One-Dimensional Kinematics

Kinematics is the branch of mechanics that describes motion — how objects move through space over time — without asking why they move. The question of why (the forces that cause motion) belongs to dynamics, which we take up in the Newton's laws chapters. Kinematics comes first because you need a precise vocabulary for motion before you can discuss its causes.

The world is in constant motion. Electrons orbit nuclei, blood cells traverse arteries, tectonic plates drift centimeters per year, and the Milky Way spirals through the cosmos. Even a "stationary" coffee mug is hurtling around the sun at $approx 30$ km/s. Kinematics provides a universal language for describing all of this motion.

In this chapter we restrict ourselves to the simplest case: motion along a straight line — called one-dimensional (1D) motion. The line may be horizontal, vertical, or inclined, but the object's entire motion stays on that single axis. We also model moving objects as particles — point-like objects with no size or shape. This idealization works whenever all parts of the object move together in the same direction at the same speed (a car driving down a highway, a ball thrown straight up).

To describe motion we first need to locate the object. We choose a reference axis — usually called the $x$-axis — with an origin (zero point) and a positive direction. Every position is then a signed number: $x = +5$ m means 5 meters in the positive direction from the origin; $x = -3$ m means 3 meters in the negative direction.

Displacement $\Delta x$ is the change in position from some initial location $x_i$ to a final location $x_f$:

Displacement is a vector quantity — it has both a magnitude and a direction, encoded in its algebraic sign. If an object moves from $x = 2$ m to $x = 8$ m, the displacement is $\Delta x = +6$ m (rightward). If it then moves back to $x = 3$ m, the second displacement is $\Delta x = -5$ m (leftward). The net displacement for the entire trip is $3 - 2 = +1$ m.

Distance is the total path length traveled — always non-negative. In the example above, the total distance is $6 + 5 = 11$ m, even though the net displacement is only 1 m. This distinction is not a technicality: it matters any time an object reverses direction.

Having defined position, we can describe how quickly a particle moves between two positions. The average velocity $\bar{v}$ over a time interval $\Delta t = t_f - t_i$ is the displacement divided by the elapsed time:

Average velocity is a vector (it has the same sign as the displacement). Its SI unit is meters per second (m/s). On a graph of position $x$ versus time $t$, the average velocity over any interval is the slope of the straight line connecting the two endpoint dots on the curve.

Average speed $s_{avg}$ is a different quantity — it uses total distance rather than displacement:

Average speed is always non-negative. For a trip that includes reversals, average speed is larger than $|\bar{v}|$. Only for one-way straight-line motion are they equal.

Example — The pickup truck problem

A truck drives 8.4 km at 70 km/h until it runs out of fuel, then the driver walks 2.0 km to a gas station, taking 30 min. What is the average velocity for the whole trip?

Total displacement: $\Delta x = 8.4 + 2.0 = 10.4$ km. Driving time: $8.4/70 = 0.12$ h. Walking time: 0.50 h. Total time: 0.62 h.

Note: average velocity is not the average of 70 km/h and 0 km/h — those quantities are weighted by time, not distance, and the weighting is unequal here.

Average velocity tells you how fast a particle moved over a whole interval but says nothing about its speed at any particular moment. Instantaneous velocity $v$ (commonly just called "velocity") is the velocity at a single instant of time. It is defined by shrinking the time interval to zero:

In the language of calculus, instantaneous velocity is the first derivative of position with respect to time. Geometrically, $v$ at any instant equals the slope of the tangent line to the $x(t)$ curve at that point.

Instantaneous speed is the magnitude $|v|$ — it strips away the directional sign and is always non-negative. Your car's speedometer reads instantaneous speed, not velocity.

Reading the x-t graph

On a position-versus-time graph:

• A steep slope means fast motion; a gentle slope means slow motion.

• A positive slope (going up to the right) means positive velocity.

• A negative slope (going down to the right) means the object is moving in the negative direction.

• A horizontal line (zero slope) means the object is stationary.

• A curved position-time graph means the velocity is changing — the object is accelerating.

Velocity can itself change with time. The rate at which velocity changes is called acceleration. Average acceleration over a time interval $\Delta t$ is:

Instantaneous acceleration is the derivative of velocity with respect to time — equivalently, the second derivative of position:

The SI unit of acceleration is $\text{m/s}^2$ (meters per second per second). On a $v$-vs-$t$ graph, acceleration at any instant is the slope of the tangent line. On an $a$-vs-$t$ graph, the area under the curve gives the change in velocity.

Sign rules: speeding up vs slowing down

The sign of acceleration indicates direction, not whether the object is speeding up or slowing down. The comparison that matters is between the signs of $v$ and $a$:

g units: Large accelerations are often expressed in multiples of $g = 9.8\text{ m/s}^2$. A fighter pilot experiencing $5g$ feels a force five times their weight. Typical safe human tolerance for sustained acceleration is 4–6g; brief spikes in crashes can reach hundreds of g.

The most practically important case in 1D kinematics is constant acceleration — when $a$ does not change with time. A car braking at a steady rate, a ball in free fall (near Earth's surface, ignoring air), and a rocket in a constant-thrust burn all approximate this case. When acceleration is constant, a complete set of five equations connects the five kinematic quantities: $x - x_0$ (displacement), $v_0$ (initial velocity), $v$ (final velocity), $a$ (acceleration), and $t$ (time).

