90,2.62);
  \draw[fill=acc!12, draw=acc, thick] (2.90,2.62) circle (1.8pt);
  \node[anchor=north] at (2.90,-0.10) {switch};
\end{tikzpicture}
$$

> **Worked example.** A sled starts with $v=1.0\ \mathrm{m\,s^{-1}}$. Its acceleration is
> $2.0\ \mathrm{m\,s^{-2}}$ for $3.0\ \mathrm s$ and then zero for
> $4.0\ \mathrm s$. The first acceleration area adds $6.0\ \mathrm{m\,s^{-1}}$,
> so the second stage begins at $7.0\ \mathrm{m\,s^{-1}}$. The first displacement is
> the velocity-trapezoid area,
>
> $$
> \Delta x_1=\frac{1.0+7.0}{2}(3.0)=12\ \mathrm m.
> $$
>
> The second displacement is $(7.0)(4.0)=28\ \mathrm m$, for a total of
> $40\ \mathrm m$. The horizontal second segment contributes no acceleration area
> but contributes substantial velocity area and therefore substantial displacement.

## Reconstructing motion from graphs

An acceleration graph does not determine a unique velocity graph by itself. Its
area gives only a change in velocity, so one initial velocity value is required to
set the vertical placement of the reconstructed graph. The same distinction appears
one level lower: integrating velocity determines a change in position, and one
initial position value sets the corresponding position graph. Derivatives remove
these constants; integrals must restore them from stated conditions.

A stepwise acceleration record is reconstructed exactly with geometry. Each
constant acceleration interval adds a rectangular signed area to the preceding
velocity. A sloping acceleration interval produces a curved velocity trace, because
the slope of the velocity graph itself is changing. The value of acceleration at a
single time fixes only the local slope of velocity, not the velocity’s height above
or below the axis.

> **Worked example.** An object has $v(0)=-2.0\ \mathrm{m\,s^{-1}}$. Its acceleration is
> $+3.0\ \mathrm{m\,s^{-2}}$ from $0$ to $2.0\ \mathrm s$, then
> $-1.0\ \mathrm{m\,s^{-2}}$ from $2.0$ to $5.0\ \mathrm s$. The first area is
> $+6.0\ \mathrm{m\,s^{-1}}$, so $v(2.0\ \mathrm s)=+4.0\ \mathrm{m\,s^{-1}}$.
> The second area is $-3.0\ \mathrm{m\,s^{-1}}$, producing a final velocity of
> $+1.0\ \mathrm{m\,s^{-1}}$. The direction reversal occurs during the first stage
> when $-2.0+3.0t=0$, at $t=0.667\ \mathrm s$.
>
> Position follows from the area under the newly drawn velocity graph. In the first
> two seconds, the signed trapezoid area is
>
> $$
> \Delta x_1=\frac{-2.0+4.0}{2}(2.0)=2.0\ \mathrm m.
> $$
>
> The negative portion before reversal is already included. Splitting it into a
> negative triangle and a positive triangle is an equivalent check, provided their
> signs are retained.

**Estimating areas from a plotted curve.**

Many laboratory graphs do not supply a convenient formula. Area is estimated from
the plotted values; the trace height alone does not supply an area. The
trapezoidal rule replaces a curved segment between adjacent times by a straight
chord. For equal intervals $\Delta t$, the displacement estimate is

$$
\Delta x\approx\sum_i\frac{v_i+v_{i+1}}{2}\Delta t.
$$

Each trapezoid retains its sign. A chord below the axis gives a negative
contribution; no separate minus sign is invented after the calculation. Narrower
intervals usually improve the geometric approximation for a smooth velocity curve,
but they do not cure a poor vertical calibration or noisy readings. A reported area
should reflect the resolution of the graph, not the number of digits on a calculator.

$$
% caption: The trapezoidal rule estimates displacement from sampled velocity by
% replacing each curved segment with a chord. The shaded trapezoids preserve signed
% area and become a better geometric approximation as the sampling intervals narrow
% for a smooth, well-resolved velocity trace.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->] (0,0) -- (6.3,0) node[right] {time};
  \draw[->] (0,0) -- (0,3.8) node[above] {$v$};
  \fill[acc!12] (0.55,0) -- (0.55,0.62) -- (2.10,1.55) -- (2.10,0) -- cycle;
  \fill[acc!12] (2.10,0) -- (2.10,1.55) -- (3.65,2.45) -- (3.65,0) -- cycle;
  \fill[acc!12] (3.65,0) -- (3.65,2.45) -- (5.20,3.05) -- (5.20,0) -- cycle;
  \draw[thick, domain=0.55:5.20, samples=100]
    plot (\x,{0.30+0.58*\x-0.010*\x^2});
  \draw (0.55,0.62) -- (2.10,1.55) -- (3.65,2.45) -- (5.20,3.05);
  \foreach \x in {0.55,2.10,3.65,5.20} {\draw[dashed] (\x,0) -- (\x,3.18);}
\end{tikzpicture}
$$

