One-Dimensional Kinematics

Slam the brakes at 20 m/s and a car needs about 50 m of road to stop. Do it at 40 m/s and it needs 200 m — twice the speed, four times the distance. That unforgiving square law is not a quirk of tires or reaction times; it falls straight out of a handful of short algebraic equations that govern every motion along a straight line whose acceleration holds constant. A dropped stone, a sprinter out of the blocks, an elevator settling to a floor, a jet on its takeoff roll — one family of formulas covers them all. One-dimensional kinematics is the full description of straight-line motion, and its entire vocabulary is three words deep. Position $x(t)$ says where you are; velocity $v = \mathrm{d}x/\mathrm{d}t$ says how fast that changes; acceleration $a = \mathrm{d}v/\mathrm{d}t$ says how fast that changes. Because only one axis is in play, vectors collapse into signed numbers, and all the directional physics lives in a plus or minus sign. The three quantities form a ladder — differentiate to climb up, integrate to climb down — and that ladder is true for any motion whatsoever, no forces required. Feed the special case $a = \text{constant}$ into that calculus and it simplifies all the way down to algebra: the four kinematic (SUVAT) equations, the computational engine of introductory mechanics. Free fall runs on them, projectile motion is two copies of them sharing one clock, and Newton's second law becomes predictive only because they convert an acceleration into an actual trajectory. And when the acceleration refuses to stay constant, the same ladder still works — you just have to climb it by honest integration.

Key takeaways

One-dimensional kinematics describes motion along a single axis using position $x(t)$ , velocity $v = \mathrm{d}x/\mathrm{d}t$ , and acceleration $a = \mathrm{d}v/\mathrm{d}t$ — all signed scalars; the definitions are calculus, true for any motion whatsoever.
For constant acceleration, five quantities — $s$ , $v_0$ , $v$ , $a$ , $t$ — are linked by four standard equations; any three known values determine the rest.
The four: $v = v_0 + at$ · $s = v_0 t + \tfrac12 at^2$ · $v^2 = v_0^2 + 2as$ · $s = \tfrac12(v_0+v)\,t$ — all four are one geometric fact, the trapezoid-shaped area under a straight $v$ – $t$ line.
Free fall is the flagship case: $a = -g$ with up positive, the same for every body, giving $h_{\max} = v_0^2/2g$ and flight time $2v_0/g$ for a vertical throw.
When $a$ is not constant, integrate: $v(t) = v_0 + \int a\,\mathrm{d}t$ , $x(t) = x_0 + \int v\,\mathrm{d}t$ ; and when $a$ depends on position, the chain-rule identity $a\,\mathrm{d}x = v\,\mathrm{d}v$ trades the clock for geometry.
Graphically: velocity is the slope of $x(t)$ , acceleration the slope of $v(t)$ , and displacement the signed area under $v(t)$ .
Key idea: position, velocity, and acceleration form a chain linked by differentiation ( $x \to v \to a$ ) and integration ( $a \to v \to x$ ); constant acceleration is just the case where the chain integrates to polynomials in $t$ — the SUVAT equations.

Where this lives today

The square law of equation (6) is written into the physical world around you. Highway engineers set speed limits, sight distances, and escape-ramp lengths from $s = v_0^2/2a$ with measured tire–road decelerations; the reason a 130 km/h motorway needs sweeping curves and long merge lanes is that quadratic term. The same arithmetic runs live inside modern cars: automatic emergency braking systems continuously compare the range to the obstacle ahead with the computed stopping distance at the current speed, and fire the brakes when the two curves are about to cross. The equations also run in the other direction — from measured acceleration to reconstructed motion, which is exactly the integral pair (3). The accelerometer in your phone integrates $a(t)$ to detect falls, count steps, and stabilize video; inertial navigation units in aircraft and spacecraft do the same double integration with laser-grade instruments. And free fall itself is now a manufactured laboratory condition: drop towers like ZARM in Bremen let sealed experiment capsules fall through a 120-metre evacuated shaft, buying about 4.7 seconds of clean weightlessness for fluid-physics and cold-atom experiments — Galileo's $\tfrac12 gt^2$ , run in reverse as an engineering specification.

The physical picture

Imagine a car on a perfectly straight road with mile-markers. Position is which marker you are at; velocity is how fast the markers tick by, and in which direction; acceleration is how fast the speedometer needle moves. These three quantities form a ladder: differentiate to climb up ( $x \to v \to a$ ), integrate to climb down ( $a \to v \to x$ ). The minus sign is everything in 1D. There are no arrows to draw, so all directional information lives in the sign relative to your chosen positive axis. A ball thrown upward has positive $v$ but negative $a = -g$ throughout its flight — it is always accelerating downward, even at the top where $v$ momentarily vanishes. "Deceleration" is not a separate concept: it just means $a$ and $v$ carry opposite signs. Now plot velocity against time for a car under steady braking, a ball in free fall, or a train pulling smoothly out of a station, and you get the same object: a straight line. The slope of that line is the acceleration. And the area trapped between the line and the time axis is the displacement, because each thin vertical strip contributes (velocity) × (sliver of time) = a sliver of distance.

