Kinematic Equations for Uniform Acceleration

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 tyres or reflexes; it falls straight out of a short family of algebraic equations that govern every motion whose acceleration holds constant. A dropped stone, a sprinter out of the blocks, a lift settling to a floor, a jet on its take-off roll — one toolkit covers them all, and it is the single most-used toolkit in your entire physics course. Five symbols run the show — displacement $s$ , initial velocity $u$ , final velocity $v$ , acceleration $a$ , time $t$ — which is why teachers call these the suvat equations. They weave the five together so tightly that knowing any three hands you the other two. And here is the open secret of exam kinematics: the equations are the easy part. What separates a 40-second solve from a five-minute struggle is selection — spotting which quantity is never mentioned and reaching directly for the one equation that omits it. This module teaches the equations, their derivations, and that selection craft as one package. Best of all, every equation here is one picture — the area under a straight velocity–time line — wearing a different algebraic costume. See the connection once and you will never need to memorise the equations again; you will be able to rebuild any of them in the margin of your exam paper.

Key takeaways

For straight-line motion with constant acceleration, five quantities — $s$ , $u$ , $v$ , $a$ , $t$ — are linked by the suvat equations; any three known values determine the rest.
The core three: $v = u + at$ · $s = ut + \tfrac12 at^2$ · $v^2 = u^2 + 2as$ , plus the often-forgotten fourth $s = \tfrac12(u+v)\,t$ .
Selection rule: each equation is missing exactly one of the five symbols — pick the equation missing the quantity you neither know nor want.
All of them are one geometric fact in disguise: displacement is the trapezium-shaped area under a straight $v$ – $t$ line.
Everything is signed along one chosen axis: "deceleration" means $a$ opposes $v$ ; a vertical throw is $a = -g$ with up positive — one sign convention, held to the last line.
Key idea: constant acceleration turns calculus into algebra — one straight line on the v–t graph, and its slope and area are the whole story.

The core insight

When acceleration is constant, velocity changes linearly with time, so the $v$ – $t$ graph is a straight line; the kinematic equations are nothing but different ways of writing the slope of that line and the area beneath it — and choosing between them is a one-step game of "which symbol is missing?".

Where this lives today

The square law of equation (3) is written into the road network. Highway engineers set speed limits, sight distances and escape-ramp lengths from $s = u^2/2a$ with measured tyre–road decelerations. The same arithmetic runs live inside modern cars: automatic emergency braking continuously compares the distance to the obstacle ahead with the computed stopping distance at the current speed, and fires the brakes the moment the two curves are about to cross. The equations also run in reverse: a lift's ride profile (accelerate, cruise, decelerate) is designed as three suvat legs stitched together — the same multi-leg decomposition you use on two-phase exam problems, running daily in every elevator controller.

The physical picture

Plot velocity against time for a car under steady braking, a ball in free fall, or a metro 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 — the two readings you mastered in the velocity–time graphs module. The suvat equations do not add any new physics to that picture; they are what happens when you compute the same trapezium-shaped area in different ways and let the algebra simplify.

Figure

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

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 trapezium formulas are dead on arrival. "Constant acceleration" is not decoration; it is the entire licence. Check it before writing a single symbol. One more picture before the algebra, because it is the payoff the overview promised. Draw braking distances to scale 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 equations and the selection game

Fix a positive direction along the line of motion and name the five quantities: $u$ is the velocity when you start the clock, $v$ the velocity a time $t$ later, $a$ the constant acceleration, $s$ the displacement accumulated in between. (AP and IB textbooks write $v_0$ and $x$ where Indian and British boards write $u$ and $s$ — same physics, different costume.) Each equation below links four of the five while leaving a different one out; that omission is the whole organising principle of fast problem-solving. Start with "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 the starting velocity:

v = u + at \qquad (1)