> **Worked example.** Velocity readings at $0$, $1$, $2$, and $3\ \mathrm s$ are $0.0$, $1.8$, $3.0$,
> and $3.6\ \mathrm{m\,s^{-1}}$. With one-second spacing, the three trapezoids give
>
> $$
> \Delta x\approx\frac{0.0+1.8}{2}+
> \frac{1.8+3.0}{2}+
> \frac{3.0+3.6}{2}=6.6\ \mathrm m.
> $$
>
> The number is an estimate of the area beneath the measured curve, not proof that
> the acceleration was constant in each interval. A position measurement over the
> same span provides an independent check on the integration and on the graph scale.

**Scale, units, and what a graph can justify.**

Graph calculations are numerical measurements. A slope is found from coordinate
differences on the axes, not from the angle a line appears to make on a page. A
velocity graph can look steep because the vertical axis has been stretched; that
visual steepness has no independent physical meaning. The axis scales convert the
chosen rise and run into $\mathrm{m\,s^{-1}}$ or $\mathrm{m\,s^{-2}}$ as required.

Two points far apart on a straight segment generally reduce the relative effect of
line thickness and reading uncertainty. They need not be plotted
data points. Grid intersections on the drawn best-fit line are often better choices,
provided they are actually on that line. Reading the slope from two noisy adjacent
markers treats measurement scatter as if it were the physical graph.

The same discipline applies to areas. An area estimated by counting squares must
include the scale represented by one square in both directions. If one horizontal
square represents $0.50\ \mathrm s$ and one vertical square represents
$0.20\ \mathrm{m\,s^{-1}}$, a square under a velocity graph represents
$0.10\ \mathrm m$, not a dimensionless unit. Negative regions should be counted
with their sign before positive and negative totals are combined.

> **Worked example.** A velocity line rises from $1.0$ to $5.0\ \mathrm{m\,s^{-1}}$ between $2.0$ and
> $6.0\ \mathrm s$. Its slope is $(5.0-1.0)/(6.0-2.0)=1.0\ \mathrm{m\,s^{-2}}$.
> Changing the printed width of the graph changes the visual angle but cannot change
> this acceleration. Over the same four seconds, the trapezoid area is
> $[(1.0+5.0)/2](4.0)=12\ \mathrm m$. Both results follow from labelled axis values,
> so they remain valid if the graph is redrawn at a different size.
>
> When a value is read from a coarse graph, its precision should match the grid and
> line thickness. Reporting $1.0000\ \mathrm{m\,s^{-2}}$ from a graph whose smallest
> division is $0.2\ \mathrm{m\,s^{-1}}$ over one second overstates the evidence. A
> rounded value or a stated reading uncertainty is the more accurate scientific claim.
> The graph supports only the precision that its scale and data warrant.
>
> State the graph operation before stating the physical conclusion. The following
> record separates a derivative, an integral, and a visual feature that does not by
> itself determine a cause.
>
> | Measured feature | Operation | Supported statement |
> | --- | --- | --- |
> | Position slope | $v\approx\Delta x/\Delta t$ | average or local velocity over the stated interval |
> | Velocity slope | $a\approx\Delta v/\Delta t$ | acceleration estimate with units of $\mathrm{m\,s^{-2}}$ |
> | Velocity area | $\Delta x\approx\sum \bar v_i\Delta t_i$ | signed displacement over the selected interval |
> | Curve shape | no numerical operation | direction change or concavity only when resolution supports it |

**Consistency across position, velocity, and acceleration.**

The three graph types describe one motion, so a proposed set can be tested without
calculating every coordinate. A horizontal velocity segment requires
a straight position segment, because a constant slope in position gives constant
velocity. A rising straight velocity segment demands a concave-up position curve
and a horizontal positive acceleration segment. A horizontal acceleration segment
does not make position horizontal; it makes the slope of the velocity graph change
at a uniform rate.