Figure

The hero picture — one straight line, and its area is the displacement

That one picture generates the entire constant-acceleration toolkit. The area under a straight line from height $v_0$ to height $v$ is a trapezoid, and you can compute a trapezoid's area three ways: average height times width, rectangle plus triangle, or by trading the width for the slope. Each bookkeeping choice is one of the kinematic equations. Nothing else is going on. The picture also announces its own boundaries. If the line is not straight — a car easing off the brakes, a raindrop approaching terminal velocity — the trapezoid formulas are dead on arrival, and you must return to the general integrals. "Constant acceleration" is not decoration; it is the entire license. One more picture before the algebra — because it is the payoff the overview promised. Draw the stopping distance to scale at highway speeds and the square law stops being a formula and becomes a road-safety argument:

Figure

Why the square law rules the road — braking distance at a = 6 m/s²

Doubling from 60 to 120 km/h doesn't double the braking distance — it quadruples it, 23 m to 93 m. Where that square comes from is exactly the time-free equation derived below.

The language of motion

Before the four famous equations, the three definitions they are built from — because these hold for every motion, constant acceleration or not. Ask "how fast is the position changing right now?" and the answer is the derivative:

v(t) = \frac{\mathrm{d}x}{\mathrm{d}t} \qquad (1)

where: $x(t)$ — position along the chosen axis, signed; $v(t)$ — instantaneous velocity, also signed. On a position–time graph this is the slope of the tangent line: a speedometer is a slope-measuring device. Note the word instantaneous — this is not "distance over time" for the whole trip, but the limit of that ratio over an ever-shorter interval, which is precisely what a derivative is. Ask the same question of the velocity and you get the second rung of the ladder:

a(t) = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}^2x}{\mathrm{d}t^2} \qquad (2)

where: $a(t)$ — instantaneous acceleration, the rate of change of velocity; equivalently the curvature of the $x(t)$ graph. A stone in free fall has $a = -9.8\ \mathrm{m/s^2}$ at every instant of its flight, whether rising, falling, or momentarily at rest at the top — the definition asks how $v$ is changing, not what $v$ is. These definitions also run backwards. Given the acceleration history, the fundamental theorem of calculus reconstructs the motion, one integral per rung:

v(t) = v_0 + \int_0^t a(t')\,\mathrm{d}t', \qquad x(t) = x_0 + \int_0^t v(t')\,\mathrm{d}t' \qquad (3)

where: $v_0,\ x_0$ — the initial velocity and position, the two constants of integration that the definitions alone cannot supply; $t'$ — a dummy integration variable. This pair is the master version of everything below: it works for any $a(t)$ , and it is literally what the accelerometer chip in your phone computes when it reconstructs how you moved. The SUVAT equations are nothing but equation (3) evaluated in the one case where the integrals are trivial.

The four equations

Fix a positive direction along the line of motion, and give names to the five quantities in play: $v_0$ is the velocity at the moment you start the clock, $v$ the velocity a time $t$ later, $a$ the constant acceleration, and $s = x - x_0$ the displacement accumulated in between. Every constant-acceleration problem you will ever meet is a story about these five numbers, and each of the four equations below links four of them while leaving a different one out. That is the whole organizing principle: identify which quantity the problem never mentions, and the equation missing that quantity is the one to use. Start with the question "how fast?". If velocity changes at a steady rate $a$ , then after $t$ seconds it has changed by exactly $at$ — that is what "steady rate" means. Stack that change on top of the starting velocity and you have the first equation:

v = v_0 + at \qquad (4)

where: $v$ — final velocity; $v_0$ — initial velocity; $a$ — acceleration, constant by assumption; $t$ — elapsed time. Read it as a straight line, $y = c + mx$ in kinematic clothing: intercept $v_0$ , slope $a$ . A car doing $20\ \mathrm{m/s}$ that gains $3\ \mathrm{m/s}$ every second is doing $20 + 3t$ — equation (4) is nothing deeper than that sentence written in symbols. It contains no $s$ : it answers "how fast?" while saying nothing about "how far?". "How far?" needs its own equation. Displacement is the area under the velocity–time line (that is the one picture behind this whole article), and for a line starting at height $v_0$ with slope $a$ that area splits into a rectangle and a triangle — the distance you would have coasted at constant $v_0$ , plus the extra bought (or lost) by the acceleration:

s = v_0 t + \tfrac12 a t^2 \qquad (5)