where: $v$ — final velocity; $u$ — initial velocity; $a$ — acceleration, constant by assumption; $t$ — elapsed time. Read it as a straight line, $y = c + mx$ in kinematic clothing: intercept $u$ , slope $a$ . A scooter doing $20\ \mathrm{m/s}$ that gains $3\ \mathrm{m/s}$ every second is doing $20 + 3t$ — equation (1) is nothing deeper than that sentence in symbols. It contains no $s$ : it answers "how fast?" while saying nothing about "how far?". "How far?" is the area question. For a line starting at height $u$ with slope $a$ , the area under the $v$ – $t$ graph splits into a rectangle and a triangle — the distance you would have coasted at constant $u$ , plus the extra bought (or lost) by the acceleration:

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

where: $s$ — displacement, signed like everything else; $ut$ — 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 ( $u = 0$ ) and (2) collapses to Galileo's $s = \tfrac12 g t^2$ . No $v$ appears — this 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 $200\ \mathrm{m}$ — find the deceleration." You could extract $t$ from (1) and grind it through (2), but that elimination can be done once and for all, leaving a direct bridge between speed and distance:

v^2 = u^2 + 2as \qquad (3)

where: every symbol as before — and no $t$ anywhere. This is the time-free equation, the one to reach for whenever the clock is never mentioned. It is also the reason stopping distance grows with the square of speed: set $v = 0$ and $s = u^2/2a$ — double the speed, quadruple the road. And multiplied by $\tfrac12 m$ it becomes the work–energy theorem in embryo, which is why examiners love it. The fourth equation is the humblest and the oldest. Because the $v$ – $t$ graph is straight, the average velocity over the trip is just the midpoint of the endpoints — pure symmetry, no calculus:

s = \tfrac12\,(u + v)\,t \qquad (4)

where: $\tfrac12(u+v)$ — the average velocity, equal to the endpoint mean only because the line is straight (the medieval mean speed theorem). Equation (4) omits $a$ : perfect when a problem gives two speeds and a time. Students forget it exists and waste minutes solving (1) and (2) simultaneously for what (4) hands over in one line. Competitive exams add a fifth family member — the displacement covered in the $n$ th second alone (not in $n$ seconds):

s_{n\text{th}} = u + \frac{a}{2}\,(2n-1) \qquad (5)

where: $s_{n\text{th}}$ — the displacement between the instants $t = n-1$ and $t = n$ (the 4th second means from $t = 3$ s to $t = 4$ s). Yes, the right-hand side looks like a velocity — that famous puzzle is resolved in the derivation below and in the FAQ. JEE-Main and NEET recycle this relation constantly because it filters students who memorised (2) from students who can subtract two copies of it. Here is the selection table — the centre of gravity of this whole module. Do not memorise it; use it until it becomes reflex:

equation	missing symbol	reach for it when
$v = u + at$	$s$	distance is neither given nor wanted
$s = ut + \tfrac12 at^2$	$v$	final speed is neither given nor wanted
$v^2 = u^2 + 2as$	$t$	time is never mentioned
$s = \tfrac12(u+v)t$	$a$	acceleration is never mentioned
$s_{n\text{th}} = u + \tfrac{a}{2}(2n-1)$	—	the question says "in the $n$ th second"

The drill: list $s, u, v, a, t$ in the margin, fill in the three you know, mark the one you want — the symbol left blank chooses the equation for you. One equation, one substitution, done.

Derivation ASlope and average velocity give the first two equations

We are deriving $v = u + at$ and $s = \tfrac12(u+v)t$ from nothing but the definition of constant acceleration and one property of straight lines. The strategy: write the definition of acceleration as a slope, rearrange, then use the fact that a linear function's average is the mean of its endpoint values. Watch how little machinery is needed — this is why you can rebuild these equations in an exam margin.

Step 1. Constant acceleration means the rate of change of velocity is uniform, so over any interval $a = \dfrac{v - u}{t}$ . Why this is legitimate: when a rate is constant, "rate" is simply total change over total time — no limiting process is needed, which is exactly what "uniform" buys us.