The same constraints can be read in reverse. A position graph with a local
maximum has zero velocity at the summit. If the graph is concave down there, the
acceleration is negative, so the velocity crosses from positive to negative. The
velocity graph should therefore pass through its axis with a negative slope, and
the acceleration graph should lie below its axis nearby. Any alternative set of
graphs violates at least one derivative relation.

$$
% caption: One constant-negative-acceleration motion in three aligned records.
% Position is concave down and reaches a maximum when velocity crosses zero; the
% velocity-line slope equals the horizontal acceleration value. Corresponding marked
% times align vertically across all panels.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \begin{scope}
    \draw[->] (0,0) -- (4.2,0) node[right] {time};
    \draw[->] (0,0) -- (0,3.3) node[above] {position};
    \draw[thick, domain=0.35:3.75, samples=100]
      plot (\x,{0.45+1.95*(\x-0.35)-0.52*(\x-0.35)^2});
    \draw[dashed] (2.00,0) -- (2.00,2.48);
  \end{scope}
  \begin{scope}[shift={(5.3,0)}]
    \draw[->] (0,0) -- (4.2,0) node[right] {time};
    \draw[->] (0,-1.7) -- (0,2.6) node[above] {$v$};
    \draw[thick] (0.35,1.85) -- (3.75,-1.20);
    \draw[dashed] (2.00,-1.7) -- (2.00,0);
    \draw[fill=acc!12, draw=acc, thick] (2.00,0) circle (1.7pt);
  \end{scope}
  \begin{scope}[shift={(10.6,0)}]
    \draw[->] (0,0) -- (4.2,0) node[right] {time};
    \draw[->] (0,-1.7) -- (0,2.6) node[above] {$a$};
    \draw[thick] (0.35,-0.88) -- (3.75,-0.88);
    \draw[dashed] (2.00,-1.7) -- (2.00,-0.88);
  \end{scope}
\end{tikzpicture}
$$

The common time scale is essential. Aligning graph panels only by their left edges
while allowing different horizontal scales can create false correspondences. Before
using a peak in $x(t)$ to locate a zero in $v(t)$, verify that both time axes share
the same origin and interval markings. A shifted clock changes the numerical time
of every feature but does not alter the velocity or acceleration values.

> **Worked example.** A position trace reaches its highest point at $t=3.0\ \mathrm s$ and is concave
> down around that point. The corresponding velocity must be zero at $3.0\ \mathrm s$
> and must change from positive to negative. If a proposed velocity graph shows a
> horizontal positive segment at $3.0\ \mathrm s$, it is incompatible with the
> position graph even if its numerical scale appears plausible. The acceleration near
> the turnaround is negative; zero acceleration would make velocity locally constant
> and could not bend the position trace into a maximum.

**Discontinuities and physical idealizations.**

Lines and steps in textbook graphs often idealize changes. A step in
acceleration is physically reasonable as a short-hand for a force changing over a
time too brief to resolve. Velocity stays continuous across that step, but its slope
changes abruptly. Position remains continuous as well. The resulting velocity graph
has a corner, and the position graph changes curvature without developing a break.

A genuine jump in velocity is a stronger idealization. It represents an impulse
whose duration is treated as zero while its velocity change remains finite. Such a
vertical segment cannot be an ordinary velocity-time trajectory at finite
acceleration; its slope would be infinite. In a measured collision, the transition
occupies a small but nonzero time interval, and the exact rounded corner depends on
the sensor’s time resolution.

$$
% caption: A step in acceleration preserves continuous velocity, while an ideal
% impulsive velocity change is drawn as a vertical jump. The latter compresses a
% brief collision into zero plotted time and therefore cannot be assigned an
% ordinary finite acceleration over the jump itself.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \begin{scope}
    \draw[->] (0,0) -- (4.8,0) node[right] {time};
    \draw[->] (0,0) -- (0,3.1) node[above] {$v$};
    \draw[thick] (0.40,0.45) -- (2.20,1.75) -- (4.20,1.75);
    \fill (2.20,1.75) circle (1.7pt);
    \node[anchor=north] at (2.20,-0.10) {$a$ step};
  \end{scope}
  \begin{scope}[shift={(6.4,0)}]
    \draw[->] (0,0) -- (4.8,0) node[right] {time};
    \draw[->] (0,0) -- (0,3.1) node[above] {$v$};
    \draw[thick] (0.40,0.62) -- (2.30,0.62);
    \draw[acc, thick] (2.30,0.62) -- (2.30,2.25);
    \draw[thick] (2.30,2.25) -- (4.20,2.25);
    \node[anchor=north] at (2.30,-0.10) {ideal impulse};
  \end{scope}
\end{tikzpicture}
$$

Position jumps require a different warning. A discontinuous plotted position can
occur when a tracking system loses and reacquires an object, when a coordinate
origin is reset, or when two separate runs have been concatenated. It is not a
kinematic path through space. A good analysis distinguishes a physical rapid change
from a change in the data-recording convention before differentiating the graph.

