--- title: One-Dimensional Motion module: Kinematics moduleNumber: 1 lessonNumber: 1 order: 101 summary: > Motion along a line already forces the two questions the whole of kinematics repeats: how fast is the object moving now, and where will it be next? Velocity and acceleration answer the first as derivatives of position; integrating them back — the signed area under a graph — answers the second. We derive the constant-acceleration equations, mark exactly where the "constant" assumption is load-bearing, and see why sign, not magnitude, is what carries direction. topics: [Kinematics] draft: false sources: - book: Tipler & Mosca ref: "Ch. 2 — Motion in One Dimension; §2-1 Displacement, Velocity, and Speed" - book: Tipler & Mosca ref: "Ch. 2 — Motion in One Dimension; §§2-2–2-4 Acceleration, Constant Acceleration, Integration" --- Kinematics specifies position, velocity, and acceleration without modelling the interactions that produce them. In one dimension, $x(t)$ is a signed coordinate in a declared reference frame; its value is not a distance from the origin. ## Position, velocity, and acceleration The function $x(t)$ gives the position of a particle at time $t$. Between two instants, its **displacement** is $$ \Delta x=x(t_2)-x(t_1). $$ Displacement can be positive, negative, or zero. It is not the total distance travelled. A runner moving from $x=0$ to $x=20\ \mathrm{m}$ and back to $x=5\ \mathrm{m}$ has displacement $+5\ \mathrm{m}$ and distance $35\ \mathrm{m}$. This distinction matters whenever the direction reverses. The average velocity over an interval is the secant slope of the position graph: $$ \bar v_x=\frac{\Delta x}{\Delta t}. $$ It reports net position change per elapsed time. Average speed instead divides total distance by elapsed time and is never negative. Neither quantity tells how the particle moved at intermediate times. $$ % caption: A position-time curve can rise, flatten, and fall. The secant line % between two instants gives average velocity; its slope ignores the details % between those instants. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (6.6,0) node[right] {t}; \draw[->, black] (0,0) -- (0,3.8) node[above] {x}; \draw[acc, very thick, domain=0.4:6.1, samples=120] plot (\x,{0.3+2.5*sin(180*(\x-0.4)/5.8)}); \draw[black, thick, dashed] (1.3,1.47) -- (5.6,1.10); \draw[fill=acc!14, draw=acc, thick] (1.3,1.47) circle (2pt); \draw[fill=acc!14, draw=acc, thick] (5.6,1.10) circle (2pt); \draw[black, dotted] (1.3,0) -- (1.3,1.47); \draw[black, dotted] (5.6,0) -- (5.6,1.10); \node[anchor=north] at (1.3,0) {t1}; \node[anchor=north] at (5.6,0) {t2}; \node[black, anchor=north] at (3.6,0.95) {chord}; \end{tikzpicture} $$ **Instantaneous velocity.** Shrinking the time interval until the two endpoints coincide gives instantaneous velocity: $$ v_x(t)=\lim_{\Delta t\to0}\frac{x(t+\Delta t)-x(t)}{\Delta t} =\frac{\d x}{\d t}. $$ The velocity is the tangent slope of $x(t)$. A positive slope means motion in the positive direction; a negative slope means motion in the negative direction; a horizontal tangent means the particle is instantaneously at rest. An object can have $v_x=0$ and still have nonzero acceleration. The peak of an upward toss is the familiar example. Velocity is a signed component. Its magnitude $|v_x|$ is speed. Saying that a car is decelerating is ambiguous unless a direction has been chosen: negative acceleration describes the direction of acceleration, while slowing down means $v_x$ and $a_x$ have opposite signs. **Acceleration and the velocity graph.** Acceleration is the rate of change of velocity, $$ a_x(t)=\frac{\d v_x}{\d t}=\frac{\d^2x}{\d t^2}. $$ It is the tangent slope on a velocity-time graph. The signed area under that graph is displacement because $\d x=v_x\,\d t$: $$ x(t_2)-x(t_1)=\int_{t_1}^{t_2}v_x(t)\,\d t. $$ Similarly, the signed area under an acceleration-time graph is the change in velocity, $\Delta v_x=\int a_x\,\d t$. These integral relations remain true for arbitrary differentiable motion, unlike the constant-acceleration formulas. $$ % caption: Velocity changes linearly under constant acceleration. The signed % area under the line is displacement: a rectangle from initial velocity plus a % triangle from the velocity increase. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->] (0,0) -- (6.2,0) node[right] {t}; \draw[->] (0,0) -- (0,4.1) node[above] {$v_x$}; \fill[acc!8] (0,0) rectangle (5.2,1.0); \fill[acc!16] (0,1.0) -- (5.2,3.25) -- (5.2,1.0) -- cycle; \draw[very thick] (0,1.0) -- (5.2,3.25); \draw[dashed] (5.2,0) -- (5.2,3.25); \draw[dashed] (0,1.0) -- (5.2,1.0); \fill (0,1.0) circle (1.8pt); \fill (5.2,3.25) circle (1.8pt); \node[anchor=east] at (0,1.0) {$v_0$}; \node[anchor=west] at (5.2,3.25) {$v$}; \node[anchor=north] at (5.2,0) {t}; \node at (2.6,0.5) {$v_0\,t$}; \node at (3.75,1.75) {$\frac{1}{2}a t^2$}; \end{tikzpicture} $$ ## Constant acceleration and vertical motion If $a_x=a$ is constant over the interval, integrating once gives $$ v=v_0+at, $$ and integrating again gives $$ x=x_0+v_0t+\frac12at^2. $$ Combining the first equation with the displacement relation yields two additional forms: $$ v^2=v_0^2+2a(x-x_0), \qquad x-x_0=\frac{v_0+v}{2}t. $$ Each relation assumes one constant acceleration. Appreciable drag, a changing propulsion force, and curved motion require the derivative or integral relations. > **Worked example.** A car travelling at $25\ \mathrm{m/s}$ brakes at constant > $a=-5.0\ \mathrm{m/s^2}$; find the stopping distance. > > With $v=0$, the time-free equation $v^2=v_0^2+2a\,\Delta x$ gives > $$ > \Delta x=\frac{v_0^2}{2|a|}=\frac{(25)^2}{2(5.0)}=62.5\ \mathrm{m}. > $$ > Doubling the initial speed quadruples this distance under the same braking force. **Vertical motion near Earth's surface.