where: $s$ — displacement, signed like everything else; $v_0 t$ — the coasting rectangle; $\tfrac12 at^2$ — the acceleration triangle. The famous $\tfrac12$ is not decoration: the velocity gain $at$ arrives gradually, so on average only half of it was available during the trip. Drop a stone ( $v_0 = 0$ ) and (5) collapses to Galileo's $s = \tfrac12 g t^2$ . This equation never mentions the final velocity $v$ — it is the tool for "how far in how long?" questions. Sometimes a problem never mentions time at all: "a car brakes from $40\ \mathrm{m/s}$ and stops in $200\ \mathrm{m}$ — what was the deceleration?" You could find $t$ from (4), feed it into (5), and grind through the algebra. But that elimination can be done once and for all, and the clock cancels out, leaving a direct bridge between speed and distance:

v^2 = v_0^2 + 2as \qquad (6)

where: every symbol as before — and no $t$ anywhere. This is the time-free equation, the one to reach for whenever the clock was never mentioned. It is also the reason stopping distance grows with the square of speed: set $v = 0$ and $s = v_0^2/2a$ — double the speed, quadruple the road. Equation (6) is the overview's braking law, derived rather than decreed; multiply it by $\tfrac12 m$ and you are staring at the work–energy theorem a few chapters early. The fourth equation is the humblest and the oldest. Because the velocity graph is a straight line, the average velocity over the trip is just the midpoint of the endpoints — no calculus, pure symmetry. Distance is average velocity times time:

s = \tfrac12\,(v_0 + v)\,t \qquad (7)

where: $\tfrac12(v_0+v)$ — the average velocity, equal to the endpoint mean only because the line is straight (the medieval mean speed theorem, proved by the Oxford Calculators three centuries before Galileo needed it). Equation (7) leaves out $a$ : perfect for problems that give you two speeds and a time but never mention acceleration. A fifth form, $s = vt - \tfrac12 at^2$ , follows by writing the trapezoid from the far end; it earns its keep when $v_0$ is the unknown. One organizing table, then, before the derivations — each equation and the quantity it omits:

equation	omits	reach for it when
$v = v_0 + at$	$s$	distance is irrelevant
$s = v_0t + \tfrac12 at^2$	$v$	final speed is irrelevant
$v^2 = v_0^2 + 2as$	$t$	time is never mentioned
$s = \tfrac12(v_0+v)t$	$a$	acceleration is never mentioned

Derivation AThe four SUVAT equations from one straight line

We are deriving all four constant-acceleration equations, (4) through (7), and the goal is to watch them fall out of a single geometric object — the straight velocity–time line — rather than to memorize them as four separate facts. The strategy: first read the slope and the average off the line, which hands us (4) and (7) almost for free; then combine those two in the two possible ways, once eliminating $v$ to get the position–time equation (5), and once eliminating $t$ to get the time-free equation (6). Everything hinges on the line being straight, which is exactly the statement $a = \text{constant}$ — watch how the endpoint-average step would collapse the moment the line bends. The same results follow from integrating equation (3) with $a$ constant; the geometry and the calculus are two tellings of one story.

Step 1. Constant acceleration means the rate of change of velocity is the same over any interval, so $a = \dfrac{v - v_0}{t}$ . Multiply through by $t$ and rearrange: $v = v_0 + at$ , which is (4). Why: when the rate of change is uniform, "rate" is just total change over total time — the $v$ – $t$ line has intercept $v_0$ and slope $a$ .

Step 2. Because $v(t)$ is a straight line, its average over the interval is the mean of the endpoints, $\bar v = \tfrac12(v_0 + v)$ , and displacement is average velocity times time: $s = \tfrac12(v_0+v)\,t$ , which is (7). Why: a linear function's average over an interval is its midpoint value — the trapezoid has the same area as a rectangle of its average height. This is the mean speed theorem, proved geometrically by the Oxford Calculators three centuries before Galileo used it.

Step 3. Substitute $v = v_0 + at$ from (4) into (7):

s = \tfrac12\big(v_0 + v_0 + at\big)\,t = v_0 t + \tfrac12 a t^2

Why: this splits the trapezoid into the green rectangle $v_0 t$ and the amber triangle $\tfrac12(at)(t)$ of the hero diagram — coasting distance plus the acceleration's contribution. This is (5).

Step 4. Solve (4) for time instead, $t = (v - v_0)/a$ , and substitute into (7):

s = \frac{(v_0+v)}{2}\cdot\frac{(v-v_0)}{a} = \frac{v^2 - v_0^2}{2a}

Why: to build a time-free relation we must spend equation (4) eliminating $t$ rather than $v$ ; the product $(v+v_0)(v-v_0)$ collapses by the difference of squares.