You only ever need two of these equations: the first and second are the fundamental ones; the rest are derived by eliminating one variable. In practice, the fastest strategy is:

1. List the five variables. Write down their values (or "unknown" for each).

2. Identify which variable is not mentioned in the problem (neither given nor asked). That is the "missing variable."

3. Select the equation in the table whose "missing quantity" column matches yours. It is the one equation where the missing variable does not appear.

Example — Braking car

A car moving at $v_0 = 30$ m/s brakes with $a = -6\text{ m/s}^2$ until it stops ($v = 0$). How far does it travel?

Known: $v_0 = 30$, $v = 0$, $a = -6$. Missing: $t$. Use $v^2 = v_0^2 + 2a\,\Delta x$:

Example — Drag race

A motorcycle accelerates from rest at $a_m = 8.4\text{ m/s}^2$ to a top speed of $v_m = 58.8$ m/s, then holds that speed. A car accelerates from rest at $a_c = 5.6\text{ m/s}^2$ without a top-speed limit. When does the car overtake the motorcycle?

The motorcycle reaches top speed at $t_m = v_m/a_m = 7.0$ s, having traveled $x_m = v_m^2/(2a_m) = 206$ m. After $t_m$, it moves at constant speed. Setting the car's position equal to the motorcycle's total position and solving the resulting quadratic gives $t \approx 16.6$ s. (This is Sample Problem 2.04 in Halliday & Resnick — an excellent exercise in setting up simultaneous equations for two-phase motion.)

The most important special case of constant acceleration is free fall near Earth's surface: any object dropped, thrown, or launched vertically, with air resistance neglected, experiences the same downward acceleration. This constant is called $g$:

The value $g \approx 9.8\text{ m/s}^2$ holds everywhere on Earth's surface to within about 0.5%. It is slightly larger at the poles ($9.832\text{ m/s}^2$) and smaller at the equator ($9.780\text{ m/s}^2$) due to Earth's rotation and shape.

Two crucial sign conventions for free-fall problems:

1. Take the positive direction as upward. Then the free-fall acceleration is $a = -g = -9.8\text{ m/s}^2$ in every equation. (Some books take downward as positive, giving $a = +g$; either works, but be consistent throughout a problem.)

2. At the highest point of a vertical trajectory, the velocity is zero but the acceleration is still $-g$, not zero. The ball is still accelerating even though it is momentarily at rest.

Example — Baseball toss

A pitcher tosses a ball straight up with initial speed $v_0 = 12$ m/s. (a) How long until it reaches maximum height? (b) What is that maximum height above the release point? (c) How long until it returns to the release point?

(a) At maximum height, $v = 0$. From $v = v_0 + at$:

(b) From $v^2 = v_0^2 + 2a\Delta y$:

(c) By symmetry, the total up-down time is exactly twice the time to the top: $t_{\text{total}} = 2 \times 1.22 = 2.45$ s. (Verify: set $\Delta y = 0$ in $\Delta y = v_0 t + \frac{1}{2}at^2$ and solve — you get $t = 0$ and $t = 2v_0/g = 2.45$ s.)

Galileo's discovery: In 1589 (or thereabouts), Galileo Galilei showed by experiment that all objects fall at the same rate regardless of mass. A feather and a hammer dropped in a vacuum reach the ground simultaneously. In air, drag complicates things — but in the absence of air resistance, the statement is exact. Apollo 15 astronaut David Scott demonstrated this on the Moon in 1971 by dropping a hammer and feather simultaneously.

Three graphs — $x(t)$, $v(t)$, and $a(t)$ — provide a complete picture of 1D motion. Each is the derivative of the one above it and the integral of the one below it.

Reading the x-t graph

A straight line on an $x$-vs-$t$ graph means constant velocity (zero acceleration). A curved line means changing velocity (nonzero acceleration). If the curve bends upward (concave up), acceleration is positive. If it bends downward (concave down), acceleration is negative. A point where the curve changes from concave up to concave down is where the sign of acceleration changes.

Reading the v-t graph

A straight line on a $v$-vs-$t$ graph means constant acceleration. The slope of that line is $a$. The area between the $v(t)$ curve and the time axis (counting area above the axis as positive, below as negative) equals the displacement:

For a trapezoidal region (constant acceleration), the area is $\frac{1}{2}(v_0 + v_f)\Delta t$ — which is exactly the fourth kinematic equation.

Graphical integration example

A rear-end collision test: a volunteer's torso accelerates from rest. The $a(t)$ graph shows roughly triangular and rectangular pulses. Integrating (finding the area) gives the change in velocity. If the torso has a triangular pulse of $50\text{ m/s}^2$ lasting 60 ms and a rectangular pulse of $50\text{ m/s}^2$ lasting 10 ms:

This graphical approach is powerful whenever you have a measured (and possibly irregular) $a(t)$ profile from sensors, such as in crash analysis, sports biomechanics, or spacecraft telemetry — situations where no simple algebraic formula for $a(t)$ exists.