Acceleration Formula Explained: Master the Three Golden Equations of Motion
JEE/NEET Physics · Motion in a Straight Line series · Part 2 of 6 · All parts →
- Acceleration = how quickly velocity CHANGES: a = (v − u)/t, in m/s²
- It’s a vector too: +a speeds you up in the + direction, −a slows you (or speeds you up backwards)
- The three equations (constant a only!): v = u + at · s = ut + ½at² · v² = u² + 2as
- Sign discipline: decide + once, then u, v, a, s all carry signs
- Any two of the five quantities (u, v, a, s, t) given → the third comes from one equation
Velocity tells you how fast; acceleration tells you how fast ‘how fast’ is changing. A car pulling away, a bike braking, a coin dropped from a roof — all three run on the same three equations. Part 2 of the Motion in a Straight Line series.
- What acceleration really is
- Average vs instantaneous acceleration
- Reading the sign of a
- The three golden equations
- Where the equations come from
- Choosing the right one
- Solved examples
- Common mistakes
- This physics in your daily life
- Practice set
- FAQ
- Recap
What Acceleration Really Is
In Part 1 we learned that velocity is the rate of change of displacement. Acceleration is the very next step up the ladder: it is the rate of change of velocity. In symbols:
a = (v − u) / t
where u is the starting velocity, v is the velocity after time t, and the answer carries the unit m/s². That unit literally reads ‘metres per second, per second’ — each second, the velocity gains (or loses) that many metres per second. If a bus accelerates at 2 m/s², then every second its speedometer climbs by 2 m/s: from 0 to 2, from 2 to 4, and so on.
Two cautions before we go further. First, acceleration is a vector: it has both magnitude and direction, just like velocity and displacement. Second — and this is where most beginners slip — zero acceleration does not mean standing still. It means velocity is unchanging. Cruising at a steady 80 km/h on a straight highway is a = 0, even though you’re covering 80 kilometres every hour. Acceleration begins the moment velocity changes: speeding up, slowing down, or (in later chapters) turning.
A useful sanity check: an object at rest has v = 0, but it may still have acceleration. A ball at the top of its throw has v = 0 for an instant, yet gravity is pulling on it the whole time — its acceleration is 9.8 m/s² downward at that very instant. Velocity can be zero while acceleration is not.
Average vs Instantaneous Acceleration
The formula a = (v − u)/t gives the average acceleration over the interval t — it smooths out everything that happened in between. The instantaneous acceleration is what a speedometer-plus-accelerometer would read at a single moment, and it is formally the limit of Δv/Δt as Δt shrinks to zero. In this chapter we restrict ourselves to constant acceleration, where average and instantaneous are the same number, which is exactly why the three golden equations work so cleanly. When acceleration varies (a rocket burning fuel, a car shifting gears), these equations no longer apply directly — you’d need calculus, which arrives in a later part of this series.
Reading the Sign of a
Because acceleration is a vector, its sign carries meaning — but the meaning depends on the direction of motion, not on ‘speeding up vs slowing down’ as most students assume. Here is the full picture, taking rightward as positive:
| Situation | Velocity sign | Acceleration sign | What happens |
|---|---|---|---|
| Moving right, speeding up | + | + | faster right |
| Moving right, braking | + | − | slows, may reverse |
| Moving left, speeding up | − | − | faster left |
| Moving left, braking | − | + | slows |
The pattern compresses into one rule: same sign = speeding up; opposite signs = braking. This is why ‘negative acceleration’ does not automatically mean slowing down — it means accelerating in the − direction. A ball thrown upward has positive (upward) velocity and negative (downward) acceleration, so it slows. The same ball falling down has negative velocity and negative acceleration — and it speeds up, with both quantities negative.
The Three Golden Equations
For constant acceleration only, three equations connect the five quantities u (initial velocity), v (final velocity), a, s (displacement), and t:
| Letter | What it means (plain words) | Value / unit |
|---|---|---|
| u | starting velocity (‘u’ for initial) | m/s, signed |
| v | velocity after time t | m/s, signed |
| a | acceleration (constant!) | m/s², signed |
| s | displacement in time t | m, signed |
| t | elapsed time | s |
Notice that each equation deliberately omits one quantity: the first has no s, the second has no v, and the third has no t. That omission is not an accident — it is the entire design of the toolkit, as we’ll see below.
Where the Equations Come From (30-Second Derivations)
You should never memorise these blind — each one is two lines of work. Equation 1 is just the definition rearranged: a = (v − u)/t → v = u + at. Equation 2 comes from the fact that under constant acceleration, average velocity = (u + v)/2, so s = average velocity × time = ((u + v)/2)·t; substitute v = u + at and you get s = ut + ½at². Equation 3 is obtained by eliminating t between the first two: from Equation 1, t = (v − u)/a; substitute into Equation 2, simplify, and v² = u² + 2as falls out. Knowing the derivations means you can rebuild any equation mid-exam if memory wobbles — and examiners love asking ‘derive the third equation of motion’ as a two-mark gimme.
Choosing the Right One
Before touching any equation, make a quick inventory: list what’s given and what’s wanted. Then apply one rule:
- No t anywhere in the problem? Use v² = u² + 2as.
- No v mentioned? Use s = ut + ½at².
- No s mentioned? Use v = u + at.
The right equation is the one that doesn’t contain the quantity nobody mentioned. This saves you the classic time-sink of solving two equations when one would do. In multi-step problems, you’ll often chain them: find v from Equation 1, then feed it into Equation 3 for the distance.
Solved Examples
Given: u = 20 m/s, v = 0 (it stops), a = −5 m/s² (opposite to motion). No s mentioned → v = u + at: 0 = 20 − 5t → t = 4 s.