## Averages, sampling, and numerical reconstruction

An average velocity is a displacement divided by elapsed time. On a velocity-time
graph it is the height of a horizontal line that would enclose the same signed area
over the chosen interval. This equal-area line need not pass through the graph at
the temporal midpoint. When velocity varies smoothly, it does match the velocity at
some time in the interval, but graph data alone may not identify that time uniquely.

Average acceleration has the analogous meaning: it is the net velocity change
divided by time. A fluctuating acceleration graph can have zero average while still
altering the velocity substantially during the interval. Positive and negative
acceleration areas then cancel only in the final velocity change. They do not
cancel the displacement accumulated while velocity was elevated or depressed.

The interval must be named. Extending an interval can change an average even when
the local graph around the original interval is unchanged. A speedometer reading is
an instantaneous speed; a route distance divided by the travel time is an average
speed. Instantaneous speed refers to one time; average speed refers to a named
interval, despite the common unit $\mathrm{m\,s^{-1}}$.

> **Worked example.** Over a four-second interval, a cart has velocity values that rise sharply during
> the first second and remain low for the next three seconds. Its velocity at the
> two-second midpoint may be $0.50\ \mathrm{m\,s^{-1}}$, yet its displacement area
> may be $6.0\ \mathrm m$, giving an average velocity of
> $1.50\ \mathrm{m\,s^{-1}}$. The average describes the whole area, not a preferred
> clock reading. For a straight velocity graph, the midpoint shortcut works because
> the two halves form matching triangles; it is a special geometric result rather
> than a definition.

**Numerical reconstruction with a table of samples.**

Graph relations also work when the source is a numerical table. A sequence of
velocity samples reconstructs position changes by summing interval areas. With
measurements at uniform time spacing, the trapezoidal estimate uses the average of
the two endpoint velocities in each interval. The result is a list of positions, not
an assertion that the hidden velocity ran linearly between every pair of
measurements.

Acceleration samples can be accumulated one layer earlier to obtain velocities,
then accumulated again to obtain positions. Errors and offsets propagate through
this process. A constant $+0.05\ \mathrm{m\,s^{-2}}$ bias in acceleration creates a
velocity error that grows linearly with elapsed time and a position error that grows
quadratically. Long reconstructed records therefore need independent position or
velocity checks, especially when acceleration comes from a sensor whose zero is
hard to calibrate.

> **Worked example.** A sensor reports velocities $1.0$, $2.0$, and $4.0\ \mathrm{m\,s^{-1}}$ at
> $t=0$, $1.0$, and $2.0\ \mathrm s$. Starting from $x=3.0\ \mathrm m$, the first
> trapezoid gives $\Delta x_1=[(1.0+2.0)/2](1.0)=1.5\ \mathrm m$. The second gives
> $\Delta x_2=[(2.0+4.0)/2](1.0)=3.0\ \mathrm m$. The reconstructed positions are
> $4.5\ \mathrm m$ and $7.5\ \mathrm m$ at the later sample times. The stated
> initial coordinate fixes absolute positions; otherwise the data yield displacements
> only.

**Full multi-interval graph interpretation.**

A complete motion-graph calculation begins with the time boundaries and the
acceleration or velocity rule on each interval. The ending velocity and position
from one interval become the initial values of the next. Velocity signs identify
direction, signed velocity areas give displacement, and absolute areas give
distance. This order keeps local derivative information separate from accumulated
integral information.

The following acceleration history has three stages: positive acceleration, zero
acceleration, and negative acceleration. It starts from a negative velocity, so the
first stage contains a direction reversal. The velocity graph must begin below its
axis, rise linearly, remain horizontal, then fall linearly. The position graph first
decreases, reaches a minimum when velocity becomes zero, then rises; during the
middle stage it is a straight rising line, and during the final stage it remains
concave down while its slope diminishes.