Step 2. Multiply through by $t$ and rearrange: $v = u + at$ . That is equation (1) — the $v$ – $t$ line with intercept $u$ and slope $a$ .

Step 3. Because $v(t)$ is a straight line, its average over the trip is the midpoint of the endpoint values: $\bar v = \tfrac12(u + v)$ . Geometrically, the trapezium under the line has the same area as a rectangle of height $\bar v$ . Why the midpoint works: a straight line spends time evenly across its range of values — for every moment it runs below the midpoint there is a matching moment the same amount above. The instant the line bends, this symmetry dies, and so does the formula.

Step 4. Displacement is average velocity times time: $s = \bar v\,t = \tfrac12(u+v)\,t$ — equation (4), the mean speed theorem of the medieval Oxford Calculators. Verified: with $u = 20$ , $a = -4$ , $t = 5$ : equation (1) gives $v = 0$ , and (4) gives $s = \tfrac12(20+0)(5) = 50$ m — matching $s = u^2/2a = 400/8 = 50$ m from the time-free route.

Go deeper · Derivation B · 1 minSubstitution and elimination give the other two costumes

Derivation BSubstitution and elimination give the other two costumes

We now manufacture $s = ut + \tfrac12 at^2$ and $v^2 = u^2 + 2as$ with no new physics — only algebra spent on equations (1) and (4). The strategy: substitute (1) into (4) to remove $v$ , then solve (1) for $t$ and substitute to remove $t$ . Each substitution deletes one symbol, which is exactly why each resulting equation has one symbol missing.

Step 1. Substitute $v = u + at$ into $s = \tfrac12(u+v)t$ : $s = \tfrac12\big(u + u + at\big)t = \tfrac12(2u + at)\,t = ut + \tfrac12 at^2$ . That is equation (2). Why the picture agrees: $ut$ is the green coasting rectangle of the hero diagram and $\tfrac12(at)(t)$ is the amber triangle — algebra and geometry telling the same story.

Step 2. To build a time-free relation, solve (1) for time instead: $t = \dfrac{v-u}{a}$ .

Step 3. Substitute into (4): $s = \dfrac{(u+v)}{2}\cdot\dfrac{(v-u)}{a} = \dfrac{v^2 - u^2}{2a}$ , since $(v+u)(v-u) = v^2 - u^2$ by the difference of squares.

Step 4. Rearrange: $v^2 = u^2 + 2as$ — equation (3). The clock has been traded away; speed and distance now talk to each other directly. Verified: multiplying (3) by $\tfrac12 m$ gives $\tfrac12 mv^2 = \tfrac12 mu^2 + (ma)s$ — the work–energy theorem in embryo, an independent physics check that the algebra is sound. Numerically: $u = 15$ , $a = -9.8$ , $s = -30$ gives $v^2 = 225 + 588 = 813$ , $\lvert v \rvert = 28.5$ m/s, matching the worked example below.

Go deeper · Derivation C · 1 minDisplacement in the nth second — the subtraction trick

Derivation CDisplacement in the nth second — the subtraction trick

We are deriving the JEE/NEET staple $s_{n\text{th}} = u + \tfrac{a}{2}(2n-1)$ : the ground covered during the $n$ th second alone. The strategy is one idea — the $n$ th second is everything covered in $n$ seconds minus everything covered in $n-1$ seconds — applied to equation (2) twice. Watch the units at the end: the resolution of the "this looks like a velocity!" puzzle is hiding in Step 3.

Step 1. Displacement in the first $n$ seconds, from (2): $s(n) = un + \tfrac12 an^2$ . Displacement in the first $n-1$ seconds: $s(n-1) = u(n-1) + \tfrac12 a(n-1)^2$ .