Step 5. Rearrange: $v^2 = v_0^2 + 2as$ , which is (6). Why: slope and area have been combined so that the clock disappears — geometry traded for history. This is the equation behind every stopping-distance calculation on every road in the world. *Two quick checks: differentiating (5) returns (4) and a second derivative equal to $a$ , confirming internal consistency; multiplying (6) by $\tfrac12 m$ gives $\tfrac12 mv^2 = \tfrac12 mv_0^2 + (ma)s$ , the work–energy theorem in embryo.*

Free fall — the flagship application

The most important constant acceleration in nature is the one under your feet. Near Earth's surface, every body dense enough for air resistance not to matter falls with the same downward acceleration, $g \approx 9.8\ \mathrm{m/s^2}$ , independent of its mass — a hammer and a feather, released together on the airless Moon in 1971, hit the dust in the same instant. Choose up as positive and free fall is simply the SUVAT machinery running with $a = -g$ : one sign convention, fixed once, handles rising, falling, and the turnaround in a single sweep. From rest, equation (5) says the distance fallen grows as the square of time: after 1, 2, 3, 4 seconds the totals stand at 1, 4, 9, 16 units. The distances covered in each successive second are therefore 1, 3, 5, 7 units — the odd numbers. Galileo discovered this signature experimentally, rolling bronze balls down inclines to dilute gravity enough for a water clock to catch it, and it remains the cleanest laboratory test that an acceleration really is constant.

Figure

Galileo's odd-number rule — free fall from rest, photographed second by second

Throw a ball straight up instead of dropping it, and the two most-asked questions — how high, how long — have answers compact enough to derive completely:

Go deeper · Derivation B · 2 minFlight time and maximum height of a vertical throw

Derivation BFlight time and maximum height of a vertical throw

The goal is the pair of results every vertical-throw problem reduces to: a ball launched straight up at speed $v_0$ rises to $h_{\max} = v_0^2/2g$ and returns to its launch height after a total flight time $t_{\text{flight}} = 2v_0/g$ . The strategy is to exploit the one moment in the flight we know something about without solving anything — the top, where the vertical velocity passes through zero — and to let each SUVAT equation contribute the piece it is best shaped for: the velocity equation (4) for the timing, the time-free equation (6) for the height. Watch the sign convention do all the work: up is positive, so $a = -g$ from launch to landing, with no case-splitting between the rising and falling halves.

Step 1. Set up with up positive: $a = -g$ , launch from $x_0 = 0$ with velocity $+v_0$ . Why one convention suffices: gravity does not care which way the ball is moving — the acceleration is $-g$ while rising, at the top, and while falling, so a single pair of equations covers the whole flight.

Step 2. At the highest point the velocity passes through zero. Equation (4) gives the time to get there: $0 = v_0 - g\,t_{\text{up}}$ , so $t_{\text{up}} = v_0/g$ . Why $v = 0$ at the top: an instant earlier the ball still moves up, an instant later it moves down; a continuously changing velocity must pass through zero in between — though the acceleration there is still $-g$ , not zero.

Step 3. The height of that point follows from the time-free equation (6) with final velocity zero:

0 = v_0^2 - 2g\,h_{\max} \;\;\Rightarrow\;\; h_{\max} = \frac{v_0^2}{2g}

Why reach for (6): the question "how high?" never mentions time, so the equation that omits $t$ goes straight to the answer — no need to compute $t_{\text{up}}$ first.

Step 4. For the total flight time, ask when the ball returns to $x = 0$ . Equation (5) gives $0 = v_0 t - \tfrac12 g t^2 = t\,(v_0 - \tfrac12 g t)$ , whose nonzero root is

t_{\text{flight}} = \frac{2v_0}{g} = 2\,t_{\text{up}}

Why the factor of exactly 2: the motion is time-symmetric — the descent retraces the ascent in reverse, so the ball spends as long falling back as it spent rising, and it lands at speed $v_0$ downward. *Two checks: for $v_0 = 10\ \mathrm{m/s}$ , the formulas give $h_{\max} = 100/(2 \times 9.8) \approx 5.1\ \mathrm{m}$ and $t_{\text{flight}} \approx 2.04\ \mathrm{s}$ ; and differentiating $x(t) = v_0 t - \tfrac12 gt^2$ confirms $v = 0$ at $t = v_0/g$ , as expected.*

Beyond constant acceleration

The SUVAT equations are a special case, and honest problems eventually leave it: air drag makes $a$ depend on velocity, a spring makes it depend on position. The definitions (1)–(3) still hold — you simply have to integrate whatever $a$ you are given. When the acceleration is handed to you as a function of position, though, the time integrals in (3) jam: you cannot integrate $a(x)$ over $t$ without already knowing $x(t)$ . The escape is a one-line change of variable that every later chapter of mechanics quietly relies on:

a\,\mathrm{d}x = v\,\mathrm{d}v \qquad (8)

where: $a$ — acceleration, here possibly a function of position; $v\,\mathrm{d}v$ — the differential of $\tfrac12 v^2$ . This identity trades the clock for geometry: it yields the speed at each position without ever solving for the schedule $t \mapsto x(t)$ . It is the kinematic seed of energy conservation — integrate both sides and the right side becomes the change in $\tfrac12 v^2$ .