** For motion over a small altitude range, gravity gives a nearly constant downward acceleration of magnitude $g=9.81\ \mathrm{m/s^2}$. If upward is positive, $$ a=-g,\qquad v=v_0-gt,\qquad y=y_0+v_0t-\frac12gt^2. $$ At the highest point of an upward toss, $v=0$ but $a=-g$, so the motion continues without a pause in the dynamical sense. For a launch and return to the same height, the ascent and descent times are equal in this model. Air resistance breaks that symmetry and lowers both the peak and the return speed. $$ % caption: Vertical motion with upward positive. Velocity reaches zero at the % maximum height, whereas acceleration stays fixed at minus g throughout. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->] (0,0) -- (0,4.4) node[above] {y}; \draw[very thick, ->] (0.7,0.65) -- (0.7,3.45); \node[anchor=south] at (0.7,3.62) {speed decreases}; \draw[fill=acc!12, draw=acc, thick] (0.7,3.45) circle (2.2pt); \draw[->, thick] (1.35,3.7) -- (1.35,2.6) node[midway, right] {g}; \draw[very thick, ->] (2.0,3.15) -- (2.0,0.45); \node[anchor=west] at (2.18,1.7) {speed increases}; \draw[dashed] (-0.2,0.3) -- (2.8,0.3); \node[anchor=west] at (2.85,0.3) {launch height}; \end{tikzpicture} $$ Dimensional consistency tests the relation. In $v^2=v_0^2+2a\Delta x$, every term has units $\mathrm{m^2/s^2}$. A formula that adds a velocity to an acceleration, or a distance to a speed, is invalid before numerical substitution. ## Motion graphs, areas, and accumulated change The derivative--integral relations for the standard graphs are $$ v_x=\frac{\d x}{\d t},\qquad a_x=\frac{\d v_x}{\d t},\qquad \Delta x=\int v_x\,\d t,\qquad \Delta v_x=\int a_x\,\d t. $$ A position graph with positive curvature has positive acceleration; a velocity graph crossing zero marks an instant of rest, not necessarily a change in direction. Direction reverses only if the velocity changes sign. Constant negative acceleration can increase speed, decrease speed, or leave speed temporarily zero, depending on the sign of velocity. A position function $x=bt^3$ has velocity $3bt^2$ and acceleration $6bt$. The acceleration is zero at $t=0$, although the position graph has a horizontal tangent there. Derivatives must be interpreted separately rather than as interchangeable descriptions of “motion.” **Piecewise motion.** Kinematic equations apply independently on intervals with different constant accelerations. A braking car that first reacts for time $t_r$ and then decelerates at magnitude $a_b$ travels $$ d=v_0t_r+\frac{v_0^2}{2a_b}. $$ The first term is reaction distance; the second is braking distance. Only the second grows quadratically with initial speed. Combining the stages into one constant-acceleration equation erases the physical distinction and gives an incorrect stopping distance. **Local gravity and model limits.** The relation $a_y=-g$ approximates a gravitational field of constant magnitude and direction. It is accurate for ordinary terrestrial heights and short times. Over orbital distances, gravity varies as $1/r^2$ and points toward Earth's centre; the constant-acceleration equations then fail. Air resistance also depends on speed and produces unequal ascent and descent times for a thrown body. **Sign audit for vertical motion.** With upward positive, $a=-g$. A body released from height $h$ has $y-y_0=-h$ at impact, so $$ v^2=0+2(-g)(-h)=2gh. $$ The positive value of $v^2$ follows from the signed displacement and acceleration. The final velocity itself is negative, $v=-\sqrt{2gh}$, because the impact motion is downward. Magnitude and signed component therefore require separate treatment in free-fall calculations. **Relative position and average quantities.** Two particles on the same line have separation $x_{A/B}=x_A-x_B$. Differentiation gives relative velocity and acceleration, $$ v_{A/B}=v_A-v_B, \qquad a_{A/B}=a_A-a_B. $$ The same equations apply to overtaking and pursuit. If $x_A=x_{A0}+v_At$ and $x_B=x_{B0}+v_Bt$, meeting occurs when $x_{A0}-x_{B0}+(v_A-v_B)t=0$. A positive solution is physically required; a negative result means the stated initial arrangement and velocities do not meet in the future interval described by the model. Average acceleration is $\bar a=\Delta v/\Delta t$. It is not generally the acceleration at the midpoint of a time interval. Likewise, average velocity equals the arithmetic mean $(v_i+v_f)/2$ only for constant acceleration. Graph areas provide the general definition. A velocity graph may have positive and negative areas that cancel in displacement, while total distance requires the sum of their absolute values. > **Worked example.** A vehicle travels at $18\ \mathrm{m\,s^{-1}}$, holds that speed > through a $0.75\ \mathrm s$ reaction interval, then brakes at > $-6.0\ \mathrm{m\,s^{-2}}$; find the total stopping distance. > > The reaction phase covers $d_r=(18)(0.75)=13.5\ \mathrm m$ at constant speed. The > braking distance follows from $0=v_0^2+2a\,\Delta x$: > $$ > d_b=\frac{18^2}{2(6.0)}=27.0\ \mathrm m. > $$ > The total is $d_r+d_b=40.5\ \mathrm m$. Combining both phases into one > constant-acceleration equation would erase the reaction motion and give the wrong > distance. Velocity discontinuities have no finite acceleration at the discontinuity. Real collisions have short but nonzero durations, so the velocity changes over a short interval. The idealized discontinuity applies when the collision duration is negligible beside the time scale under study. **Graph interpretation and inverse problems.** Kinematic graphs retain signed quantities. On an $x$--$t$ graph, a steep negative slope denotes a large speed in the negative direction; a flat segment denotes zero instantaneous velocity. On a $v$--$t$ graph, a negative ordinate denotes motion in the negative direction, not negative speed. The total distance over an interval is $$ d=\int_{t_i}^{t_f}|v_x(t)|\,\d t, $$ whereas displacement omits the absolute value. A velocity graph that crosses the time axis must be split at its zero when distance is calculated from area. Given an acceleration function, integration introduces a constant fixed by an initial condition. For $a_x=6t\ \mathrm{m\,s^{-3}}$ and $v_x(0)=2\ \mathrm{m\,s^{-1}}$, $$ v_x(t)=2+\int_0^t6t'\,\d t'=2+3t^2. $$ If $x(0)=1\ \mathrm m$, a second integration gives $$ x(t)=1+\int_0^t(2+3t'^2)\,\d t'=1+2t+t^3. $$ The constants are physical information. Omitting them changes the initial position or velocity and therefore describes a different motion. > **Worked example.