How to use the Acceleration Calculator
- Enter the initial and final velocity in matching units.
- Enter the time taken for that change.
- Read acceleration in m/s², and in g where useful for comparison.
- Use the force field to find the force required for that acceleration on a known mass.
How the calculation works
Average acceleration is the change in velocity divided by the elapsed time. It is a vector, so deceleration is simply negative acceleration in the direction of travel, and a car cornering at constant speed is still accelerating because its direction is changing. Standard gravity is 9.80665 m/s², which is why quoting an acceleration in g gives an intuitive sense of scale.
Newton's second law connects acceleration to cause: F = m × a. That is what turns an acceleration requirement into a force, and therefore into a motor, actuator or brake specification. For braking distance the useful form is derived from the kinematic equation v² = u² + 2as, which shows distance rising with the square of speed — doubling speed quadruples the stopping distance at the same deceleration.
a = (v − u) / t ; F = m × a ; s = ut + ½at² ; v² = u² + 2asSource: Newtonian kinematics and Newton's second law; standard gravity 9.80665 m/s² per CGPM.
Worked example
A 1,480 kg car brakes from 27 m/s (about 97 km/h) to rest in 3.4 seconds.
- a = (0 − 27) / 3.4 = −7.94 m/s².
- In g: 7.94 / 9.807 = 0.81 g.
- Braking force = 1,480 × 7.94 = 11,751 N.
- Distance = v²/2a = 27² / (2 × 7.94) = 45.9 m.
Deceleration of about 7.9 m/s² (0.81 g), needing roughly 11.8 kN of braking force over 46 m.
Frequently asked questions
What does an acceleration in g actually mean?+
It expresses the acceleration as a multiple of standard gravity, 9.80665 m/s². One g sideways in a corner means the lateral force equals the vehicle's weight, which is around the limit for road tyres.
Is deceleration a different quantity?+
No, it is acceleration opposite to the direction of motion, so it appears as a negative value. The magnitude is what matters for calculating force and stopping distance.
Why is my calculated 0-60 time faster than the real one?+
The simple formula assumes constant acceleration and perfect traction. Real launches lose time to wheelspin, gearshifts and a torque curve that varies with engine speed.
Does this work for objects in free fall?+
Yes, using 9.81 m/s² as the acceleration, provided air resistance is negligible. Once drag matters — a skydiver, a light object — velocity approaches a terminal value and acceleration falls towards zero.
Last reviewed August 31, 2026. We review this page whenever the underlying formula, tax year, published rate or standard changes.