# What is the acceleration?

Computes average acceleration from the starting speed, the final speed and the time, or any one of the four from the other three, with the distance covered and the acceleration in g.

- Page: https://www.acalculator.org/physics/acceleration-calculator
- JSON spec: https://www.acalculator.org/physics/acceleration-calculator.json
- Version: fc40e31cdd01

## Default answer

Example with the default inputs (Starting speed 0 mph, Final speed 60 mph, Time 8 s): Going from 0 mph to 60 mph in 8 s is an average acceleration of 3.3528 m/s².

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| u | Starting speed | The speed at the start, along the direction of travel. |
| v | Final speed | The speed at the end, along the same direction. |
| t | Time | The time taken to change from the starting speed to the final speed. |
| a | Acceleration (m/s²) | The average acceleration in metres per second squared; negative when the object slows down. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| u | Starting speed | The speed at the start, along the direction of travel. |
| v | Final speed | The speed at the end, along the same direction. |
| t | Time | The time taken to change from the starting speed to the final speed. |
| a | Acceleration | The average acceleration: the change in speed divided by the time, in m/s². |
| g | Acceleration in g | The acceleration divided by standard gravity, 9.80665 m/s². |
| fts2 | Acceleration in ft/s² | The acceleration in feet per second squared: m/s² ÷ 0.3048. |
| distance | Distance covered | The distance covered at constant acceleration: (u + v) ÷ 2 × t. |

## Method

a = (v − u) ÷ t; v = u + a × t; u = v − a × t; t = (v − u) ÷ a; distance = (u + v) ÷ 2 × t.

## Assumptions

- Motion is along a straight line, and the speeds are in the direction of travel (0 or more).
- The acceleration is the average over the time; the distance assumes it is constant.
- Units are converted exactly: 1 mph = 0.44704 m/s, 1 ft = 0.3048 m, standard gravity g = 9.80665 m/s² (NIST SP 811).

## Worked examples

1. u = 0, v = 26.8224, t = 8 gives a = 3.3528, g = 0.34189, distance = 107.2896. Source: OpenStax University Physics Volume 1, section 3.4, Motion with Constant Acceleration (https://openstax.org/books/university-physics-volume-1/pages/3-4-motion-with-constant-acceleration, retrieved 2026-10-01); NIST SP 811: 1 mph = 0.44704 m/s.
2. u = 70, v = 0, a = -5 gives t = 14, distance = 490. Source: OpenStax University Physics Volume 1, section 3.4, Motion with Constant Acceleration, equation v = v₀ + at (https://openstax.org/books/university-physics-volume-1/pages/3-4-motion-with-constant-acceleration).
3. u = 10, a = 2, t = 5 gives v = 20, distance = 75, fts2 = 6.56168. Source: OpenStax University Physics Volume 1, section 3.4, Motion with Constant Acceleration (https://openstax.org/books/university-physics-volume-1/pages/3-4-motion-with-constant-acceleration).

## FAQ

### How do I calculate acceleration?

Subtract the starting speed from the final speed and divide by the time: a = (v − u) ÷ t. A car going from 0 to 60 mph (26.8224 m/s) in 8 seconds accelerates at 26.8224 ÷ 8 = 3.3528 m/s².

### What are the units of acceleration?

Speed per unit of time. In SI units it is metres per second squared (m/s²): the speed changes by that many metres per second every second. The calculator also shows feet per second squared and g.

### What does negative acceleration mean?

The object slows down: the final speed is lower than the starting speed. Braking from 70 m/s to a stop in 14 seconds is an acceleration of −5 m/s², often called a deceleration of 5 m/s².

### What is acceleration in g?

The acceleration divided by standard gravity, 9.80665 m/s². 1 g is the acceleration of a dropped object near the Earth’s surface, ignoring air. 0 to 60 mph in 8 seconds is about 0.34 g.

### How do I find the time from the acceleration?

Rearrange the formula: t = (v − u) ÷ a. Fill the two speeds and the acceleration, and leave the time empty.

### How far does the object travel while it accelerates?

At constant acceleration, the distance is the average of the two speeds times the time: d = (u + v) ÷ 2 × t. From 0 to 26.8224 m/s in 8 seconds that is 13.4112 × 8 = 107.29 m.

### Is this the same as instantaneous acceleration?

Only when the acceleration is constant. Otherwise the result is the average acceleration over the time, which says nothing about the peak.

## Sources

- OpenStax, University Physics Volume 1, section 3.3, Average and Instantaneous Acceleration (a = Δv ÷ Δt), CC BY 4.0, retrieved 2026-10-01. https://openstax.org/books/university-physics-volume-1/pages/3-3-average-and-instantaneous-acceleration
- OpenStax, University Physics Volume 1, section 3.4, Motion with Constant Acceleration (v = v₀ + at, average velocity (v₀ + v) ÷ 2), CC BY 4.0, retrieved 2026-10-01. https://openstax.org/books/university-physics-volume-1/pages/3-4-motion-with-constant-acceleration
- NIST Special Publication 811, Guide for the Use of the International System of Units, Appendix B (mile per hour = 0.44704 m/s, foot = 0.3048 m, standard acceleration of gravity = 9.80665 m/s²). https://www.nist.gov/pml/special-publication-811