** A ball is thrown straight up at $v_0=14\ \mathrm{m\,s^{-1}}$; > find the time to the top and the rise. Take upward positive, so $v=14-9.8t$. > > The top is where $v=0$: > $$ > t_{\rm top}=\frac{14}{9.8}=1.43\ \mathrm s. > $$ > The rise is $\Delta y=v_0t_{\rm top}-\tfrac12 g t_{\rm top}^2=10.0\ \mathrm m$. > Velocity is zero only at that instant; acceleration stays $-9.8\ \mathrm{m\,s^{-2}}$, > so the ball begins descending immediately afterward. The particle model neglects size, orientation, and deformation. It is appropriate for a train's schedule position but not for wheel rotation or braking stress. Force analysis is required to determine the interaction producing an acceleration. **Reading position and velocity graphs together.** Position and velocity graphs form a derivative pair. A rising position curve has positive velocity; a falling curve has negative velocity. The steepness gives speed, while curvature indicates whether velocity is changing. A concave-up position graph has positive acceleration, and a concave-down graph has negative acceleration. The statements refer to the signed coordinate system rather than an informal description such as “moving forward.” A local maximum of position corresponds to zero velocity; zero acceleration additionally requires the velocity graph to have zero slope at that same instant. Turning-point and vertical-toss problems turn on this difference. $$ % caption: Matched position and velocity records for one-dimensional motion. The % position maximum coincides with zero velocity, while the nonzero velocity-graph % slope at that marker gives nonzero acceleration. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (4.8,0) node[right] {time}; \draw[->, black] (0,0) -- (0,3.2) node[above] {position}; \draw[acc, thick, domain=0.30:4.50, samples=100] plot (\x,{0.35+1.90*(\x-0.30)-0.452*(\x-0.30)^2}); \draw[dashed, black] (2.4,0) -- (2.4,2.35); \begin{scope}[shift={(6.1,0)}] \draw[->, black] (0,0) -- (4.8,0) node[right] {time}; \draw[->, black] (0,-1.7) -- (0,2.4) node[above] {$v$}; \draw[acc, thick] (0.3,1.7) -- (4.5,-1.35); \draw[dashed, black] (2.4,-1.55) -- (2.4,0); \end{scope} \end{tikzpicture} $$ **Area as accumulated change.** The signed area under a velocity-time graph is displacement. Rectangles, triangles, and trapezoids apply only when the graph has the corresponding straight-line segments. For a curved graph, the area is an integral or a numerical approximation. The same reasoning applies to acceleration-time graphs: their signed area gives velocity change, not displacement. $$ \Delta x=\int_{t_i}^{t_f}v(t)\,\d t, \qquad \Delta v=\int_{t_i}^{t_f}a(t)\,\d t. $$ $$ % caption: Positive and negative areas under a velocity graph contribute opposite % signed displacements. Their algebraic sum gives net displacement, whereas their % absolute areas must be added to obtain total distance travelled. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (6.0,0) node[right] {time}; \draw[->, black] (0,-2.0) -- (0,3.0) node[above] {$v$}; \fill[acc!12] (0.5,0) -- (1.7,2.0) -- (3.0,0) -- cycle; \fill[black] (3.0,0) -- (4.2,-1.4) -- (5.4,0) -- cycle; \draw[acc, thick] (0.5,0) -- (1.7,2.0) -- (3.0,0) -- (4.2,-1.4) -- (5.4,0); \node[acc, anchor=south] at (1.7,2.25) {forward}; \node[black, anchor=north] at (4.2,-1.72) {back}; \end{tikzpicture} $$ > **Worked example.** A cart has velocity increasing linearly from $-2\ \mathrm{m\,s^{-1}}$ to > $4\ \mathrm{m\,s^{-1}}$ over $3.0\ \mathrm s$. Its acceleration is > > $$ > a=\frac{4-(-2)}{3.0}=2.0\ \mathrm{m\,s^{-2}}. > $$ > > Displacement is the trapezoid area, > > $$ > \Delta x=\frac{-2+4}{2}(3.0)=3.0\ \mathrm m. > $$ > > The cart reverses direction at $t=1.0\ \mathrm s$, when velocity is zero. Total > distance is not $3.0\ \mathrm m$: it is the sum of the magnitude of the negative > and positive triangular areas. The graph therefore separates net position change > from distance travelled without any additional motion equation. **Displacement and accumulated distance.** Displacement is a signed coordinate difference, $$ \Delta x=x_f-x_i. $$ It records net change in position and can be zero after substantial travel. Distance is the total path length and is nonnegative. For a differentiable one-dimensional trajectory, $$ d=\int_{t_i}^{t_f}|v(t)|\,\d t. $$ The absolute value is required whenever velocity changes sign. A position-time graph can return to its initial level, producing zero displacement, while its path has a nonzero total variation. This distinction affects average velocity and average speed: average velocity is $\Delta x/\Delta t$, whereas average speed is $d/\Delta t$. > **Worked example.** A cart moves from $x=2\ \mathrm m$ to $x=14\ \mathrm m$, reverses, and stops at > $x=5\ \mathrm m$. Its displacement is > > $$ > \Delta x=5-2=3\ \mathrm m. > $$ > > Its distance is > > $$ > d=|14-2|+|5-14|=12+9=21\ \mathrm m. > $$ > > If the trip takes $7.0\ \mathrm s$, average velocity is > $0.43\ \mathrm{m\,s^{-1}}$, whereas average speed is > $3.0\ \mathrm{m\,s^{-1}}$. The difference is caused by the reversal, not by a > change of time unit or coordinate origin. **Average and instantaneous velocity.** Average velocity over a finite interval is the slope of a secant line on a position-time graph: $$ \bar v=\frac{x(t_2)-x(t_1)}{t_2-t_1}. $$ Instantaneous velocity is the limiting slope as the interval shrinks: $$ v(t)=\lim_{\Delta t\to0}\frac{x(t+\Delta t)-x(t)}{\Delta t} =\frac{\d x}{\d t}. $$ The secant gives a net rate across an interval; the tangent gives the local rate at one instant. A curved position trace can have a zero secant slope while its tangent slope is nonzero at every interior point. Conversely, a horizontal tangent can occur at a local maximum or minimum even when the object has nonzero acceleration. > **Worked example.