Step 2. Subtract: $s_{n\text{th}} = s(n) - s(n-1) = u\big[n-(n-1)\big] + \tfrac12 a\big[n^2 - (n-1)^2\big] = u + \tfrac12 a\,(2n-1)$ , using $n^2-(n-1)^2 = 2n-1$ . Why the odd numbers appear: from rest ( $u = 0$ ) the formula gives $s_{n\text{th}} = \tfrac{a}{2}(2n-1)$ — successive seconds cover distances in the ratio $1:3:5:7$ , Galileo's odd-number signature, now derived instead of observed.

Step 3. Read the units honestly. Each term multiplies a time interval of exactly one second that the subtraction has hidden: properly, $s_{n\text{th}} = \big[u + \tfrac{a}{2}(2n-1)\big] \times (1\ \mathrm{s})$ . The formula is numerically correct when $u$ and $a$ are in SI units and the interval is one second — it is a displacement wearing a velocity's clothes. Verified: $u = 10$ m/s, $a = 4$ m/s²: the formula gives 24 m for the 4th second and 32 m for the 6th; direct subtraction of (2) gives $s(4)-s(3) = 72 - 48 = 24$ m and $s(6)-s(5) = 132 - 100 = 32$ m — exact match, and these are precisely the numbers of the competitive worked example below.

Limiting cases

case	result	meaning
$a = 0$	$v = u$ , $s = ut$	uniform motion recovered — the trapezium flattens into a rectangle
$u = 0$ (from rest)	$v = at$ , $s = \tfrac12 at^2$	pure triangle; distances in successive seconds run $1:3:5:7$
$v = 0$ (braking to rest)	$s = u^2/2\lvert a\rvert$	stopping distance — quadratic in speed, the road-safety law
$a = -g$ , up positive	$h_{\max} = u^2/2g$ , flight time $2u/g$	vertical throw: rise and fall are time-symmetric
$t$ very small	$s \approx ut$	the triangle term $\tfrac12 at^2$ is second-order — acceleration needs time to matter
$a$ varies (drag, bounce)	all five equations invalid	the $v$ – $t$ line bends — split the motion or leave suvat behind

Common trapUsing suvat when the acceleration is not constant

This is the highest-scoring wrong move in kinematics. The equations are a constant-acceleration privilege: apply them across a bounce, a gear change, a rocket burning fuel, or a raindrop with air resistance, and every answer is meaningless — even if the algebra runs smoothly. The endpoint-mean rule shows the failure most nakedly: idle at rest for 9 s, then cruise at 10 m/s for 1 s. True average velocity: $10\ \mathrm{m}/10\ \mathrm{s} = 1$ m/s. The formula $\tfrac12(u+v)$ says 5 m/s — wrong by a factor of five, because velocity did not change linearly. Before writing any suvat symbol, ask: is $a$ genuinely constant over this stretch? If it jumps, split the motion at the jump, give each leg its own $u$ , $v$ , $a$ , $t$ , and carry the final velocity of one leg in as the initial velocity of the next.

Worked examples

Board level — braking with a unit conversion (CBSE/ICSE pattern)

A car is travelling at 54 km/h when the driver applies the brakes, producing a uniform deceleration of magnitude 2.5 m/s². Find (a) the time taken to stop, (b) the distance covered before stopping.

Step 1. Convert first, always: $u = 54 \times \tfrac{5}{18} = 15$ m/s. Choose the direction of motion positive, so $a = -2.5$ m/s² and the car stops when $v = 0$ .

Step 2. List the five: $u = 15$ , $v = 0$ , $a = -2.5$ , $t = ?$ , $s = ?$ . For (a), $s$ is not needed → equation (1): $0 = 15 - 2.5t$ , so $t = \boxed{6.0\ \mathrm{s}}$ .

Step 3. For (b), time is now optional — the cleanest route never touches it. Equation (3): $0 = 15^2 + 2(-2.5)s$ , so $s = 225/5 = \boxed{45\ \mathrm{m}}$ .

Step 4. Cross-check with (4): $s = \tfrac12(15+0)(6.0) = 45$ m ✓. The redundancy of the suvat family is free error-catching — use it whenever seconds permit.