> **Worked example.** Take the first interval as $0\leq t<2.0\ \mathrm s$ with
> $a=+2.0\ \mathrm{m\,s^{-2}}$ and initial velocity
> $v_0=-2.0\ \mathrm{m\,s^{-1}}$. The velocity change is
> $+4.0\ \mathrm{m\,s^{-1}}$, so the interval ends at
> $+2.0\ \mathrm{m\,s^{-1}}$. The zero crossing occurs one second after the start,
> not at the interval boundary. The signed displacement over this first interval is
> the symmetric trapezoid area,
>
> $$
> \Delta x_1=\frac{-2.0+2.0}{2}(2.0)=0.
> $$
>
> Zero displacement does not mean no travel. The negative triangle from $0$ to
> $1.0\ \mathrm s$ has area $-1.0\ \mathrm m$ and the positive triangle from
> $1.0$ to $2.0\ \mathrm s$ has area $+1.0\ \mathrm m$. The distance in the first
> stage is therefore $2.0\ \mathrm m$, while the object returns to its starting
> coordinate at the stage boundary.
>
> During the second interval, from $2.0$ to $5.0\ \mathrm s$, let $a=0$. Velocity
> remains $+2.0\ \mathrm{m\,s^{-1}}$ and the position graph is a straight line.
> The velocity rectangle gives
>
> $$
> \Delta x_2=(2.0)(3.0)=6.0\ \mathrm m.
> $$
>
> For the third interval, from $5.0$ to $7.0\ \mathrm s$, let
> $a=-1.0\ \mathrm{m\,s^{-2}}$. The velocity falls from
> $+2.0\ \mathrm{m\,s^{-1}}$ to zero. Its triangular area is
>
> $$
> \Delta x_3=\frac{2.0+0}{2}(2.0)=2.0\ \mathrm m.
> $$
>
> The total displacement is $0+6.0+2.0=8.0\ \mathrm m$, and total distance is
> $2.0+6.0+2.0=10.0\ \mathrm m$. The final velocity is zero at a position different
> from the starting coordinate. Different graph features determine the three results:
> endpoint height gives final velocity, signed area gives displacement, and the
> magnitudes of all travelled segments give distance.

**Audit questions for a graph-based answer.**

Before finalizing a result, check four points:

- the velocity slope matches the acceleration in every interval;
- the velocity area reproduces the stated position change;
- a direction reversal carries a velocity sign change;
- reported units match the operation, slope or area.

These checks locate most sign errors without a second solution method.

## What motion graphs determine

Even a complete set of one-dimensional graphs does not explain the cause of the
motion. A velocity line with negative slope establishes negative acceleration; it
does not by itself identify friction, gravity, a braking force, or a motor setting.
Several physical systems can produce the same kinematic record. Force claims require
additional information about the object and its interactions. Keeping this boundary
clear prevents a graph-reading exercise from importing assumptions that were never
measured.

Derivative and integral claims follow from graph data; force claims require
interaction data. State that evidential boundary explicitly so a kinematic
conclusion does not acquire an unmeasured mechanism.

| Graph evidence | Quantitative inference | Unsupported inference |
| --- | --- | --- |
| negative slope of $v(t)$ | negative acceleration in the chosen axis | which force produced it |
| zero crossing of $v(t)$ | instantaneous reversal of direction | zero net force |
| signed area under $v(t)$ | displacement over the interval | total distance without sign analysis |
| curve between samples | interpolation model | unmeasured instantaneous detail |

Reference frame and coordinate choice determine the graph labels and signed values.
Reversing the positive axis reverses the signs of position, velocity, and acceleration, while
distance and speed remain nonnegative. Translating the position origin shifts the
whole position graph vertically but leaves its slopes and curvature unchanged. A
frame moving at constant speed changes a velocity graph’s vertical placement but
leaves the acceleration graph unchanged. These transformations alter labels and
values, not the underlying observed motion.

Sampling and line thickness set further limits. A smooth curve drawn through sparse
points is an interpolation, not direct evidence about every intervening instant.
Reports should distinguish a measured feature from a modelled feature: “the sampled
velocities support a decreasing trend” is warranted more directly than a precise
acceleration history below the time resolution. State those limits alongside the
calculation.

> **Worked example.** If a velocity in one coordinate system is $+4.0\ \mathrm{m\,s^{-1}}$, reversing
> the axis reports $-4.0\ \mathrm{m\,s^{-1}}$. A measured acceleration of
> $-1.5\ \mathrm{m\,s^{-2}}$ becomes $+1.5\ \mathrm{m\,s^{-2}}$ in the reversed
> axis. The velocity graph’s slope still has the displayed acceleration sign, and
> the physical speed change is unchanged. A correct graph interpretation always names
> the sign convention before assigning a directional meaning to an area or slope.

