Audited 26 May 2026·Last updated 27 Jul 2026·7 citations·Tier 1·0 uses

Acceleration Calculator

Free acceleration calculator — solve a = Δv/Δt and the four SUVAT kinematic equations for acceleration, final velocity, time, or distance.

Acceleration Calculator

Solve for
Speed at the start of the interval, in metres per second. Use 0 for an object starting from rest. To convert from km/h divide by 3.6, from mph divide by 2.237.
m/s
Speed at the end of the interval, in metres per second. Leave blank when solving for final velocity.
m/s
Rate of change of velocity, in metres per second squared. Use 9.80665 for free fall in vacuum near Earth's surface. Leave blank when solving for acceleration.
m/s²
Duration of the interval, in seconds. Must be greater than 0 when solving for acceleration. Leave blank when solving for time.
s
Acceleration
2.7
Constant (or average) acceleration a — the rate of change of velocity per unit time, equal to (v − u) / t. Positive values mean speeding up in the direction of motion; negative values mean slowing down (deceleration).
Final velocity
27 m/s
Initial velocity
0 m/s
Time elapsed
10 s
Distance travelled
135 m

Background.

This acceleration calculator implements the full set of one-dimensional kinematic equations for motion under constant acceleration — the so-called SUVAT equations of classical mechanics — and lets you solve for whichever of acceleration, final velocity, time, or distance you do not already know. At its core sits the definition every physics student learns first: acceleration is the rate of change of velocity with time, written a = Δv/Δt, or in its expanded form a = (v − u) / t, where u is the initial velocity, v is the final velocity, and t is the time elapsed. From that one definition and the assumption that a is constant over the interval, four equations follow algebraically: v = u + at, s = ut + ½at², v² = u² + 2as, and s = ½(u + v)t — together known as the SUVAT equations after the five quantities they relate. The calculator routes your inputs through this system and returns a self-consistent set of values in SI units (metres per second for velocities, metres per second squared for acceleration, seconds for time, metres for distance).

Acceleration is one of the most physically loaded quantities in elementary physics because it links three otherwise-separate stories: kinematics (the geometry of motion described in this calculator), dynamics (Newton's second law, F = ma, which says any net force on a mass produces a proportional acceleration), and gravitation (the universal claim that every freely-falling object near Earth's surface accelerates downward at g ≈ 9.80665 m/s² in vacuum, regardless of its mass — Galileo's famous Pisa observation).

The same equations describe a sports car launching from a traffic light (0 to 100 km/h in 4 seconds is about 7 m/s², roughly 0.7 g), a feather falling in a vacuum chamber (9.80665 m/s², the same as a bowling ball), a Formula 1 driver braking from 300 km/h to 80 km/h before a hairpin (about −5 g of deceleration sustained over two seconds), and the Saturn V rocket lifting off (1.2 g at liftoff, climbing to about 4 g at first-stage burnout).

Below the widget you will find the derivations of all four SUVAT equations from a = Δv/Δt, a worked example for car braking distance, a table of real-world acceleration benchmarks measured in g-units (where 1 g = 9.80665 m/s² exactly, the NIST-defined standard acceleration of gravity), the connection to Newton's second law and how force converts to acceleration once mass is known, an introduction to projectile motion as two-dimensional kinematics with g acting only on the vertical component, the human-physiology limits (sustained 5 g is fighter-pilot territory; 50 g for a tenth of a second is survivable in a modern crash structure; 100 g is generally fatal), and an honest discussion of where constant-acceleration kinematics breaks down — variable forces, air resistance, relativistic speeds, and the non-uniform g-field at altitude.

The calculator is a thin wrapper over a pure-function solver registered at acceleration.solve and exercised by 19 unit tests covering boundary cases (zero initial velocity, zero acceleration in solveFor='finalVelocity' mode, negative accelerations for deceleration, free fall with g = 9.80665, very small time intervals, very large velocities), so the answer you read here is the same answer used elsewhere in the Quanta physics toolset.