Competitive level — the nth-second filter (JEE-Main pattern)

A body moving with uniform acceleration covers 24 m in the 4th second and 32 m in the 6th second of its motion. Find its initial velocity, its acceleration, and the distance covered in the first 10 seconds.

Step 1. Translate "in the $n$ th second" — this is equation (5), twice: $u + \tfrac{a}{2}(7) = 24$ and $u + \tfrac{a}{2}(11) = 32$ .

Step 2. Subtract the equations: $\tfrac{a}{2}(11-7) = 8$ , so $2a = 8$ and $a = \boxed{4\ \mathrm{m/s^2}}$ .

Step 3. Back-substitute: $u = 24 - \tfrac{4}{2}(7) = 24 - 14 = \boxed{10\ \mathrm{m/s}}$ .

Step 4. First 10 seconds is ordinary equation (2): $s = 10(10) + \tfrac12(4)(100) = 100 + 200 = \boxed{300\ \mathrm{m}}$ .

Step 5. Sanity check: 4th second directly, $s(4)-s(3) = (40+32)-(30+18) = 72 - 48 = 24$ m ✓. Timing note: 60–90 seconds. The whole question is recognising "in the 4th second" as formula (5); the two linear equations then subtract in one line. Rebuilding via $s(n) - s(n-1)$ from scratch is the safe 3-minute fallback if the formula deserts you.

Competitive level — thrown up from a cliff edge, one signed sweep (NEET/AP pattern)

A stone is thrown vertically upward at 15 m/s from the edge of a cliff 30 m above a beach. Find the maximum height above the launch point, the total time to reach the beach, and the impact speed. Take g = 9.8 m/s², up positive.

Step 1. Set $a = -9.8$ m/s² and keep one sign convention for the whole flight — no splitting into "up" and "down" phases needed.

Step 2. Maximum height: at the top $v = 0$ and time is not mentioned → equation (3): $0 = 15^2 + 2(-9.8)s$ , so $s = 225/19.6 = \boxed{11.5\ \mathrm{m}}$ above the launch point.

Step 3. Total time: the beach is below the launch, so the final displacement is $s = -30$ m. Equation (2): $-30 = 15t - 4.9t^2$ , i.e. $4.9t^2 - 15t - 30 = 0$ .

Step 4. Quadratic formula: $t = \dfrac{15 + \sqrt{225 + 588}}{9.8} = \dfrac{15 + \sqrt{813}}{9.8} = \dfrac{15 + 28.5}{9.8} = \boxed{4.44\ \mathrm{s}}$ . The negative root is a fictitious pre-launch time — discard it and say so.

Step 5. Impact: equation (1): $v = 15 - 9.8(4.44) = -28.5$ m/s — downward, speed $\boxed{28.5\ \mathrm{m/s}}$ . Cross-check with (3), signs intact: $v^2 = 15^2 + 2(-9.8)(-30) = 813$ ✓. Timing note: about 2 minutes, and almost all of it is the quadratic. The mark-winning habit is Step 3's signed displacement $s = -30$ m — students who split the flight into three legs take twice as long and triple their sign-error surface.

Exam traps

Signs flipped mid-solution. The biggest mark-killer. Choose a positive direction in line 1, write $u$ , $a$ , $s$ with signs once, never re-negotiate. "Deceleration of 4 m/s²" means $a = -4$ m/s² only because motion was chosen positive — the sign belongs to the axis, not the word.

suvat under varying acceleration. If the problem has air resistance, a bounce, an engine cutting out, or "acceleration proportional to…", the $v$ – $t$ line bends and every suvat answer is invalid. Split at the change, or recognise a calculus question (see the calculus-in-kinematics module).

km/h fed into SI equations. Convert before listing: km/h $\times \tfrac{5}{18}$ = m/s (54 km/h = 15 m/s, 72 km/h = 20 m/s). A speed left in km/h inside $v^2 = u^2 + 2as$ inflates the answer by a factor of $(18/5)^2 \approx 13$ — option setters know this and plant the wrong answer.