**Stationary points, extrema, and inflection points.**

Several graph features have related but distinct kinematic meanings. A stationary
point on a position graph has zero tangent slope, so its velocity is zero at that
instant. It may be a maximum, a minimum, or neither. A local maximum requires the
position graph to rise before the point and fall after it; velocity changes from
positive to negative. A local minimum requires the reverse sign change. A flat
shoulder can have zero velocity without reversing direction when the slope touches
zero and retains its sign.

An extremum on a velocity graph is different. Its slope is zero, so acceleration
is zero there. The object may still be moving rapidly. A speed maximum is found
from the magnitude of velocity, which means a positive velocity maximum or a
negative velocity minimum can each represent a largest speed. Treating every
horizontal tangent as “the object stops” confuses the graph’s vertical quantity with
the derivative of that quantity.

An inflection point on a position graph marks a change in concavity. It is commonly
associated with acceleration changing sign, but data with a sharp corner or coarse
sampling may not support a precise statement about the instantaneous acceleration.
A smooth trace has a velocity-graph local extremum at the same time as the position
curve changes concavity. The three features align by time, not by their
vertical graph values.

**Classifying a zero-velocity instant.**

At $t=2.0\ \mathrm s$, a position graph has a horizontal tangent. Immediately
before and after that time, the graph rises. The velocity is zero at the marked
instant but positive on both sides, so the object pauses without reversing. If the
position graph is concave up at the pause, acceleration is positive and the velocity
touches zero from above only in an idealized limiting sense; a smooth physical trace
with positive velocity on both sides instead has a flat local slowdown whose detailed
shape must be resolved by the data. A velocity sign change establishes whether a
turnaround occurred; a horizontal tangent alone does not establish one.

**Cumulative area as a new graph.**

Fixing an initial position turns a velocity graph into a cumulative-displacement
function. Define $D(t)$ as the signed area under velocity from the start time to
time $t$. Then $x(t)=x_0+D(t)$. The value of $D$ rises while velocity is positive,
falls while velocity is negative, and has a horizontal tangent when velocity is
zero. The cumulative-area graph is the reconstructed position graph apart from the
vertical offset $x_0$.

The cumulative graph clarifies interval areas. The displacement from $t_1$ to
$t_2$ equals $D(t_2)-D(t_1)$. It equals the height of $D$ at $t_2$ only when the
chosen start time is $t_1$. A large positive area accumulated before $t_1$ can place
the graph high above zero while leaving the later interval displacement unchanged.
Definite-integral limits belong to the physical statement and keep a graph offset
from being mistaken for an interval result.

$$
% caption: The cumulative signed area beneath velocity becomes the position change
% from the stated start time. Positive velocity makes the cumulative curve rise;
% negative velocity makes it fall. The same zero of velocity appears as a horizontal
% tangent on the accumulated-displacement graph.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \begin{scope}
    \draw[->, black] (0,0) -- (5.7,0) node[right] {time};
    \draw[->, black] (0,-1.7) -- (0,2.8) node[above] {$v$};
    \draw[acc, thick] (0.40,1.65) -- (2.55,1.65) -- (3.40,0) -- (4.35,-1.15) -- (5.25,0);
    \fill[acc!12] (0.40,0) -- (0.40,1.65) -- (2.55,1.65) -- (3.40,0) -- cycle;
    \fill[black] (3.40,0) -- (4.35,-1.15) -- (5.25,0) -- cycle;
  \end{scope}
  \begin{scope}[shift={(7.1,0)}]
    \draw[->, black] (0,0) -- (5.7,0) node[right] {time};
    \draw[->, black] (0,0) -- (0,3.3) node[above] {position change};
    \draw[acc, thick, domain=0.40:5.25, samples=100]
      plot (\x,{0.38+2.40*sin(180*(\x-0.40)/4.85)+0.324*(\x-0.40)});
    \draw[dashed, black] (3.40,0) -- (3.40,2.72);
  \end{scope}
\end{tikzpicture}
$$

> **Worked example.** Suppose the cumulative displacement relative to $t=0$ is $D(2.0\ \mathrm s)=
> 5.4\ \mathrm m$ and $D(6.0\ \mathrm s)=3.1\ \mathrm m$. The displacement from
> $2.0$ to $6.0\ \mathrm s$ is $3.1-5.4=-2.3\ \mathrm m$, even though both plotted
> heights are positive. The negative result states that the coordinate decreased
> during this later interval; it does not state that the object lies on the negative
> side of the original coordinate origin.