** For $x(t)=4t^2-3t+1$ in metres with $t$ in seconds, > > $$ > v(t)=8t-3, > \qquad a(t)=8\ \mathrm{m\,s^{-2}}. > $$ > > Velocity vanishes at $t=0.375\ \mathrm s$. The acceleration is still positive, > so the object changes from negative to positive velocity at a local minimum of > position. Solving $v=0$ is a turning-point test only after the sign of velocity on > the two sides is checked. ## Constant-acceleration derivations and inverse problems With $a$ constant, one integration of $\d v/\d t=a$ gives the velocity, and a second integration of $\d x/\d t=v_0+at$ gives the position: $$ \int_{v_0}^{v}\d v=\int_0^t a\,\d t\ \Rightarrow\ v=v_0+at, \qquad x-x_0=\int_0^t(v_0+at')\,\d t'=v_0t+\frac12at^2. $$ Eliminating time through $a\,\d x=v\,\d v$ gives the time-free form $$ v^2=v_0^2+2a(x-x_0). $$ Select the form containing the known and requested variables. Drag, changing propulsion, and a changing gravitational field require derivative, integral, or numerical methods. > **Worked example.** A vehicle has speed $24\ \mathrm{m\,s^{-1}}$ and stops in $72\ \mathrm m$ under > constant acceleration. The time-free equation gives > > $$ > 0=(24)^2+2a(72), > \qquad a=-4.0\ \mathrm{m\,s^{-2}}. > $$ > > Stopping time follows from $0=24-4.0t$, giving $t=6.0\ \mathrm s$. The negative > sign identifies acceleration opposite the chosen positive velocity direction; it > does not mean the magnitude of acceleration is negative. **Inverse graph problems.** An inverse graph problem infers a motion function from slopes or areas. A constant positive slope on a velocity graph identifies constant positive acceleration. A zero area under velocity over an interval identifies zero displacement, not zero distance. A discontinuity in velocity represents an idealized impulse; its derivative is not a finite ordinary acceleration at that instant. The physical collision has a short but nonzero duration that must be resolved if force is required. **Local versus average behavior.** An average quantity can conceal reversals and pauses. A runner returning to the starting point has zero average velocity but may have a substantial average speed. Zero average acceleration likewise does not imply constant velocity: velocity may increase and later decrease by equal amounts. Instantaneous quantities are limits of averages over shrinking intervals, so they require a differentiable position function or an experimentally resolved time interval. Speed decreases whenever the velocity value moves toward zero, whether the $v_x(t)$ graph lies above or below the time axis; "deceleration" names decreasing speed, not a fixed sign of acceleration. **Graph reconstruction from measured velocity.** A velocity graph and an initial position determine position through accumulated signed area: $$ x(t)=x(t_0)+\int_{t_0}^{t}v(t')\,\d t'. $$ The integration constant is physical information. Without $x(t_0)$, the velocity history determines only changes in position, not an absolute coordinate. A constant positive velocity segment reconstructs as a straight rising position segment; a zero-velocity segment reconstructs as a horizontal position segment; and a negative velocity segment reconstructs as a falling position segment. > **Worked example.** A body begins at $x(0)=3\ \mathrm m$. Its velocity is > $2\ \mathrm{m\,s^{-1}}$ from $0$ to $4\ \mathrm s$, zero from $4$ to > $6\ \mathrm s$, and $-1\ \mathrm{m\,s^{-1}}$ from $6$ to $10\ \mathrm s$. > The displacement contributions are $8\ \mathrm m$, $0$, and $-4\ \mathrm m$. > Thus > > $$ > x(10)=3+8-4=7\ \mathrm m. > $$ > > Total distance is $12\ \mathrm m$, whereas displacement is $4\ \mathrm m$. > The difference arises from the final negative-velocity interval. ## Piecewise and sampled motion Distinct motion stages require distinct initial conditions. For an interval that begins at $t_i$ with $x_i$ and $v_i$ and has constant acceleration $a_i$, $$ v(t)=v_i+a_i(t-t_i), \qquad x(t)=x_i+v_i(t-t_i)+\frac12a_i(t-t_i)^2. $$ Position and velocity are continuous between ordinary stages. Acceleration can change abruptly when a force changes, but a velocity jump represents an idealized impulse and requires a collision model rather than a finite acceleration segment. $$ % caption: Reaction, braking, and rest are separate kinematic stages. Velocity % remains continuous at each boundary while its slope changes when braking starts % and ends, so one average acceleration cannot reconstruct the full distance history. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (6.0,0) node[right] {time}; \draw[->, black] (0,0) -- (0,3.5) node[above] {$v$}; \draw[acc, thick] (0.4,2.5) -- (2.0,2.5) -- (4.5,0) -- (5.5,0); \draw[dashed, black] (2.0,0) -- (2.0,2.5); \draw[dashed, black] (4.5,0) -- (4.5,2.5); \node[anchor=north] at (1.2,-0.15) {reaction}; \node[anchor=north] at (3.25,-0.15) {braking}; \node[anchor=north] at (5.0,-0.15) {rest}; \end{tikzpicture} $$ **Numerical differentiation of position data.** Velocity inferred from positions is a difference quotient. Over a sampling interval $\Delta t$, $$ v\left(t+\frac{\Delta t}{2}\right)\approx \frac{x(t+\Delta t)-x(t)}{\Delta t}. $$ Short sampling intervals improve time resolution but amplify position noise in the subtraction. Acceleration, as a second difference, is more sensitive still. A fitted position curve can reduce random noise but may obscure a short physical impulse. > **Worked example.** Position readings are $x(1.0)=4.00\ \mathrm m$ and > $x(1.2)=4.62\ \mathrm m$. The midpoint velocity estimate is > > $$ > v(1.1)\approx\frac{4.62-4.00}{0.20}=3.1\ \mathrm{m\,s^{-1}}. > $$ > > If each position has uncertainty $\pm0.02\ \mathrm m$, the difference can have > an uncertainty near $0.04\ \mathrm m$, producing a velocity uncertainty near > $0.2\ \mathrm{m\,s^{-1}}$. Measurement resolution therefore constrains the > meaningful digits in a graph-derived velocity. **Constant acceleration from relative increments.