$u$ and $v$ swapped. "Comes to rest" means $v = 0$ , not $u = 0$ ; "starts from rest" means $u = 0$ . Underline the phrase in the paper. In equation (3) the swap silently flips the sign of $2as$ and yields a beautifully wrong positive number.

The nth-second formula's units. $s_{n\text{th}} = u + \tfrac{a}{2}(2n-1)$ looks like a velocity — it is a displacement per hidden one-second interval. Don't "fix" it and don't distrust it: with SI inputs it delivers metres (full story in the derivation and FAQ).

Which g? CBSE, JEE and NEET numericals usually take $g = 10$ m/s² when the paper says so (and it usually does); AP, IB, IGCSE and A-Level mark schemes use $g = 9.8$ m/s². Read the cover instruction — with $g = 10$ , a 45 m fall takes exactly 3 s; with 9.8 it takes 3.03 s, and option grids are built to distinguish the two.

Forgetting equation (4). Two speeds and a time, no acceleration: one line of $s = \tfrac12(u+v)t$ . Solving (1) for $a$ and feeding (2) gets the same answer in five error-prone lines.

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 in Two New Sciences (1638). Every equation on this page is his geometry translated into symbols he never had — when you subtract $s(n-1)$ from $s(n)$ and the odd numbers fall out, you are re-running his experiment in algebra.

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, plotting 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. NCERT Physics XI Part 1, Ch. 3 (Motion in a Straight Line) — the CBSE/NEET canonical treatment · H.C. Verma, Concepts of Physics Vol. 1, Ch. 3 · Halliday, Resnick & Walker, Ch. 2 · Galileo, Two New Sciences, Third Day. The full calculus-first treatment with variable acceleration is the deep dive One-Dimensional Kinematics (UG, aq-mech tier).

Provenance. All equations are standard textbook results; derivations follow the v–t-graph route of the v3 canon pilot this module extends. A verifier should recompute: equation (3) by eliminating $t$ between (1) and (4); $n$ th-second values $s_4 = 24$ m and $s_6 = 32$ m from $u = 10$ , $a = 4$ (and by direct subtraction $72-48$ , $132-100$ ); the braking example ( $54$ km/h $= 15$ m/s, $t = 6.0$ s, $s = 45$ m three ways); the cliff problem ( $h_{\max} = 225/19.6 = 11.5$ m, $4.9t^2 - 15t - 30 = 0 \Rightarrow t = (15+\sqrt{813})/9.8 = 4.44$ s, impact $\sqrt{813} = 28.5$ m/s); the bar-chart distances $u^2/2a$ at $a = 6$ m/s² (5.8, 23.1, 52.1, 92.6 m for 30/60/90/120 km/h); and the warning's counterexample (1 m/s true vs 5 m/s endpoint-mean).

Prerequisites. Acceleration — the quantity these equations assume constant · Velocity–Time Graphs — the straight line whose slope and area generate everything here. Leads to. Free Fall & Motion Under Gravity — suvat with $a = -g$ · the Chase, Catch & Meet and Stopping Distance archetypes — suvat under exam pressure.

How do I stop trying all the equations one by one and wasting time?

Use the missing-symbol drill, every single time. Write the five letters $s\ u\ v\ a\ t$ in the margin, fill in the three the problem gives, mark the one it asks for. Exactly one letter is left untouched — neither given nor wanted — and the equation that omits that letter is your equation, guaranteed. Example: "brakes from 20 m/s, stops in 50 m, find the deceleration" — you have $u, v, s$ , want $a$ , so $t$ is the orphan → $v^2 = u^2 + 2as$ . The drill takes ten seconds and replaces the try-everything panic completely. It also flags under-determined problems: only two knowns means information is missing or a second phase of motion supplies it.

The nth-second formula gives units of velocity — is it dimensionally wrong?