**Cross-checking measured graph data.**

Experimental graph sets should be checked in both directions. A slope estimated
from the position trace predicts local velocity. A signed area from the velocity
trace predicts position change. A slope from the velocity trace predicts
acceleration, and an area from the acceleration trace predicts velocity change.
Exact agreement is not expected when curves are drawn through independent noisy
measurements, but disagreement larger than the stated reading uncertainty points to
a timing offset, a calibration error, or an inappropriate smoothing method.

The comparison must be made on matching intervals. A centered velocity difference
derived from position values at $t-\Delta t$ and $t+\Delta t$ belongs at time $t$.
An acceleration computed from two successive velocity samples belongs naturally at
the midpoint between them. Shifting these derived values to an endpoint can create
an apparent lag between the velocity and acceleration graphs even when the physical
measurements are consistent.

Unit cancellation provides an initial audit. Position differences divided by time
give velocity; velocity differences divided by time give acceleration; velocity
times time gives position change. An incorrect time unit can still yield a
plausible-looking number, so display the unit cancellation in a written calculation.
A millisecond clock treated as seconds introduces factors of one thousand into a
slope or area.

> **Worked example.** A motion tracker gives positions $x_0=0.00\ \mathrm m$, $x_1=0.52\ \mathrm m$, and $x_2=1.28\ \mathrm m$ at equally spaced times $t=0$, $0.40$, and $0.80\ \mathrm s$.
>
> The two interval-average velocities are
> $$
> v_{1/2}=\frac{0.52-0.00}{0.40}=1.30\ \mathrm{m\,s^{-1}},
> \qquad
> v_{3/2}=\frac{1.28-0.52}{0.40}=1.90\ \mathrm{m\,s^{-1}}.
> $$
> Their difference over the $0.40\ \mathrm s$ separation of the midpoint times estimates the acceleration at $t=0.40\ \mathrm s$,
> $$
> a\approx\frac{1.90-1.30}{0.40}=1.5\ \mathrm{m\,s^{-2}}.
> $$
> The centered velocity estimate at $0.40\ \mathrm s$ uses the outer two samples,
> $$
> v(0.40\ \mathrm s)\approx\frac{1.28-0.00}{0.80}=1.60\ \mathrm{m\,s^{-1}},
> $$
> midway between the two interval averages because their change is linear across this three-point record.

Three measured positions admit many smooth acceleration histories. Additional
readings and residuals against a fitted model establish whether a
constant-acceleration approximation is justified over a longer interval.

**A graph-reading protocol.**

An unfamiliar motion graph should first be identified by its plotted quantity and
units. Mark the time interval before extracting a slope or area. Identify whether a zero
is a zero of position, velocity, or acceleration; each has a different physical
meaning. Use the graph relation appropriate to the vertical quantity, preserve signs
throughout the computation, and state whether the result is instantaneous, average,
or an estimated integral. These few steps prevent a visually neat calculation from
answering the wrong kinematic question.

## Time shifts and finite differences

The clock origin is a convention. Replacing $t$ with $T=t-t_0$ shifts every event
left or right on a graph but does not change any measured velocity or acceleration.
The slope of a position graph is unchanged by a horizontal translation, and the
area under velocity between two physical events is unchanged when the same events
are expressed with a new time coordinate. Confusion arises only when numerical time
labels from one origin are mixed with intervals measured from another.

Position origins behave differently. Replacing $x$ with $X=x-x_0$ shifts the
position graph vertically but leaves velocity, acceleration, displacement, and
distance unchanged. A vertical offset can make every displayed position positive
without changing whether an object reverses direction. Direction is determined by
the slope of the position trace or the sign of velocity, never by whether the line
lies above the plotted horizontal axis.

Velocity offsets need physical interpretation. A constant vertical shift in a
velocity graph can arise when the reference frame moves at constant speed. It changes
the displacement reported by that frame and may change whether the velocity crosses
zero, while acceleration is unchanged. A vertical shift caused by an uncorrected
sensor zero is instead an error: integrating it creates a position error that grows
with time. Graph transformations should therefore be labelled as coordinate changes
or calibration corrections, not treated as interchangeable drawing operations.

> **Worked example.** A cart has $x=4.0\ \mathrm m$ at the chosen position origin and travels with
> constant $v=1.5\ \mathrm{m\,s^{-1}}$. Shifting the coordinate origin by
> $+4.0\ \mathrm m$ reports an initial coordinate of zero and leaves the velocity
> graph unchanged. Shifting to a frame moving at $1.0\ \mathrm{m\,s^{-1}}$ in the
> same direction instead reports velocity $0.50\ \mathrm{m\,s^{-1}}$. In both
> descriptions acceleration is zero. The differing velocity values are not a
> contradiction because they answer questions in different reference frames.

