Acceleration Calculator

Solve for acceleration, velocity change (Δv), or time using a = Δv / t. Change any value to recalculate instantly, with unit conversion and a step-by-step walkthrough.

Acceleration
10m/s²
Felt as g-force1.02 g
0.01g0.1g1g10g

Comparable to free fall (1 g).

How is this calculated?

a = Δv / t

1. Velocity change in SI: 100 mps = 100 m/s
2. Time in SI: 10 s = 10 s
3. Acceleration = Δv ÷ time = 10 m/s²

About the Acceleration Calculator

Acceleration is the rate at which velocity changes, and because velocity is a vector, an object accelerates whenever it speeds up, slows down, or changes direction. This is the point most commonly missed: a car rounding a bend at a constant 40 km/h is accelerating, because its direction is changing even though the speedometer reads steady. Acceleration is measured in metres per second squared, a unit that reads awkwardly but means something simple — how many metres per second the velocity gains with each passing second. It sits at the centre of Newtonian mechanics because Newton's second law makes it the direct link between the forces acting on a body and the motion that results. Every kinematics problem involving constant acceleration reduces to four equations, commonly called the SUVAT equations, and choosing the right one is largely a matter of identifying which quantity the question does not mention.

Mathematical Formula & Logic

Acceleration is the rate of change of velocity with respect to time. 1. Average acceleration: a = Δv / Δt = (v − u) / t Where u is initial velocity, v is final velocity, and t is elapsed time. 2. Instantaneous acceleration: a = dv/dt = d²s/dt² 3. The four constant-acceleration (SUVAT) equations: v = u + a·t (no displacement) s = u·t + ½·a·t² (no final velocity) v² = u² + 2·a·s (no time) s = ((u + v) / 2) · t (no acceleration) Select the equation that omits the quantity you neither know nor need. 4. From Newton's second law: a = F_net / m 5. SI unit: metres per second squared (m/s²) Standard gravity g = 9.80665 m/s² (a defined constant)

Step-by-Step Example

A car travelling at 25 m/s brakes to a complete stop over a distance of 62.5 m. Find its acceleration and the time taken: Finding acceleration — time is unknown, so use the equation without t 1. Known: u = 25 m/s, v = 0 m/s, s = 62.5 m 2. Apply v² = u² + 2·a·s 3. 0² = 25² + 2·a·(62.5) 4. 0 = 625 + 125·a 5. a = −625 / 125 = −5 m/s² The negative sign indicates deceleration, meaning the acceleration points opposite to the direction of motion. Finding the time — now use the equation containing t 6. Apply v = u + a·t 7. 0 = 25 + (−5)·t 8. t = 25 / 5 = 5 seconds Cross-check with a third equation 9. s = u·t + ½·a·t² = 25(5) + ½(−5)(25) = 125 − 62.5 = 62.5 m. Confirms the original distance. Comparing against gravity 10. As a fraction of standard gravity: 5 / 9.80665 = 0.51 g, which is a firm but entirely routine braking effort for a car with good tyres on dry tarmac. Why stopping distance punishes speed so hard Rearranging v² = u² + 2·a·s for the distance needed to stop gives s = u² / (2·a). The speed is squared and the braking rate is not, so distance grows with the square of how fast you were going. Holding the same 5 m/s² from the example above: 30 km/h (8.333 m/s) needs 6.94 m. 60 km/h (16.667 m/s) needs 27.78 m. 90 km/h (25 m/s) needs 62.50 m. 120 km/h (33.333 m/s) needs 111.11 m. Doubling the speed does not double the distance, it quadruples it — 6.94 to 27.78, and 27.78 to 111.11, each exactly four times the last. This is the single most useful consequence of the SUVAT equations in ordinary life, and it is why speed limits fall disproportionately near schools and junctions rather than in proportion to the hazard. Note that this is braking distance alone. Total stopping distance also includes the reaction distance travelled before the brakes are touched, and that part is linear in speed — roughly one second of travel, so 8.3 m at 30 km/h and 33.3 m at 120 km/h. The linear term dominates at low speed and the squared term dominates at high speed, which is why the difference between 100 and 120 km/h matters far more than the difference between 30 and 50. The same arithmetic read forwards Acceleration figures quoted for cars are the same equation with the sign flipped. A 0 to 100 km/h time of 5.0 seconds is 27.778 / 5 = 5.556 m/s², or 0.567 g. A 0 to 60 mph time of 6.0 seconds is 4.470 m/s², or 0.456 g. Both sit close to the 0.51 g braking figure above, which is not a coincidence: on dry tarmac both accelerating and braking are limited by the same thing, the grip available between four tyre contact patches and the road.

Reference Data & Values

situationacceleration m_s2as fraction_of_g
Lift starting upward1.00.10 g
Car accelerating briskly3.00.31 g
Emergency braking (dry road)8.00.82 g
Free fall (standard gravity)9.806651.00 g
Sports car 0–100 km/h in 3 s9.260.94 g
Fighter jet manoeuvre88.09.0 g
Gravity on the Moon1.620.17 g

Frequently Asked Questions

It means the velocity changes by that many metres per second during every second that passes. An acceleration of 3 m/s² means an object moving at 10 m/s will be moving at 13 m/s one second later, 16 m/s after two seconds, and so on. The unit reads as metres per second, per second, which is why the second appears squared in the denominator. Writing it out that way usually makes it click faster than reading the squared notation directly.
Negative acceleration means the acceleration vector points in the negative direction of your chosen coordinate system, which is not always the same thing as slowing down. When an object moves in the positive direction and acceleration is negative, it does decelerate. But an object already moving in the negative direction with negative acceleration is speeding up while moving backwards. The reliable test is to compare the signs: when velocity and acceleration share a sign the object speeds up, and when they differ it slows down.
Yes, whenever it changes direction. Uniform circular motion is the classic case: a satellite in a circular orbit or a car on a roundabout at a steady speed has a continuously changing velocity vector because the direction changes at every instant, so it is accelerating even though a speedometer would read constant. This centripetal acceleration always points toward the centre of the circle and has magnitude v²/r, where r is the radius of the path.
G-force expresses acceleration as a multiple of standard gravity, which is defined as exactly 9.80665 m/s², making it an intuitive way to describe forces the human body experiences. An acceleration of 19.6 m/s² is 2 g. Trained pilots in high-performance aircraft can tolerate around 9 g briefly with the aid of a pressure suit, while sustained exposure well below that causes loss of consciousness as blood is forced away from the brain. Roller coasters typically peak around 4 to 6 g for very short intervals.
Identify the one quantity among displacement, initial velocity, final velocity, acceleration and time that the problem neither gives you nor asks for, then pick the equation that omits it. If time is neither known nor wanted, use v² = u² + 2as. If final velocity is the missing one, use s = ut + ½at². This elimination approach turns what looks like four formulas to memorise into a single decision, and it works for every constant-acceleration problem.
No. All four SUVAT equations assume acceleration is uniform throughout the interval, and applying them to varying acceleration gives wrong answers. For non-uniform acceleration you need calculus, integrating the acceleration function to obtain velocity and integrating again for displacement. A common workaround in practice is to split the motion into segments over which acceleration is approximately constant and apply the equations to each segment separately, which is exactly how the worked example above treats its two phases.