What is acceleration calculator?

Acceleration is the rate at which an object's velocity changes with time. Mathematically it is the time derivative of velocity, a = dv/dt; for the constant-acceleration case this calculator handles, the derivative reduces to the simple ratio a = (v − u) / t, called the average acceleration over the interval. The SI unit is the metre per second squared (m/s²), which reads as 'metres per second, per second' — every second that passes, the velocity changes by that many metres per second. Because velocity is a vector (it has both magnitude and direction), acceleration is also a vector: an object speeding up in a straight line, slowing down in a straight line, or going around a corner at constant speed are all accelerating. Three things are easy to confuse and worth pinning down. First, acceleration is not the same as velocity — a car cruising at 100 km/h on a straight motorway has high velocity but zero acceleration. Second, the sign of acceleration tells you direction, not whether the object is speeding up or slowing down: an object moving in the +x direction with negative acceleration is decelerating, but an object moving in the −x direction with negative acceleration is speeding up. Third, the everyday term 'g-force' is shorthand for acceleration expressed in units of standard gravity (1 g = 9.80665 m/s² exactly, per the 3rd CGPM in 1901 and NIST's current SI tables); it is a measure of acceleration, not a force, despite the name. When you feel pressed back in your seat as a car accelerates, your body is being accelerated by the seat's normal force, and you sense that force as a g-load. In the limit of zero acceleration (uniform velocity) you feel weightless relative to your surroundings; in freefall (a = g downward), you also feel weightless because there is no contact force resisting gravity — this is why astronauts in low Earth orbit float, despite being inside a strong gravitational field. The calculator handles the constant-a case; for variable acceleration you would need calculus or a numerical integrator.

How to use this calculator.

  1. Choose which quantity you want the calculator to solve for in the 'Solve for' menu — acceleration, final velocity, time, or distance.
  2. Enter the three other quantities in SI units: velocities in m/s, acceleration in m/s², time in seconds. To convert km/h to m/s divide by 3.6; to convert mph to m/s divide by 2.237.
  3. Leave the field corresponding to your unknown blank — the calculator ignores it and computes the missing value from the others.
  4. Read the primary output (your solved-for quantity) along with the full set: acceleration, initial velocity, final velocity, time, and distance, all returned as a self-consistent set.
  5. For a 0 to 100 km/h car test, set solveFor = 'acceleration', initialVelocity = 0, finalVelocity = 27.78 (which is 100 km/h in m/s), and time = your measured time in seconds.
  6. For a free-fall problem, set acceleration = 9.80665 (or −9.80665 if you take 'up' as positive) and use solveFor = 'finalVelocity', 'time', or 'distance' depending on what you want to know.
  7. For deceleration / braking problems, enter acceleration as a negative number (e.g. −7.5 m/s² for hard ABS braking on dry tarmac) and solveFor = 'distance' to get the braking distance, or solveFor = 'time' to get the stopping time.

The formula.

a = (v − u) ⁄ t

The five SUVAT quantities are u (initial velocity), v (final velocity), a (acceleration), t (time), and s (displacement / distance). From the constant-acceleration assumption and the definition a = (v − u) / t, four independent equations follow:

v = u + at (1) — final velocity from initial velocity plus impulse-like Δv s = u·t + ½·a·t² (2) — displacement from time, with the quadratic ½at² being the area under the v-t triangle v² = u² + 2·a·s (3) — energy-flavoured form, useful when t is unknown s = ½·(u + v)·t (4) — displacement as average velocity times time

Each equation omits exactly one of the five SUVAT variables, so given any three you can always solve for the remaining two. The calculator chooses internally which equation to apply based on your 'solveFor' selection:

solveFor = 'acceleration' uses (1) rearranged: a = (v − u) / t, then derives s from (3) solveFor = 'finalVelocity' uses (1) directly: v = u + at, then derives s from (3) when a ≠ 0 or s = u·t when a = 0 solveFor = 'time' uses (1) rearranged: t = (v − u) / a, then derives s from (3) solveFor = 'distance' uses (3) rearranged: s = (v² − u²) / (2a)