** Constant acceleration can be identified without first solving a differential equation. Over equal time intervals $\Delta t$, velocity changes by equal increments $$ \Delta v=a\Delta t. $$ Position increments are not equal because the velocity itself changes. For equal intervals starting from rest, successive displacement increments are proportional to the odd integers $1,3,5,\ldots$. This follows from $$ \Delta x_n=\frac12a(n\Delta t)^2- \frac12a[(n-1)\Delta t]^2 =\frac12a(2n-1)(\Delta t)^2. $$ Uniform acceleration requires the displacement increments to follow the linear odd-integer relation within measurement resolution. Increasing increments alone do not establish that condition. > **Worked example.** An object starts from rest and travels $0.80\ \mathrm m$ in its first second under > constant acceleration. The relation $\Delta x_1=\tfrac12a(1\ \mathrm s)^2$ gives > > $$ > a=1.60\ \mathrm{m\,s^{-2}}. > $$ > > During the second second it travels three times $0.80\ \mathrm m$, or > $2.40\ \mathrm m$; during the third it travels $4.00\ \mathrm m$. These values are > interval distances, not positions from the origin. The position after three seconds > is their sum, $7.20\ \mathrm m$, which agrees with $x=\tfrac12at^2$. ## Free fall and relative motion Near Earth's surface, vertical motion can be treated as one-dimensional motion with approximately constant acceleration. With upward positive, $$ a=-g, \qquad v=v_0-gt, \qquad y=y_0+v_0t-\frac12gt^2. $$ The sign choice is arbitrary but must remain fixed. A released object has $v_0=0$ and negative velocity after release. A thrown-up object has positive initial velocity, zero velocity at its top, and negative velocity during descent. The acceleration remains negative throughout all three stages in the ideal model. > **Worked example.** A tool is dropped from rest at height $45\ \mathrm m$; find the > fall time and impact velocity. Take the ground as $y=0$ with upward positive. > > The tool reaches the ground when $0=45-\tfrac12 g t^2$: > $$ > t=\sqrt{\frac{2(45)}{9.8}}=3.03\ \mathrm s, > \qquad v=-gt=-29.7\ \mathrm{m\,s^{-1}}. > $$ > The negative sign marks downward motion. Reporting $29.7\ \mathrm{m\,s^{-1}}$ > without a direction gives speed, not the signed velocity the coordinate choice > requires. **Relative one-dimensional motion.** For particles A and B on the same coordinate axis, $$ x_{A/B}=x_A-x_B, \qquad v_{A/B}=v_A-v_B, \qquad a_{A/B}=a_A-a_B. $$ An interception time is obtained by setting relative position to zero. If relative velocity is constant, $$ t_{\rm meet}=-\frac{x_{A/B}(0)}{v_{A/B}}. $$ A future meeting requires $t_{\rm meet}>0$. A negative value places the coincidence before the initial instant or corresponds to separation under the stated constant-velocity model. > **Worked example.** A runner starts $60\ \mathrm m$ behind a cyclist. The runner moves at > $8.0\ \mathrm{m\,s^{-1}}$ and the cyclist at $5.0\ \mathrm{m\,s^{-1}}$ in the > same direction. Relative position is initially $-60\ \mathrm m$ and relative > velocity is $3.0\ \mathrm{m\,s^{-1}}$. Therefore > > $$ > t=\frac{60}{3.0}=20\ \mathrm s. > $$ > > The runner's displacement by then is $160\ \mathrm m$ and the cyclist's is > $100\ \mathrm m$; the $60\ \mathrm m$ difference matches the initial separation. > Back-substitution checks the relative-motion solution. **Limits of one-dimensional representation.** One coordinate is adequate only when the motion is constrained to a line or when transverse motion is irrelevant to the question. A vehicle on a curved road cannot be fully described by one fixed Cartesian coordinate because its direction changes. A particle moving along a known curved track can use arc length as a one-dimensional coordinate, but its acceleration then has tangential and normal components. The scalar kinematic relations describe the coordinate along the track; full spatial acceleration also requires its normal component. ## Measurement and numerical reconstruction Kinematic quantities are frequently inferred from instrument readings. Position may be obtained from a ruler, camera calibration, encoder, or GPS coordinate; velocity and acceleration are then computed from differences and derivatives. The uncertainty of a derived quantity depends on both measurement resolution and the time interval used in the calculation. For average velocity $\bar v=\Delta x/\Delta t$, a first-order independent-error estimate is $$ \frac{\delta\bar v}{|\bar v|}\approx \frac{\delta(\Delta x)}{|\Delta x|} +\frac{\delta(\Delta t)}{\Delta t}. $$ When initial and final positions are measured independently with similar uncertainty $\delta x$, the displacement uncertainty can approach $2\delta x$ under a conservative bound. A very short time interval may make the displacement comparable with this uncertainty and produce a poor velocity estimate even if the time clock is precise. A longer interval reduces fractional position-difference uncertainty but averages over changes in the underlying velocity. > **Worked example.** Positions $x_1=(1.20\pm0.02)\ \mathrm m$ and > $x_2=(2.00\pm0.02)\ \mathrm m$ are separated by > $\Delta t=(0.400\pm0.002)\ \mathrm s$; find the average velocity and its uncertainty. > > The central value is $\bar v=0.80/0.400=2.00\ \mathrm{m\,s^{-1}}$. With a > conservative displacement uncertainty of $0.04\ \mathrm m$, the fractional error is > $$ > \frac{\delta\bar v}{\bar v}\approx > \frac{0.04}{0.80}+\frac{0.002}{0.400}=0.055, > $$ > so $\bar v=(2.00\pm0.11)\ \mathrm{m\,s^{-1}}$. The reported digits follow the derived > uncertainty, not the calculator display. **Numerical integration from acceleration samples.** Acceleration data sampled at finite intervals can be integrated numerically. With uniform sample spacing $\Delta t$, the forward Euler update is $$ v_{n+1}=v_n+a_n\Delta t, $$ $$ x_{n+1}=x_n+v_n\Delta t. $$ The method uses acceleration and velocity at the beginning of each interval. For constant acceleration, its velocity update is exact at the sample points, but its position update has a finite step error because it uses the beginning velocity rather than the interval-average velocity. A trapezoidal position update, $$ x_{n+1}=x_n+\frac{v_n+v_{n+1}}{2}\Delta t, $$ is exact for a velocity that changes linearly during the interval. > **Worked example.