Go deeper · Derivation C · 2 minTrading time for position — the chain-rule identity at work

Derivation CTrading time for position — the chain-rule identity at work

We are deriving the identity (8), $a\,\mathrm{d}x = v\,\mathrm{d}v$ , and then putting it to work on the most important position-dependent acceleration in physics: the linear restoring law $a = -\omega^2 x$ of a mass on a spring. The strategy is pure change of variables — acceleration is defined as a time derivative, but the chain rule lets us route that derivative through position, and the resulting differential relation separates cleanly whenever $a$ is a function of $x$ alone. The payoff is a speed–position relation obtained without ever finding $x(t)$ , which is fortunate: for the spring the full schedule $x(t)$ is a sine curve, yet the speed-versus-position answer is a one-line integral. Watch what the result turns out to be — a conserved quantity, discovered by kinematics alone, several chapters before anyone says the word "energy."

Step 1. Start from the definition and apply the chain rule, differentiating through the position:

a = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}v}{\mathrm{d}x}\,\frac{\mathrm{d}x}{\mathrm{d}t} = v\,\frac{\mathrm{d}v}{\mathrm{d}x}

Why this is legal: along a moving trajectory the velocity can be regarded as a function of position, $v(x(t))$ , wherever the motion does not reverse; the chain rule then splits the time derivative into a position derivative times $\mathrm{d}x/\mathrm{d}t = v$ .

Step 2. Rearrange into differential form: $a\,\mathrm{d}x = v\,\mathrm{d}v$ , which is (8). Why it is useful: if $a = a(x)$ , the left side involves only $x$ and the right side only $v$ — the variables have separated, and each side integrates on its own.

Step 3. Apply it to the restoring law $a = -\omega^2 x$ : substitute and integrate both sides from the initial state $(x_0, v_0)$ to a general state $(x, v)$ :

\int_{v_0}^{v} v'\,\mathrm{d}v' = -\,\omega^2 \int_{x_0}^{x} x'\,\mathrm{d}x'

Why these limits: both sides track the same stretch of motion, so the integration bounds must correspond — start together, end together.

Step 4. Evaluate the elementary integrals and multiply by 2:

v^2 = v_0^2 - \omega^2\,(x^2 - x_0^2)

where: $\omega$ — the oscillator's angular frequency; the equation delivers the speed at every position of the oscillation, fastest at the center and zero at the turning points. Why this is a conservation law: rearranged, $v^2 + \omega^2 x^2$ keeps the same value throughout the motion — kinematics has discovered that $\tfrac12 v^2 + \tfrac12\omega^2 x^2$ is conserved, the oscillator's energy per unit mass in disguise. *Two checks: differentiating $v^2 + \omega^2 x^2 = \mathrm{const}$ with respect to $t$ gives $2v\dot v + 2\omega^2 x v = 0$ , i.e. $a = -\omega^2 x$ , recovering the starting law; and setting $a = \mathrm{const}$ in (8) reproduces SUVAT III, $v^2 = v_0^2 + 2as$ , confirming that the identity contains the constant-acceleration case.*

Limiting cases

Case	Result	Meaning
$a = 0$	$v = v_0$ , $s = v_0 t$	uniform motion — the trapezoid flattens into a rectangle; the kinematic content of Newton's first law
$v_0 = 0$ (start from rest)	$s = \tfrac12 at^2$ , $v = at$	pure triangle; distance $\propto t^2$ , the odd-number regime
$v = 0$ (braking to rest)	$s = v_0^2 / 2\lvert a\rvert$	stopping distance — quadratic in speed, the road-safety law
$a = -g$ , up positive	$h_{\max} = v_0^2/2g$ , $t_{\text{flight}} = 2v_0/g$	a vertical throw: rise and fall are time-symmetric
$a = a(x)$	$v\,\mathrm{d}v = a(x)\,\mathrm{d}x$	SUVAT retired; the chain-rule identity gives $v(x)$ — the gateway to energy methods and oscillations
real air, long falls	constant- $a$ algebra fails	drag makes $a$ velocity-dependent; integrate the actual $a(v)$ instead

Common trap“Average velocity is always (v₀ + v)/2”

The endpoint mean is a constant-acceleration privilege, not a law of nature. Try it on a trip where you idle at rest for 9 s and then cruise at 10 m/s for 1 s: true average velocity is $10\ \mathrm{m}/10\ \mathrm{s} = 1\ \mathrm{m/s}$ , but the endpoint mean says $5\ \mathrm{m/s}$ — off by a factor of five. The formula works only when velocity changes linearly, so the time spent near each velocity is evenly distributed. The same fine print covers all four equations: apply them across a bounce, a gear change, or any interval where $a$ jumps, and the answers are meaningless — split the motion at the jump and treat each constant- $a$ leg separately. And a second signed-number trap rides along: velocity and displacement are signed while speed and distance are not, so an out-and-back trip has zero displacement and zero average velocity but a perfectly nonzero average speed. Keep the two vocabularies apart and half the "paradoxes" of kinematics evaporate.