Two guards are baked into the implementation. First, when solving for acceleration the calculator refuses to divide by zero time (an instantaneous velocity change implies infinite acceleration, which is unphysical for a finite mass under finite force). Second, when solving for time or distance the calculator refuses to operate with a = 0 — under zero acceleration, time and distance are not determined by the kinematic system alone; you would instead use the trivial s = u·t for constant-velocity motion. In both cases an InvalidInputError is raised rather than silently returning Infinity or NaN. The equations apply only to constant acceleration along a straight line; for two-dimensional projectile motion you apply the same equations independently to the horizontal component (a_x = 0, constant velocity) and the vertical component (a_y = −g for free fall under Earth's gravity), then combine the results with vector addition.

A worked example.

Example

Consider the canonical car-magazine test: a mid-range sports sedan claims '0 to 100 km/h in 5.4 seconds'. To turn that marketing line into a physical acceleration, first convert 100 km/h to SI units: 100 / 3.6 = 27.78 m/s. Set solveFor = 'acceleration', initialVelocity = 0 (starting from rest), finalVelocity = 27.78, time = 5.4. The calculator returns a = (27.78 − 0) / 5.4 = 5.14 m/s², which converts to roughly 0.52 g — meaning the launch is just over half the strength of Earth's gravity, comfortably below what a roller-coaster delivers in its drops (typically 2 to 4 g) and well below what a fighter pilot tolerates in a hard turn (5 to 9 g). The same calculation also gives you the distance covered during the launch: from v² = u² + 2as, we get s = (27.78² − 0²) / (2 × 5.14) = 75.0 m, which matches the half-base × height area of the v-t triangle: s = ½ × 5.4 × 27.78 = 75.0 m. So this car covers three-quarters of a football pitch reaching 100 km/h. If you wanted to know how long it would take to reach 200 km/h (55.56 m/s) at the same acceleration — assuming traction and engine torque held up, which they would not in practice — you would switch to solveFor = 'time' with a = 5.14: t = 55.56 / 5.14 = 10.8 seconds, exactly double the 100 km/h time, because acceleration is linear in v at constant a. This is also why 0-to-60 times collapse for performance cars: the Bugatti Chiron does 0 to 100 km/h in 2.4 s, which is a = 11.6 m/s² ≈ 1.18 g — sustained, which feels considerably more aggressive than a roller coaster's transient spikes.

initial Velocity0
final Velocity27.78
time5.4
solve Foracceleration

Frequently asked questions.