** A cart begins with $v_0=1.0\ \mathrm{m\,s^{-1}}$. During a > $0.50\ \mathrm s$ interval, its measured acceleration is > $2.0\ \mathrm{m\,s^{-2}}$. The updated velocity is > > $$ > v_1=1.0+(2.0)(0.50)=2.0\ \mathrm{m\,s^{-1}}. > $$ > > Euler position update gives $\Delta x=0.50\ \mathrm m$. The trapezoidal update > gives > > $$ > \Delta x=\frac{1.0+2.0}{2}(0.50)=0.75\ \mathrm m. > $$ > > For this constant-acceleration interval, $0.75\ \mathrm m$ agrees with the exact > kinematic expression. The discrepancy quantifies the beginning-velocity Euler error > for the chosen step. ## Applied one-dimensional models Inverse free-fall problems infer an unknown height, launch speed, or elapsed time from a measured event. The sign convention should be fixed before rearranging an equation. With upward positive, a dropped object from height $h$ to ground has $\Delta y=-h$ and $a=-g$, giving $$ v_f^2=2gh. $$ The square root gives impact speed. The signed impact velocity is negative. A time-of-flight measurement can instead infer height through $$ h=\frac12gt^2. $$ Time and impact speed provide independent checks when both are available. Their disagreement indicates drag, timing offset, release velocity, or measurement error. > **Worked example.** A dropped object has flight time $(2.50\pm0.02)\ \mathrm s$. The timing estimate > is > > $$ > h=\frac12(9.8)(2.50)^2=30.6\ \mathrm m. > $$ > > The fractional timing contribution is $2(0.02/2.50)=0.016$, giving approximately > $\pm0.5\ \mathrm m$ from timing uncertainty alone. If a speed sensor reports > $24.5\ \mathrm{m\,s^{-1}}$, its implied height is > $v^2/(2g)=30.6\ \mathrm m$, consistent with the ideal model at the stated > precision. **Frame changes in one-dimensional motion.** One-dimensional kinematics can be described from any inertial frame. If frame $S'$ moves at constant speed $V$ in the positive direction relative to frame $S$, the Galilean transformation is $$ x'=x-Vt, \qquad v'=v-V, \qquad a'=a. $$ Position and velocity values depend on the chosen frame, while acceleration is the same in all inertial frames. A car traveling at $25\ \mathrm{m\,s^{-1}}$ relative to road has zero velocity relative to a passenger moving with the car, yet both frames agree on its acceleration during braking. The transformation changes the numerical description, not the physical event. > **Worked example.** A car at $24\ \mathrm{m\,s^{-1}}$ brakes at > $-3.0\ \mathrm{m\,s^{-2}}$ while a truck ahead holds $18\ \mathrm{m\,s^{-1}}$; find > how far the car gains on the truck before their speeds match. > > In the truck frame the car starts at relative velocity $6.0\ \mathrm{m\,s^{-1}}$ > with relative acceleration $-3.0\ \mathrm{m\,s^{-2}}$, reaching zero relative > velocity at $t=6.0/3.0=2.0\ \mathrm s$. The relative displacement is > $$ > \Delta x_{C/T}=6.0(2.0)+\tfrac12(-3.0)(2.0)^2=6.0\ \mathrm m. > $$ > Subtracting the two road-frame positions gives the same $6.0\ \mathrm m$; working in > the truck frame removes the common motion from the algebra. **Stopping-distance limits and model assumptions.** The constant-deceleration stopping formula $$ d_b=\frac{v_0^2}{2a_b} $$ uses positive braking magnitude $a_b$. It assumes constant acceleration, level motion, and a braking force independent of speed. Air drag, grade, brake fade, anti-lock control, and changing road friction alter the acceleration and require a force or numerical model. The quadratic dependence on initial speed follows directly from the formula, which also gives a constant-acceleration reference calculation. On a downhill grade with slope angle $\alpha$, gravitational acceleration adds a downslope term $g\sin\alpha$. If braking produces a constant uphill acceleration magnitude $a_b$, the net deceleration magnitude is $$ a_{\rm net}=a_b-g\sin\alpha. $$ The stopping calculation is meaningful only when $a_b>g\sin\alpha$. Otherwise the brakes cannot overcome the downslope gravitational component in this simplified model. > **Worked example.** A vehicle at $22\ \mathrm{m\,s^{-1}}$ has braking magnitude > $5.0\ \mathrm{m\,s^{-2}}$ on a $6.0^\circ$ downhill grade. The net deceleration > magnitude is > > $$ > a_{\rm net}=5.0-9.8\sin6.0^\circ=3.98\ \mathrm{m\,s^{-2}}. > $$ > > Stopping distance is therefore > > $$ > d=\frac{22^2}{2(3.98)}=60.8\ \mathrm m. > $$ > > On level ground the same brake magnitude would predict $48.4\ \mathrm m$. > The difference follows from the modelled gravity component, not from a change in > the initial speed. **Meeting constraints with acceleration.** When one object accelerates and another moves at constant velocity, their meeting condition is a quadratic rather than a linear relative-motion equation. If object A starts at $x_A(0)$ with $v_A(0)$ and constant acceleration $a_A$, while B starts at $x_B(0)$ with constant velocity $v_B$, then $$ x_A(0)+v_A(0)t+\frac12a_At^2=x_B(0)+v_Bt. $$ The roots are possible meeting times. A positive root must be checked against the physical interval and any constraints on the acceleration stage. Two positive roots can represent two crossings when the accelerating object passes the other body and later returns under a direction change. $$ % caption: An accelerating trajectory and a constant-velocity trajectory can % intersect zero, one, or two times. Their intersections are roots of the relative- % position quadratic; only positive times within the stated motion stages are % physical meetings. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (5.8,0) node[right] {time}; \draw[->, black] (0,0) -- (0,3.7) node[above] {position}; \draw[acc, thick, domain=0.4:4.0, samples=90] plot (\x,{0.25+0.22*(\x-0.4)+0.175*(\x-0.4)^2}); \draw[black, thick] (0.4,1.05) -- (4.8,2.7); \draw[fill=acc!12, draw=acc, thick] (3.03,2.04) circle (1.8pt); \node[anchor=north west] at (3.16,1.92) {meeting}; \end{tikzpicture} $$ > **Worked example.