Worked examples

1 · Braking to rest (confidence builder)

A car travels at $v_0 = 20\ \mathrm{m/s}$ when the driver brakes with constant deceleration of magnitude $4.0\ \mathrm{m/s^2}$ . Find the stopping time and the stopping distance. Choose the direction of motion as positive, so $a = -4.0\ \mathrm{m/s^2}$ and the car stops when $v = 0$ .

Step 1. Stopping time from (4): $0 = 20 + (-4.0)\,t$ , so $t = 20/4.0 = 5.0\ \mathrm{s}$ .

Step 2. Stopping distance from the time-free (6): $0 = 20^2 + 2(-4.0)\,s$ , so $s = 400/8.0 = \boxed{50\ \mathrm{m}}$ in $\boxed{5.0\ \mathrm{s}}$ .

Step 3. Cross-check with (7): $s = \tfrac12(20 + 0)(5.0) = 50\ \mathrm{m}$ ✓ — average velocity 10 m/s for 5 s.

Step 4. Read the scaling. At $40\ \mathrm{m/s}$ the same brakes give $s = 1600/8.0 = 200\ \mathrm{m}$ : doubling speed quadruples the distance, because (6) puts $v_0$ in squared.

2 · Thrown up from a cliff edge (exam level)

A stone is thrown straight up at $15\ \mathrm{m/s}$ from the edge of a cliff $30\ \mathrm{m}$ above a beach. Find the maximum height above the launch point, the total time until it hits the beach, and the impact speed. (Take $g = 9.8\ \mathrm{m/s^2}$ , up positive, so $a = -9.8\ \mathrm{m/s^2}$ .)

Step 1. Maximum height from (6), with $v = 0$ at the top: $0 = 15^2 + 2(-9.8)\,s$ , so $s = 225/19.6 = 11.5\ \mathrm{m}$ above the launch point.

Step 2. Total flight time from (5), with final displacement $s = -30\ \mathrm{m}$ (the beach is below the launch): $-30 = 15t - 4.9t^2$ , i.e. $4.9t^2 - 15t - 30 = 0$ .

Step 3. Quadratic formula: $t = \dfrac{15 + \sqrt{15^2 + 4(4.9)(30)}}{2(4.9)} = \dfrac{15 + \sqrt{813}}{9.8} = \dfrac{15 + 28.5}{9.8} = 4.44\ \mathrm{s}$ . The negative root is a fictitious pre-launch time and is discarded.

Step 4. Impact velocity from (4): $v = 15 - 9.8(4.44) = -28.5\ \mathrm{m/s}$ — the sign says "downward," the magnitude is the speed.

Step 5. Cross-check with the time-free (6): $v^2 = 15^2 + 2(-9.8)(-30) = 225 + 588 = 813$ , so $\lvert v\rvert = \sqrt{813} = 28.5\ \mathrm{m/s}$ ✓.

Answers: $h_{\max} = 11.5\ \mathrm{m}$ above the launch, $t = 4.44\ \mathrm{s}$ , impact speed $28.5\ \mathrm{m/s}$ . Notice one set of signed equations handled up, over, and down in a single sweep — no need to split the flight into pieces.

3 · The well and the echo (exam level, chained motions)

A stone is dropped into a well. The splash is heard $3.00\ \mathrm{s}$ after release. Taking the speed of sound as $343\ \mathrm{m/s}$ , find the depth of the well — and the error made by ignoring the sound's travel time.

Step 1. Two legs share the 3.00 s: the stone falls for $t_1$ , then the sound climbs back for $t_2$ , with $t_1 + t_2 = 3.00$ .

Step 2. Express the same depth both ways: falling from rest, $d = \tfrac12 g t_1^2 = 4.9\,t_1^2$ ; the returning sound gives $d = 343\,t_2 = 343\,(3.00 - t_1)$ .

Step 3. Equate: $4.9\,t_1^2 = 343(3.00 - t_1)$ , i.e. $4.9\,t_1^2 + 343\,t_1 - 1029 = 0$ .

Step 4. Solve the quadratic: $t_1 = \dfrac{-343 + \sqrt{343^2 + 4(4.9)(1029)}}{2(4.9)} = \dfrac{-343 + \sqrt{137817}}{9.8} = \dfrac{-343 + 371.2}{9.8} = 2.88\ \mathrm{s}$ .

Step 5. Depth: $d = 4.9 \times (2.88)^2 = 4.9 \times 8.30 = \boxed{40.7\ \mathrm{m}}$ . Cross-check the sound leg: $t_2 = 40.7/343 = 0.119\ \mathrm{s}$ , and $2.88 + 0.12 = 3.00$ ✓.