What is the formula for acceleration?
The defining formula is a = (v − u) / t, where u is the initial velocity, v is the final velocity, and t is the time elapsed. In SI units, velocities are in metres per second (m/s) and acceleration comes out in metres per second squared (m/s²). This formula gives the average acceleration over the interval; for motion at constant acceleration it also gives the instantaneous acceleration at every moment. Three sibling equations form the complete SUVAT set: v = u + at, s = ut + ½at², and v² = u² + 2as, where s is the displacement. Together these four equations let you solve any constant-acceleration problem given any three of the five quantities (u, v, a, t, s).
How do I convert acceleration in m/s² to g-force?
Divide by 9.80665. The constant 1 g = 9.80665 m/s² is the standard acceleration of free fall, defined exactly by the 3rd General Conference on Weights and Measures (CGPM) in 1901 and published in NIST's SI tables. So an acceleration of 5 m/s² is 5 / 9.80665 = 0.51 g, and 30 m/s² is 30 / 9.80665 = 3.06 g. For reference: a sneeze peaks at about 3 g, a sports-car launch is 0.5 to 1.2 g, a roller-coaster drop is 2 to 4 g, a fighter pilot in a hard turn sustains 5 to 9 g, the Apollo astronauts hit 6.5 g during reentry, and Formula 1 drivers regularly experience 5 g of lateral acceleration through fast corners and 5 to 6 g of deceleration under heavy braking.
What does negative acceleration mean? Is it the same as deceleration?
Negative acceleration means the acceleration vector points in the negative direction of your chosen coordinate axis. Whether that corresponds to slowing down (deceleration) or speeding up depends on the direction of motion. An object moving in the +x direction with negative acceleration is decelerating; an object moving in the −x direction with negative acceleration is speeding up (the acceleration is pushing it further in the negative direction). 'Deceleration' is a colloquial term for any acceleration that reduces an object's speed, regardless of sign. In this calculator, if you enter braking from 30 m/s to 0 m/s over 4 s, you will get a = −7.5 m/s² — the negative sign reflects that the velocity is decreasing, while the magnitude (7.5 m/s², about 0.76 g) tells you how hard the brakes are working.
How is acceleration related to Newton's second law?
Newton's second law states that the net force on an object equals its mass times its acceleration: F = m·a, or equivalently a = F/m. This means acceleration is the direct kinematic signature of a force — if you know the mass of an object and the net force acting on it, you know its acceleration exactly, and from there the SUVAT equations give you everything else (velocity at any later time, distance covered, etc.). Three consequences follow. First, in the absence of any net force (Newton's first law, the special case F = 0), a = 0 and the object continues at constant velocity. Second, equal forces on different masses produce different accelerations — push a shopping trolley and a parked car with the same hand-force and the trolley accelerates a hundred times more. Third, in free fall the gravitational force on a mass m is F = m·g, so by Newton's second law a = (m·g) / m = g — the mass cancels out, which is why all objects fall at the same rate in vacuum regardless of mass. Galileo's Pisa demonstration and Apollo 15's hammer-and-feather drop on the Moon both verify this directly.
What is the acceleration of free fall, and is it really the same for all objects?
The standard acceleration of gravity is g = 9.80665 m/s², defined exactly. In vacuum, every freely-falling object — a feather, a hammer, a bowling ball, a human — accelerates at exactly this value near Earth's sea level, regardless of its mass. This was demonstrated cleanly in 1971 by Apollo 15 commander David Scott on the lunar surface (where there is no air): a hammer and a falcon feather, released simultaneously, hit the ground at the same instant. In Earth's atmosphere, air resistance complicates the picture — a feather has a much higher drag-to-mass ratio than a bowling ball and reaches its terminal velocity in fractions of a second, while the bowling ball keeps accelerating much longer. The 9.80665 value also varies slightly with latitude (about 9.78 at the equator, 9.83 at the poles, due to centrifugal effects from Earth's rotation and the equatorial bulge) and with altitude (it drops by about 0.3% per 10 km of altitude). The defined 9.80665 value is the agreed standard for engineering and metrology, not a measurement at any specific location.
How do I calculate stopping distance from initial speed and deceleration?
Use the third SUVAT equation, v² = u² + 2as, set the final velocity v = 0 (you have stopped), and solve for s: s = u² / (2|a|), where |a| is the magnitude of the deceleration. For a car braking from 30 m/s (108 km/h) with maximum ABS deceleration of about 8 m/s² on dry tarmac, s = 900 / 16 = 56.25 m of braking distance — and that is on top of about 22 m of thinking distance (1 second of reaction at 30 m/s). On wet tarmac the deceleration drops to maybe 5 m/s² and braking distance grows to s = 900 / 10 = 90 m. On ice the deceleration can be as low as 1 m/s² and stopping from 30 m/s takes 450 m, longer than four football pitches. The quadratic dependence on initial speed is the single most important fact about braking: doubling your speed quadruples your braking distance, which is why urban speed limits matter for pedestrian survival. NHTSA crash data consistently shows that pedestrian fatality risk rises sharply above 30 mph (48 km/h) precisely because braking and impact energy both scale with v².
What does the calculator do with projectile motion problems?
The calculator handles only one-dimensional constant-acceleration kinematics. For a projectile — say a ball thrown at an angle — you decompose the problem into two independent one-dimensional problems and run the calculator twice. For the horizontal component, acceleration is zero (ignoring air resistance), so the ball moves at constant horizontal velocity v_x = u·cos(θ) and the horizontal distance is just v_x times the time of flight. For the vertical component, the acceleration is −g (taking up as positive), the initial vertical velocity is v_y = u·sin(θ), and the SUVAT equations give you the time to apex (t = v_y / g), the maximum height (h = v_y² / 2g), and the total time of flight (2·t for level ground). The horizontal range is then R = (u² / g)·sin(2θ), maximised at θ = 45°. Use this calculator for the vertical or horizontal slice individually, and combine the results with basic trigonometry.
What are typical 0 to 60 mph times converted to acceleration in m/s²?
60 mph is 26.82 m/s, so a = 26.82 / t where t is the 0-to-60 time in seconds. A typical compact sedan with 0-60 in 9 s has a = 2.98 m/s² ≈ 0.30 g. A hot hatch at 0-60 in 6 s has a = 4.47 m/s² ≈ 0.46 g. A sports car at 0-60 in 4 s has a = 6.71 m/s² ≈ 0.68 g. A supercar at 0-60 in 3 s has a = 8.94 m/s² ≈ 0.91 g. The Tesla Model S Plaid at 0-60 in 1.99 s gives a = 13.5 m/s² ≈ 1.37 g — over one full Earth gravity of sustained horizontal acceleration, which is why journalists describe the launch as a 'roller coaster you cannot get off'. The upper limit for street tyres on dry tarmac is roughly 1.0 to 1.2 g of grip, which is why no production car beats about 2 seconds to 60 mph without resorting to launch control and very specific surface conditions. For drag racers with sticky tyres and downforce, peak accelerations of 4 to 5 g are achievable over the first second.