** A bicycle is $20\ \mathrm m$ ahead and moves steadily at > $4.0\ \mathrm{m\,s^{-1}}$. A runner starts from rest at the origin with constant > acceleration $1.0\ \mathrm{m\,s^{-2}}$. Meeting requires > > $$ > \frac12(1.0)t^2=20+4.0t. > $$ > > The positive root is > > $$ > t=4+\sqrt{56}=11.5\ \mathrm s. > $$ > > At this time the runner position is $66.0\ \mathrm m$, matching the bicycle > position $20+4.0(11.5)=66.0\ \mathrm m$ to rounding precision. **Velocity relaxation under linear resistance.** Constant acceleration is a local model. Once a resistive force changes markedly with speed, the velocity graph is curved and the constant-acceleration equations no longer extrapolate reliably. For modest speeds, a one-dimensional model can use a resistive force proportional to velocity. For an object sliding along the positive axis after its driving force has been removed, $$ m\frac{\d v}{\d t}=-bv, $$ where $b$ has units of $\mathrm{kg\,s^{-1}}$. Dividing by $m$ identifies the relaxation time $\tau=m/b$. Separation of variables gives $$ \frac{\d v}{v}=-\frac{\d t}{\tau}, \qquad v(t)=v_0e^{-t/\tau}. $$ The velocity remains positive for every finite time in this idealized model and approaches zero asymptotically. The form can describe a damped sensor or a low-speed fluid experiment. A wheel that locks, sticks, or reverses requires a contact model with additional force regimes. Integrating the velocity gives the distance travelled from the release point: $$ x(t)-x_0=v_0\tau\left(1-e^{-t/\tau}\right). $$ With no additional driving force, linear resistance gives the finite limiting distance $v_0\tau$. Differentiate this expression to recover $v_0e^{-t/\tau}$; its initial slope is $v_0$. A position curve that crosses the limiting distance is incompatible with this model. > **Worked example.** A small cart has $v_0=1.20\ \mathrm{m\,s^{-1}}$ and a measured relaxation time > $\tau=2.00\ \mathrm s$. Its speed after $3.00\ \mathrm s$ is > > $$ > v=(1.20)e^{-3.00/2.00}=0.268\ \mathrm{m\,s^{-1}}. > $$ > > The corresponding displacement is > > $$ > \Delta x=(1.20)(2.00)(1-e^{-1.50})=1.86\ \mathrm m. > $$ > > The limiting distance is $2.40\ \mathrm m$, so the untravelled remainder after > three seconds is $0.54\ \mathrm m$. Treating the initial deceleration as constant > would instead predict a straight velocity line and, beyond its zero crossing, > would generate an unphysical reversal. **A steady drive with linear resistance.** With a constant applied force $F_0$ along the positive axis, the same resistance law becomes $$ m\frac{\d v}{\d t}=F_0-bv. $$ The velocity tends to the terminal value $v_T=F_0/b$, while the time constant remains $\tau=m/b$. For release from rest, integration yields $$ v(t)=v_T\left(1-e^{-t/\tau}\right), \qquad x(t)-x_0=v_T\left[t-\tau\left(1-e^{-t/\tau}\right)\right]. $$ At early times, $e^{-t/\tau}\approx1-t/\tau$, so $v\approx(F_0/m)t$ and the motion follows the constant-acceleration approximation. At late times, the velocity graph flattens and position becomes nearly linear with slope $v_T$. Data covering only a short interval can therefore fit either constant acceleration or the early part of a relaxation curve. > **Worked example.** A motorized cart settles near $0.80\ \mathrm{m\,s^{-1}}$ under an applied force > of $0.40\ \mathrm N$. The fitted resistance coefficient is > $b=F_0/v_T=0.50\ \mathrm{kg\,s^{-1}}$. If $\tau=1.6\ \mathrm s$, the inferred > mass is $m=b\tau=0.80\ \mathrm{kg}$. These values require force measurements to > be expressed in the same coordinate direction; reversing the axis reverses the > signed force and velocity but leaves the positive resistance coefficient unchanged. **Centered differences and the cost of noisy data.** Measured positions are recorded at discrete times, whereas velocity and acceleration are derivatives defined at an instant. A forward difference, $$ v(t_i)\approx\frac{x_{i+1}-x_i}{\Delta t}, $$ associates its answer most naturally with the midpoint of the interval. For evenly spaced data, a centered estimate uses information on both sides of the requested time: $$ v(t_i)\approx\frac{x_{i+1}-x_{i-1}}{2\Delta t}, \qquad a(t_i)\approx\frac{x_{i+1}-2x_i+x_{i-1}}{(\Delta t)^2}. $$ The second expression is the discrete curvature of the position record. A point above the chord joining its neighbors gives a negative numerator and therefore negative acceleration in the positive-coordinate convention. A point below the chord gives positive curvature. The chord provides a direct sign check on the subtraction. Differentiation magnifies position noise. If each independent position reading has standard uncertainty $\sigma_x$, the approximate velocity uncertainty from the centered difference is $\sigma_v=\sigma_x/(\sqrt2\,\Delta t)$. The acceleration estimate is more sensitive still, with $$ \sigma_a\approx\frac{\sqrt6\,\sigma_x}{(\Delta t)^2}. $$ Reducing $\Delta t$ improves temporal resolution but can enlarge the random uncertainty in the derived acceleration. Select the interval from the required time resolution and the expected position noise. A smooth fit to the positions, followed by differentiation of the fitted curve, can be preferable when physical acceleration is expected to vary gradually. > **Worked example.** Positions sampled every $0.20\ \mathrm s$ are $0.180$, $0.240$, and > $0.340\ \mathrm m$. The centered velocity at the middle reading is > > $$ > v\approx\frac{0.340-0.180}{0.40}=0.400\ \mathrm{m\,s^{-1}}, > $$ > > while the acceleration estimate is > > $$ > a\approx\frac{0.340-2(0.240)+0.180}{(0.20)^2} > =1.00\ \mathrm{m\,s^{-2}}. > $$ > > The acceleration belongs to the middle time, not the final time. A report that > labels it as the acceleration at the third reading shifts the inferred physical > event by one sampling interval. **Fitting a quadratic position record.