**Finite-difference slopes and sampling limits.**

Recorded positions arrive at discrete clock times, whereas the derivative
$v(t)=\d x/\d t$ refers to an instant. A difference quotient bridges those two
descriptions. For equally spaced position samples separated by $\Delta t$, the
forward quotient

$$
v_{i+1/2}\approx\frac{x_{i+1}-x_i}{\Delta t}
$$

is the average velocity across the interval. Its natural time location is the
midpoint $t_{i+1/2}$, not the left-hand sample time. A backward quotient has the
same meaning on the preceding interval. Assigning either one to an endpoint hides
the half-sample timing shift and can make a derived velocity record appear to lag
or lead a position record.

The centered quotient for a smooth position function

$$
v_i\approx\frac{x_{i+1}-x_{i-1}}{2\Delta t}
$$

uses readings equally far before and after $t_i$. Its estimate belongs at $t_i$.
The symmetric geometry also removes the first-order curvature error that affects a
one-sided quotient. That improvement assumes comparable spacing and a smoothly
varying path; it does not turn sparse or noisy positions into an exact velocity
measurement. The position samples still supply only the secant information shown
in the figure, while a tangent remains a local limiting slope.

The second derivative is more demanding. At equal time spacing, a central
position estimate for acceleration is

$$
a_i\approx\frac{x_{i+1}-2x_i+x_{i-1}}{(\Delta t)^2}.
$$

It measures how successive position increments differ. Equal increments give zero
in the numerator and therefore zero estimated acceleration. Increments that grow
by the same amount from one interval to the next give a constant acceleration.
The estimate is assigned to the middle time $t_i$ because it compares the intervals on
either side of that time. A calculation that reports it at $t_{i+1}$ shifts the
entire acceleration sequence by one sample.

Measurement uncertainty changes the choice of time interval. Suppose independent
position readings each have uncertainty $\sigma_x$. The standard uncertainty of a
two-point velocity difference is approximately

$$
\sigma_v\approx\frac{\sqrt{2}\,\sigma_x}{\Delta t},
$$

and the corresponding three-point acceleration difference has uncertainty of order
$\sqrt{6}\,\sigma_x/(\Delta t)^2$. Halving the sample interval therefore doubles
the noise scale in a velocity difference and roughly quadruples it in a position-
derived acceleration. Faster sampling captures short changes in motion, yet a very
short interval can produce a derivative dominated by digitization, marker jitter,
or camera-pixel error. The interval must be chosen against both the motion time
scale and the instrument resolution.

Smoothing may lower random scatter before differentiation, but it also changes the
record. A moving average blurs sharp starts, stops, and collision features across
several sample times. Polynomial fits and spline fits supply differentiable curves,
but their derivative depends on the fitting window and model order. A graph or
report should identify whether a velocity trace came from raw differences, a
filtered record, or a fitted trajectory. That distinction matters most near an
event whose duration is comparable with the smoothing window.

Consider position readings separated by $0.20\ \mathrm s$:
$x_{i-1}=1.06\ \mathrm m$, $x_i=1.42\ \mathrm m$, and
$x_{i+1}=1.90\ \mathrm m$. The centered velocity at the middle reading is

$$
v_i\approx\frac{1.90-1.06}{0.40}=2.10\ \mathrm{m\,s^{-1}}.
$$

The second-difference acceleration is

$$
a_i\approx\frac{1.90-2(1.42)+1.06}{(0.20)^2}
=3.0\ \mathrm{m\,s^{-2}}.
$$

Both values belong to the time of $x_i$. The calculation uses three readings but
does not establish a constant-acceleration law beyond that neighborhood. A longer
set of points, plotted residuals, and stated instrument uncertainty determine
whether the local estimate represents a stable trend or a measurement fluctuation.

Sampling also imposes an aliasing limit. If an object reverses direction between two
position readings, nearly equal endpoints can conceal a substantial intervening
excursion. A short secant would then be mistaken for a small velocity even though
the object travelled far during the interval. Video analysis needs a frame rate that
resolves the relevant change of motion and a clock synchronized with the image
samples. No differentiation procedure can recover a feature absent from the record.

[^tipler22]: Tipler and Mosca, _Physics for Scientists and Engineers_, 6th ed., §§2-1–2-2, 2-4.