Step 6. Ignoring the sound's travel time would use the full 3.00 s for the fall: $d = 4.9 \times 9.00 = 44.1\ \mathrm{m}$ — an overestimate of $3.4\ \mathrm{m}$ , about 8%. The physics point: the measurement is a chain of two motions, and the fast leg still matters at the few-percent level.

Origins · 1 minGalileo Galilei and how this was discovered

Galileo Galilei

1564–1642 · Pisa, Padua, Florence

Galileo built uniform-acceleration kinematics with no algebra, no calculus, and no clock better than dripping water. His move was to dilute gravity: bronze balls rolling down gently inclined grooves accelerate slowly enough to time, yet obey the same law as free fall. Measuring distances in successive equal intervals, he found the odd-number signature 1 : 3 : 5 : 7 and inferred that distance from rest grows as the square of time — publishing the full theory of "naturally accelerated motion" in Two New Sciences (1638), written while under house arrest. He also insisted, against two thousand years of Aristotle, that heavy and light bodies share the same acceleration once air is taken out of the story. Every equation on this page is his geometry, translated into symbols he never had.

History

1335

— William Heytesbury and the Oxford Calculators at Merton College state the mean speed theorem: a uniformly accelerating body covers the same ground as one moving steadily at the endpoint average.

1350

— Nicole Oresme in Paris proves it geometrically, drawing velocity against time and reading distance as area — the hero diagram of this article, six centuries early.

1604

— Galileo, in a letter to Paolo Sarpi, states that distance in free fall grows as the square of time.

1638

— Two New Sciences publishes the inclined-plane experiments, the odd-number rule, and the full theory of uniformly accelerated motion.

1687

— Newton's Principia explains why the acceleration is constant: a constant force, via

F = ma

, produces it — kinematics acquires its dynamical engine.

Going further

Texts. Morin, Introduction to Classical Mechanics, Ch. 2 · Taylor, Classical Mechanics, Ch. 1 · Kleppner & Kolenkow, An Introduction to Mechanics, Ch. 1 · Landau & Lifshitz, Mechanics (Vol. 1), §1–4 · Goldstein, Classical Mechanics, Ch. 1. SUVAT III follows by eliminating $t$ between SUVAT I and II. The free-fall formulas give $h_{\max}=v_0^2/2g\approx5.1\,\mathrm{m}$ and $t_{\text{flight}}=2v_0/g\approx2.04\,\mathrm{s}$ for $v_0=10\,\mathrm{m/s}$ , $g=9.81\,\mathrm{m/s^2}$ ; the SHM energy relation $v^2+\omega^2 x^2=\mathrm{const}$ is confirmed by differentiating to recover $a=-\omega^2 x$ . The value $g\approx9.81\,\mathrm{m/s^2}$ is the standard near-surface gravitational acceleration. Worked-example numbers on this page use $g = 9.8\ \mathrm{m/s^2}$ : the braking results ( $t = 5.0$ s, $s = 50$ m from 20 m/s; 200 m from 40 m/s), the cliff throw ( $h_{\max} = 11.5$ m, $t = 4.44$ s from $4.9t^2 - 15t - 30 = 0$ , impact speed $\sqrt{813} = 28.5$ m/s), the well depth ( $40.7$ m with $t_1 = 2.88$ s, $t_2 = 0.119$ s), and the bar-chart braking distances $v_0^2/2a$ at $a = 6\ \mathrm{m/s^2}$ (5.8, 23.1, 52.1, 92.6 m for 30/60/90/120 km/h).

Builds on: Position, displacement, and reference frames · Differential and integral calculus. Leads to: Velocity and acceleration · Projectile motion · Vectors and coordinate systems in kinematics · Newton's second law · Gravitational force near Earth's surface.

Why can't I use the SUVAT equations when air resistance matters?

Because air resistance makes the acceleration depend on velocity — the faster you fall, the harder the drag pushes back — so $a$ is no longer constant, and "constant" is the entire license behind all four equations. Every one of them was derived from a straight velocity–time line; drag bends that line, flattening it toward terminal velocity, and the trapezoid formulas silently compute the area of the wrong shape. The definitions themselves survive perfectly well: $v = \mathrm{d}x/\mathrm{d}t$ and $a = \mathrm{d}v/\mathrm{d}t$ hold for any motion. You just have to integrate the actual $a(v)$ — which is exactly what equation (3) is for. Rule of thumb: dense, slow, short falls are safely SUVAT; feathers, parachutes, and long falls are not.

At the top of a throw the velocity is zero — why isn't the acceleration zero too?