** For motion that is plausibly constant-acceleration over a stated interval, the most informative position model is a quadratic in time, $$ x(t)=c_0+c_1t+c_2t^2. $$ Comparison with $x=x_0+v_0t+\tfrac12at^2$ identifies $c_0=x_0$, $c_1=v_0$, and $a=2c_2$. The fitted coefficients therefore have direct kinematic meaning. Shifting the time origin changes $c_0$ and $c_1$, but the curvature coefficient $c_2$ remains the same. The invariance matters when two instruments start their clocks at different moments. Three exact position readings at distinct times determine one quadratic. Real records contain scatter, so a least-squares fit uses all readings and exposes the remaining discrepancies as residuals, $$ r_i=x_{i,\mathrm{measured}}-x(t_i). $$ Residuals randomly distributed above and below zero support the model to the precision of the measurements. A residual sequence that first rises and then falls systematically is evidence that the assumed quadratic misses a changing acceleration. The curvature itself may be quite small on a broad position plot; the residual plot can identify the model failure first. > **Worked example.** A tracker returns $x(t)=0.146+0.318t+0.742t^2$, with position in metres and time > in seconds. The intercept records a position of $0.146\ \mathrm m$ at the clock > origin, and the initial velocity is $0.318\ \mathrm{m\,s^{-1}}$. The acceleration > is twice the quadratic coefficient: > > $$ > a=2(0.742)=1.484\ \mathrm{m\,s^{-2}}. > $$ > > Units provide an immediate audit. Since $c_2t^2$ must have units of length, > $c_2$ has units $\mathrm{m\,s^{-2}}$, the same units as half an acceleration. > Reporting $0.742\ \mathrm{m\,s^{-1}}$ for this coefficient would be > dimensionally inconsistent even before a numerical comparison is made. **Sampling rate and hidden motion between readings.** A discrete record does not display everything that happened between sampling instants. If a position sensor reports once every $\Delta t$, a short reversal that begins and ends between two readings can be missed entirely. A single average velocity over the interval remains correct for the net displacement, but it does not establish that the instantaneous velocity held the same sign throughout the interval. Distance travelled is especially vulnerable: it cannot be recovered from widely spaced positions without a justified model for the path between them. A cart can move forward, pause, and roll back to the same marked position during one interval. Its displacement is zero, yet it has accumulated distance and changed velocity twice. Additional timing marks or a higher-rate record distinguish these possibilities. Kinematic analysis must retain the information actually observed and state any interpolation or path model. > **Worked example.** Two readings separated by $0.50\ \mathrm s$ both give $x=1.00\ \mathrm m$. The > average velocity is therefore zero, yet the cart may have been stationary, or it > may have travelled in either direction and returned. If an intermediate camera > frame at $0.25\ \mathrm s$ records $x=1.20\ \mathrm m$, the path contains at > least $0.20\ \mathrm m$ outward and $0.20\ \mathrm m$ back, so the travelled > distance is at least $0.40\ \mathrm m$. Thus $0.40\ \mathrm m$ is a lower bound; > a finer record could resolve further reversals inside either half-interval. **Event times from sampled position records.** Many experiments require the time of an event: a cart reaches a photogate, a runner crosses a marked line, or a moving stage enters a permitted region. The event condition has the form $x(t)=x_\star$, where $x_\star$ is the coordinate of the boundary. A position table will rarely contain the exact crossing time. It usually gives two adjacent readings that bracket the event, one on each side of the boundary. Over a short interval with no appreciable curvature, linear interpolation estimates the crossing time. If $x_i **Worked example.** A tracker records $x=1.84\ \mathrm m$ at $t=2.10\ \mathrm s$ and > $x=2.05\ \mathrm m$ at $t=2.20\ \mathrm s$. A gate is fixed at > $x_\star=2.00\ \mathrm m$. Linear interpolation gives > > $$ > t_\star=2.10+ > \frac{2.00-1.84}{2.05-1.84}(0.10) > =2.176\ \mathrm s. > $$ > > The final digits do not imply millisecond accuracy. If the positions are only known > to $\pm0.01\ \mathrm m$, the uncertainty in the fraction of the interval may > already dominate that last digit. A faster sampling rate, a narrow optical gate, or > a fitted local trajectory can improve the event timing; extra decimal places cannot. **Timing uncertainty near a coordinate boundary.** Position uncertainty translates into time uncertainty through the local speed at the crossing. When the motion is locally monotonic and the boundary coordinate has uncertainty $\sigma_{x_\star}$, a first-order estimate is $$ \sigma_t\approx\frac{\sqrt{\sigma_x^2+\sigma_{x_\star}^2}}{|v(t_\star)|}. $$ The estimate grows when the object approaches a boundary slowly. A photogate can locate a fixed edge very well yet still yield a broad time estimate for a nearly stalled cart because a small coordinate ambiguity corresponds to a long interval of time. Near a turnaround, linearized timing formulas become unreliable: velocity approaches zero and the same coordinate may be reached on two distinct branches of the trajectory. In that case the two roots and their separate uncertainty ranges should be reported from a fitted position model or from additional resolved frames. **Reporting a crossing time.** An event time should carry the same model qualification as the trajectory from which it was obtained. “The gate was crossed at $2.18\ \mathrm s$ by linear interpolation of the two adjacent frames” states both the measurement and the assumption. Report the interpolation method and bracketing frames with the time value. When a boundary has finite width, the reported event also needs a convention: leading edge, centre, or trailing edge. Those conventions differ by a real travel time and cannot be repaired later by rounding. The coordinate of the boundary, the direction of crossing, and the clock reference complete a reproducible one-dimensional event statement.