✔
Answer: 4 s
No t given → v² = u² + 2as: 0 = 400 + 2(−5)s.
s = 40 m.
Doubling the speed would quadruple this — the u² inside is why highway speeds kill. ✔
Answer: 40 m
Stage 1: u = 0, so s = ½(2)(10²) = 100 m, and v = 2 × 10 = 20 m/s at the end of the stage.
Stage 2: the 20 m/s becomes the new u. Braking: 0 = 400 − 2(4)s → s = 50 m.
Total = 150 m — always split multi-stage problems at the velocity handover: the final velocity of one stage is the initial velocity of the next. ✔
Answer: 150 m
At the top, v = 0. No t involved → v² = u² + 2as with a = −g: 0 = 400 − 2(10)s → s = 20 m. Notice we never needed to know how long the flight took — the t-free equation handled it in one line. ✔
Answer: 20 m
- Using the equations when a isn’t constant. They hold ONLY for uniform acceleration — check before plugging in.
- Mixing units. km/h must become m/s (÷3.6) before entering any equation with metres. 72 km/h = 20 m/s, not 72.
- Sign chaos. Braking car moving +: u = +20, a = −5 — both signs must appear, or the answer silently flips. Choose your positive direction once, at the start, and stick to it.
- Stopping-distance intuition. Twice the speed = FOUR times the stopping distance (u² law) — never ‘twice’.
- Dropping to rest vs being at rest. ‘Comes to rest’ means v = 0; ‘starts from rest’ means u = 0. Mixing these up in stage problems is a classic lost mark.
This Physics in Your Daily Life
- ‘0 to 100 in 3 seconds’ car ads are acceleration marketing: 100 km/h in 3 s ≈ 9.3 m/s² — about the same as free fall, which is why fast launches feel like a dropping lift.
- Yellow-light dilemma at crossings — ‘can I stop?’ is v² = u² + 2as solved live by every driver: braking distance grows as speed SQUARED.
- Plane takeoff — a runway is sized for ~2–3 m/s² over 30–40 s: the same s = ut + ½at² with lives at stake.
- Lift journeys — the stomach-flutter at launch is your body feeling ~1.5 m/s² of extra acceleration; cruises are a = 0 and feel like nothing.
- Train metro codes — smooth ±1 m/s² limits are chosen so standing passengers don’t stumble: acceleration, not speed, is what topples people.
- Safety crumple zones — a crash at fixed speed is survivable only if the stopping TIME (and distance) is stretched, because a = Δv/t: longer t means smaller a. Airbags are Equation 1, engineered.
A rickshaw wallah negotiating fare by speed is missing the point — passengers care about the JERK of the launch and the lurch of the brake. Speed is how the world slides past; acceleration is what your body actually feels pressed into the seat. You never feel speed in a smooth flight; you feel every change of it.
0→100 km/h in 3 s: velocity changes by 27.8 m/s in 3 s → a ≈ 9.3 m/s², one g. At 40 m/s² (fighter jet, crash): velocity changes by a whole highway speed EVERY second. The m/s² unit says it directly: ‘this many m/s of velocity, gained every second’.
Draw velocity against time: acceleration is the SLOPE of that line. Flat = cruising. Upward tilt = speeding up. Downward = braking. A curving slope = changing acceleration (jerk). Every motion story is one graph, and a is its tilt — we’ll read these graphs properly in Part 3.
Practice set (answers hidden — try first)
(NEET-level) Rest to 30 m/s at 3 m/s²: time =
(JEE Main-level) u = 0, a = 4, t = 5: displacement =
(NEET-level) v² = u² + 2as with v=0, u=15, s=22.5: |a| =
(Concept) A body moves at constant 80 km/h in a straight line. Its acceleration:
(JEE Main-level) Stopping distance at 20 m/s with a = −5 is 40 m. At 40 m/s:
(JEE Advanced-level) A particle covers 40 m in the 5th second starting from rest with constant a. Find a.
(Concept) Can a body have zero velocity but non-zero acceleration?
Frequently Asked Questions
Can acceleration be negative while speed increases?
Why are the three equations invalid for changing acceleration?
Do I use s or distance in v² = u² + 2as?
- a = (v − u)/t, m/s² — rate of velocity change
- same signs = speed up, opposite = brake
- v = u + at · s = ut + ½at² · v² = u² + 2as (constant a only)
- pick the equation missing your unknown
- stopping distance ∝ speed squared
- 🔁 a = Δv/Δt, vector, m/s²
- 🔁 three equations, constant a only
- 🔁 equation choice = skip the unknown
- 🧠 Chant: ‘u-vat, suat, v-u-2as’ — the three tools.
- 🧠 Signs: ‘same sign speeds, opposite brakes’.
- 🧠 Doubling speed quadruples braking distance — the u² law.
- 🧠 Multi-stage problems: split at the velocity handover.
- 🏠 Daily: ‘0–100 in 3 s’ ads ≈ free-fall launch feel.
- 🏠 Daily: metro’s ±1 m/s² keeps standing riders upright.
Quick revision
- Acceleration = how quickly velocity CHANGES: a = (v − u)/t, in m/s²
- It’s a vector too: +a speeds you up in the + direction, −a slows you (or speeds you up backwards)
- The three equations (constant a only!): v = u + at · s = ut + ½at² · v² = u² + 2as
- Sign discipline: decide + once, then u, v, a, s all carry signs
- Any two of the five quantities (u, v, a, s, t) given → the third comes from one equation
- What acceleration really is
Have a doubt on this topic?