References& sources.

  1. [1]NIST Reference on Constants, Units, and Uncertainty — Standard acceleration of gravity g_n = 9.80665 m/s² (exact by definition, 3rd CGPM 1901). National Institute of Standards and Technology, CODATA 2018 recommended values.
  2. [2]Halliday, D., Resnick, R. & Walker, J. (2014). Fundamentals of Physics, 10th ed., Wiley. Chapter 2 'Motion Along a Straight Line', sections 2-4 through 2-7, on average and instantaneous acceleration, constant-acceleration kinematics, and free fall. Chapter 4 extends the same equations to two-dimensional projectile motion.
  3. [3]Feynman, R. P., Leighton, R. B. & Sands, M. (1964). The Feynman Lectures on Physics, Volume I, Chapter 8 'Motion' and Chapter 9 'Newton's Laws of Dynamics'. Feynman's derivation of velocity and acceleration as time derivatives of position, and his treatment of F = ma as the operational definition of force.
  4. [4]NASA Glenn Research Center — Beginner's Guide to Aeronautics: 'Newton's Laws of Motion' and 'Free Fall'. NASA educational reference on kinematics, the connection F = ma, and free-fall acceleration in vacuum versus with air resistance.
  5. [5]NHTSA (National Highway Traffic Safety Administration). 'Traffic Safety Facts: Speeding'. United States Department of Transportation. Crash data demonstrating the v² dependence of braking distance and pedestrian fatality risk as a function of impact speed.
  6. [6]Bureau International des Poids et Mesures (BIPM). The International System of Units (SI), 9th ed., 2019. Definitions of the metre, the second, and derived units including m/s². Section 2.3.1 covers the standard acceleration of free fall.
  7. [7]Galilei, Galileo (1638). Discorsi e Dimostrazioni Matematiche, intorno à due nuove scienze (Discourses and Mathematical Demonstrations Relating to Two New Sciences). Leiden: Elsevier. Galileo's original derivation of the constant-acceleration law for falling bodies, including the s ∝ t² relationship that follows from a = constant.

In this category

Embed

Quanta Pro

Paid features are coming later.

  • All 313 calculators remain free
  • No billing is enabled
Coming soon