Because acceleration measures how velocity is changing, not what it is. An instant before the top the ball moves up; an instant after, down. The velocity is passing through zero at a perfectly steady rate of $9.8\ \mathrm{m/s}$ per second — which is precisely a nonzero acceleration. If the acceleration actually hit zero at the top, the velocity would stop changing there, and a ball with $v = 0$ and $a = 0$ would simply hang in the air forever. The graph makes it obvious: the $v$ – $t$ line for a throw is one unbroken straight line of slope $-g$ that happens to cross zero at the apex — the line's height is zero there, but its slope never changes.

My average velocity came out zero even though I was moving the whole time. What went wrong?

Nothing — you have just met the difference between displacement and distance. Average velocity is defined as displacement over time, and displacement is the signed change in position: end where you started, and it is zero no matter how far you wandered. Your intuition was tracking average speed, which is total distance over time and is never zero for genuine motion. The two vocabularies answer different questions: "where did you end up?" versus "how much ground did you cover?". On a velocity–time graph the distinction is visual: areas below the axis subtract from displacement but add to distance.

Where does the ½ in s = v₀t + ½at² actually come from?

From the fact that the velocity gain arrives gradually rather than all at once. Over a time $t$ the acceleration adds $at$ to your velocity, but you did not have that extra $at$ for the whole trip — at the start you had none of it, at the end all of it, and on average exactly half. So the extra distance is (average extra velocity) × (time) $= \tfrac12 at \times t = \tfrac12 at^2$ . Geometrically it is the area of the triangle on the hero diagram: base $t$ , height $at$ , area $\tfrac12(at)(t)$ . Calculus says the same thing a third way: $\int_0^t at'\,\mathrm{d}t' = \tfrac12 at^2$ . Three languages, one factor of one-half.

Is v² = v₀² + 2as secretly energy conservation?

Very nearly — it is energy conservation before anyone has defined energy. Multiply through by $\tfrac12 m$ : $\tfrac12 mv^2 = \tfrac12 mv_0^2 + (ma)s$ . The left side is the final kinetic energy, the first term on the right the initial kinetic energy, and $(ma)s = Fs$ is the work done by the constant net force. That is the work–energy theorem, derived here by pure kinematics. This is also why the equation has no clock in it: energy accounting never cares how long something took, only between which two states it happened. The chain-rule identity $a\,\mathrm{d}x = v\,\mathrm{d}v$ generalizes the same bookkeeping to forces that vary with position — integrate it and you get energy methods for springs, gravity at altitude, and everything beyond.

Does a negative acceleration always mean the object is slowing down?

No — and this misreading causes more lost marks than any other sign error. Negative acceleration means the acceleration points along the negative axis, nothing more. Whether the object speeds up or slows down depends on the relative signs of $a$ and $v$ : opposite signs mean slowing (a car braking while moving in the positive direction), same signs mean speeding up (a ball already falling, with up positive, has $v < 0$ and $a < 0$ and gains speed every second). "Deceleration" is everyday language, not physics: the falling ball has negative acceleration and is emphatically not decelerating. Fix your axis first, then read signs against it — the words will take care of themselves.

When should I reach for a dx = v dv instead of the SUVAT set?

Whenever the acceleration is handed to you as a function of position — a spring's $a = -\omega^2 x$ , gravity varying with altitude, a magnetic brake that weakens with distance. SUVAT is dead there (it demands constant $a$ ), and the time integrals of equation (3) jam because you cannot integrate $a(x)$ over $t$ without already knowing $x(t)$ . The identity $a\,\mathrm{d}x = v\,\mathrm{d}v$ sidesteps the clock entirely: both sides separate, each integrates on its own, and out comes the speed at every position. You lose the schedule — it never tells you when the object reaches a given point — but for turning points, maximum speeds, and escape conditions, position is all you need. As a bonus, setting $a$ constant in it reproduces SUVAT III, so there is nothing to unlearn.

How did Galileo verify constant acceleration without a stopwatch?

By slowing gravity down and counting instead of timing. A ball rolling down a gentle incline is still driven by gravity alone — just by the diluted component along the slope — so the motion unfolds leisurely enough for a water clock (weighing the water that dripped during each interval) to resolve. Galileo marked the ball's position at equal ticks and found the spacings grew as 1 : 3 : 5 : 7 — the odd numbers. That pattern is the fingerprint of constant acceleration: totals growing as $1, 4, 9, 16$ , i.e. as $t^2$ , force the interval-by-interval differences to be successive odd numbers. The same signature appeared at every tilt of the ramp, so he extrapolated to the vertical: free fall is uniformly accelerated motion. It remains the cleanest way to test for constant acceleration with primitive equipment.

One-Dimensional Kinematics

Where this lives today

The physical picture

The language of motion

The four equations

Free fall — the flagship application

Beyond constant acceleration

Limiting cases

Worked examples

1 · Braking to rest (confidence builder)

2 · Thrown up from a cliff edge (exam level)

3 · The well and the echo (exam level, chained motions)

Going further

Before this

Where this leads