Physics I: Mechanics › Kinematics › full formula sheet

v² = v₀² + 2aΔx

The time-free kinematic equation

How to solve constant-acceleration problems without ever knowing the time — and why the equation is really an energy statement in disguise.

Notation on this page: v₀ is the initial velocity, v the final velocity, a the (constant) acceleration, and Δx = x − x₀ the displacement. Signs follow your chosen coordinate system — pick one direction as positive and stick with it.

Where it comes from

Most constant-acceleration problems hand you three of the five SUVAT quantities (displacement, initial velocity, final velocity, acceleration, time) and ask for a fourth. But some problems never mention time at all — how fast is a cart going after traveling 6 m with a = 3 m/s²? — and dragging time into it is a detour.

The naive route: find t first with v = v₀ + at, then plug it into Δx = v₀t + ½at². Watch what happens on a concrete case: v₀ = 5 m/s, a = 3 m/s², Δx = 6 m. Find v.

the long way
6 = 5t + ½·3·t²  ⇒  t = (−5 + √61)/3 ≈ 0.94 s
First solve a quadratic for t — a variable nobody asked about.
v = 5 + 3·((−5 + √61)/3) = 5 − 5 + √61 = √61 ≈ 7.81 m/s
Then the ×3 cancels the /3, and 5 − 5 cancels. The detour computed √61 — and then uncomputed half its own work.
the time-free way
v² = 25 + 2·3·6 = 61  ⇒  v = √61 ≈ 7.81 m/s
One line. The quadratic formula was secretly computing v² all along — the time-free equation skips the disguise.

So the equation is a shortcut past time. But it is more than a shortcut — it has a physical soul. Multiply both sides by m/2:

½m·(v² = v₀² + 2aΔx)
=
½mv² − ½mv₀² = (ma)Δx = FΔx
Change in kinetic energy equals force times distance — the work-energy theorem. This equation is the work-energy theorem wearing kinematics clothes.

The intuition: acceleration acting over a distance changes speed; time is just the middleman the two variables use to talk. Its roots go back to Galileo’s inclined-plane experiments (early 1600s), which established that uniformly accelerated bodies cover distance proportional to t² — the relation this equation is built from.

Derivation

Start with the two basic constant-acceleration equations and eliminate t between them. That is the entire strategy: t appears in both but is wanted in neither.

t
=
(v − v₀)/a
Step 1 — isolate t. From v = v₀ + at. (We divide by a, so a ≠ 0 here; the a = 0 case is trivially v = v₀.)
Δx
=
v₀·(v−v₀)/a + ½a·[(v−v₀)/a]²
Step 2 — substitute. Put that t into Δx = v₀t + ½at². Time is now gone — everything is in terms of v, v₀, a, Δx.
=
v₀(v−v₀)/a + (v−v₀)²/(2a)
Step 3 — tidy the second term. ½a·(v−v₀)²/a² = (v−v₀)²/(2a): one power of a cancels top and bottom.
=
[2v₀(v−v₀) + (v−v₀)²] / (2a)
Step 4 — common denominator. Multiply the first term’s top and bottom by 2 so both terms sit over 2a. This is where the 2 in 2aΔx comes from.
=
(v−v₀)·(v + v₀) / (2a)
Step 5 — factor. Both numerator terms share (v−v₀): (v−v₀)·[2v₀ + (v−v₀)] = (v−v₀)(v + v₀).
=
(v² − v₀²) / (2a)
Step 6 — difference of squares. (v−v₀)(v+v₀) = v² − v₀². Multiply both sides by 2a and rearrange: v² = v₀² + 2aΔx. ∎

Why does a ≠ 0 not matter? If a = 0, velocity never changes: v = v₀, and the equation reads v² = v₀² — true, and Step 1’s division is never needed. The derivation covers the moving case; the trivial case covers itself.

How to use it

The procedure, every time:

  1. List knowns and unknowns SUVAT-style. Write down v₀, v, a, Δx, t — circle what you have. If t is missing and nobody asks for it, this is your equation.
  2. Check that a is constant. Gravity near Earth’s surface, steady braking, a smooth incline — yes. Springs, air drag, rockets burning fuel — no.
  3. Fix your sign convention FIRST. Pick one direction as positive and write it down: “up is positive” or “right is positive”. Then v₀, v, a, and Δx each get a sign from that choice — before you compute.
  4. Plug in and solve. Four variables, one equation — it solves for whichever one is unknown.
  5. Handle the ± square root. The equation returns v², so v = ±√(v₀² + 2aΔx). Pick the sign matching the object’s actual direction of motion in your coordinate system.

When NOT to use it

If time is given or asked for, the other equations are easier: v = v₀ + at when t appears, Δx = v₀t + ½at² when you know t. If acceleration varies (a spring, drag near terminal velocity, a rocket), this equation is simply wrong — or at best an approximation. And if a = 0, just write v = v₀.

Rearranged forms

a = (v² − v₀²) / (2Δx)solve for acceleration — braking and takeoff problems
Δx = (v² − v₀²) / (2a)solve for stopping / runway distance
v₀ = ±√(v² − 2aΔx)solve for the initial speed — “how fast was it going before?”
Common mistake: plugging in the position x instead of the displacement Δx = x − x₀. If the motion doesn’t start at x = 0, x alone is wrong — always subtract where it started.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic plug-in: v₀ = 0, a = 3 m/s², Δx = 6 m

  1. Inventory. Known: v₀ = 0, a = +3 m/s², Δx = +6 m. Unknown: v. Time: not given, not asked — reach for the time-free equation.
  2. Write the equation. v² = v₀² + 2aΔx.
  3. Substitute. v² = 0² + 2·3·6 = 36. (Why no sign fuss? Everything moves in the +x direction, so all signs are positive.)
  4. Square-root with judgment. v = ±6 m/s — pick +6 m/s, matching the direction of motion. Done.
Common mistake: writing v = 36 m/s — forgetting the square root. The equation gives v²; the answer needs √. A units check catches it: v² is in m²/s², v is in m/s.