It is dimensionally disguised, not wrong. When you subtract $s(n-1)$ from $s(n)$ , every term secretly multiplies the width of the interval you subtracted over — exactly one second. Written honestly, $s_{n\text{th}} = \big[u + \tfrac{a}{2}(2n-1)\big]\times(1\ \mathrm{s})$ , which is a perfectly respectable displacement. Because the interval is numerically 1, textbooks drop it and the formula looks like a velocity. Practical consequences: feed it SI values (m/s and m/s²) and read the answer in metres; never use it for "distance in the nth minute" without rebuilding via $s(n)-s(n-1)$ ; and if an examiner asks whether the equation is dimensionally consistent, the answer is "yes, once the suppressed 1-second factor is restored" — that exact conceptual question has appeared in JEE papers.

Where did time go in v² = u² + 2as?

It was deliberately eliminated. Between $v = u + at$ and $s = \tfrac12(u+v)t$ you have two equations sharing the symbol $t$ ; solve one for $t$ , substitute into the other, and $t$ cancels out of existence, leaving a relation among the four survivors. Nothing physical was lost — any motion satisfying (1) and (4) automatically satisfies (3). What you gained is directness: questions about speed-versus-distance (stopping distances, speed after falling a height) no longer route through the clock. The physics meaning surfaces when you multiply by $\tfrac12 m$ : it becomes $\tfrac12 mv^2 - \tfrac12 mu^2 = mas$ — kinetic energy gained equals force times distance. The time-free equation is the work–energy theorem wearing kinematic clothes.

My textbook writes v₀ and x instead of u and s — which is correct for my exam?

Both are correct physics; use your board's dialect fluently and recognise the other. CBSE, ICSE, JEE, NEET and A-Level papers write $u, v, a, s, t$ (hence "suvat"). AP Physics 1 and IB papers write $v_0$ for initial velocity and $x - x_0$ or $\Delta x$ for displacement — their sheets print $v = v_0 + at$ and $x = x_0 + v_0t + \tfrac12at^2$ . The translation is mechanical: $u \leftrightarrow v_0$ , $s \leftrightarrow \Delta x$ . One caution: AP's sheet works with positions rather than displacement, so remember to subtract; in Indian exams the marking scheme expects $s$ .

Why is average velocity (u+v)/2 — and when is that illegal?

It is the geometry of a straight line: velocity spends as much "time-weight" below the midpoint as above it, so the endpoint mean is the true time-average — precisely the mean speed theorem of the 1330s. The formula is illegal the moment the $v$ – $t$ graph bends, i.e. whenever acceleration varies. Brutal counterexample: rest for 9 s, then 10 m/s for 1 s — endpoint mean 5 m/s, truth 1 m/s. In two-phase problems the rule works within each constant- $a$ piece separately, never across the join. When in doubt, retreat to the definition — total displacement over total time — which cannot betray you.

Do I take g as 9.8 or 10 m/s²?

Follow your paper, not your habit. JEE-Main, NEET and most CBSE/ICSE numericals instruct "take $g = 10\ \mathrm{m/s^2}$ " — their options are computed with 10 (a 20 m/s throw rises exactly 20 m). AP, IB and A-Level mark schemes use 9.8 (the same throw rises 20.4 m). Read the instruction line before question 1, and never mix values mid-solution — that produces an answer on no option grid. Unstated in a JEE/NEET paper: 10 is the safe default. In an IB/AP paper: 9.8 always.

What if the problem gives me only two of the five quantities?

Then either the problem is unsolvable — or, far more often, a phrase is hiding the third value. "Starts from rest": $u = 0$ . "Comes to rest": $v = 0$ . "Dropped": $u = 0$ and $a = g$ downward. "At the highest point": $v = 0$ there. "Uniform velocity": $a = 0$ . Train yourself to translate English into symbol values as you read — competitive papers deliberately bury one datum in wording. If after translation you still have only two knowns, look for a second phase of motion whose final state feeds yours as its initial state; multi-leg problems chain suvat legs precisely this way.