Example 2 — braking distance: car at 20 m/s stops with a = −4 m/s²

  1. Inventory and convention. v₀ = +20 m/s (moving right, right is positive), v = 0 (stops), a = −4 m/s² (deceleration opposes motion — negative). Unknown: Δx.
  2. Rearrange first. Δx = (v² − v₀²) / (2a) — solving for the unknown before plugging numbers keeps the algebra clean.
  3. Substitute. Δx = (0² − 20²) / (2·(−4)) = (−400) / (−8) = 50 m. Two negatives make a positive — as they must: distance is positive.
  4. Sanity check. 50 m at 20 m/s is about 2.5 seconds of braking at 4 m/s² — plausible for a hard stop. ✓
Common mistake: writing a = +4 m/s² (“it’s just 4”). Then Δx = (−400)/8 = −50 m — a negative distance, which is the equation telling you the signs are inconsistent. The minus on a is not decoration.

Example 3 — runway takeoff: reach 80 m/s at a = 4 m/s² from rest

  1. Inventory. v₀ = 0 (from rest), v = 80 m/s, a = +4 m/s². Unknown: Δx, the runway length needed.
  2. Rearrange. Δx = (v² − v₀²) / (2a).
  3. Substitute — square first. v² = 80² = 6400 (not 1600 — square the whole 80). Δx = 6400 / (2·4) = 6400 / 8 = 800 m.
  4. Sanity check. 800 m runways are short for jets — this constant-a model says 80 m/s needs about 20 s of acceleration. Real takeoffs vary a, so treat it as an estimate. ✓
Common mistake: computing 80² as 1600 (squaring 8, not 80) or writing Δx = 6400/4 = 1600 m (dropping the factor of 2). The 2 in 2aΔx is load-bearing — it came from clearing denominators in Step 4 of the derivation.

Example 4 — judgment call: ball thrown upward at 15 m/s, max height?

This one is a sign-convention trap. Set the convention before touching the equation.

  1. Convention: up is positive. So v₀ = +15 m/s, and gravity points down: a = −9.8 m/s². (Why negative? The convention decides the sign, not your intuition about “gravity is positive 9.8”.)
  2. What does “max height” mean? The instant the ball stops rising: v = 0. Unknown: Δx (how far it rose).
  3. Solve. 0² = 15² + 2·(−9.8)·Δx ⇒ 225 = 19.6·Δx ⇒ Δx = 225/19.6 ≈ 11.5 m.
  4. Check the trap. With a = +9.8 (wrong sign), you get Δx = −11.5 m — the ball “rises” downward. The equation answered faithfully; the convention was betrayed.
Common mistake: answering “11.5 m above the ground” when the ball was thrown from a height. Δx is displacement, not position: thrown from a 2 m balcony, the max height above ground is 2 + 11.5 = 13.5 m. Always ask: “11.5 m from where?”

Memorization tips

  • Chant it: “v-squared equals v-naught-squared plus two-a-delta-x.” The rhythm matches the formula — say it three times and it sticks.
  • The units check (your 5-second self-check): 2aΔx has units (m/s²)·m = m²/s² = (m/s)² — exactly v²’s units. If a rearranged form breaks the units, it’s wrong.
  • The missing-variable trick: of the five SUVAT quantities, this is the one with no t. “Which variable is missing?” identifies every kinematic equation — and tells you when to reach for this one.
  • The energy twin: it’s the work-energy theorem in disguise (½mv² − ½mv₀² = FΔx). If you remember one, you can rebuild the other by multiplying or dividing by m/2.
  • The ± rule: the equation answers in squares; you answer in signs. v = ±√(…) — then the direction of motion picks the sign. Never leave the ± undecided.
  • The derivation anchor: if you blank on the formula, re-derive it: t = (v−v₀)/a, substitute into Δx = v₀t + ½at², clear denominators, factor, difference of squares. Sixty seconds, and it’s back.

Final challenge

Five mixed questions — computations, rearrangements, and the traps, all in one. Score 5/5 and the time-free equation is yours.

← Back to the Physics I formula sheet

How to learn a formula here

  1. Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
  2. Work the examples with the answers covered, then uncover one step at a time and compare.
  3. Finish with the final challenge — five mixed questions including the classic traps.
  4. Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

When do I use v² = v₀² + 2aΔx?

Use it for motion with constant acceleration when time is unknown or not asked for. If time appears in the problem — given or asked — equations like v = v₀ + at are usually easier.

Is Δx the same as x?

No. Δx is displacement: x − x₀. If the object doesn’t start at x = 0, you must subtract the starting position. Using x itself is one of the most common errors.

Do I take the positive or negative square root for v?

The equation gives v², so v = ±√(v₀² + 2aΔx). Choose the sign matching the object’s direction of motion in your coordinate system — and fix that coordinate system before you compute.

When does the equation fail?

When acceleration isn’t constant: springs, air drag near terminal velocity, rockets burning fuel. It was derived by assuming a is constant, so varying acceleration breaks it.

How is it related to kinetic energy?

Multiply both sides by m/2 and you get ½mv² − ½mv₀² = (ma)Δx = FΔx — the work-energy theorem. The time-free equation is the work-energy theorem wearing kinematics clothes.

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