Acceleration Calculator
Acceleration from a change in speed over a time or a distance, with the full motion solved.
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Your details
Result
- Acceleration
- 5.5556 m/s²
- Change in speed
- 27.7778 m/s
- Distance covered
- 69.444 m
- Average speed
- 13.889 m/s
0.567 g
From 0 to 27.778 m/s
The mean of start and finish, because the acceleration is constant
From 0 to 27.778 m/s in 5 s…
a = (v − u) ÷ t
v − u = 27.778 − 0 = 27.7778
27.7778 ÷ 5
a = 5.5556 m/s²
Formula — written step by step
0 to 100 km/h in 5 seconds…
a = (v − u) ÷ t
v = 100 km/h = 27.777778 m/s
v − u = 27.777778 − 0
27.777778 ÷ 5
a = 5.5556 m/s²
u is the starting speed, v the final speed and t the time taken. That works out at about 0.57 g — the acceleration itself is modest; it is the sustained nature of it that makes a car quick.
A worked example
Braking distance from a speed and a rate
- Starting speed
- 30 m/s
- Final speed
- 0
- Distance to stop
- 75 m
- a = (v² − u²) ÷ 2s
- v² − u² = 0 − 900 = −900
- a = −900 ÷ 150
- a = −6 m/s², or 0.61 g
A deceleration of 6 m/s²
Around 6 m/s² is realistic hard braking on dry asphalt for a car with good tyres; the physical ceiling is set by the friction available, not the brakes. Doubling the starting speed to 60 m/s at the same rate needs 300 m — four times the distance, because the distance depends on the square of the speed.
How to use this calculator
- Enter the starting and final speeds and pick their unit — both are converted to metres per second.
- Choose whether you know the time taken or the distance covered; the calculator uses a different equation of motion for each.
- A negative result means deceleration, and is labelled as such.
- These equations assume the acceleration is constant. For real journeys with varying rates, they give an average rather than an instantaneous value.
Acceleration is a rate of change
Acceleration measures how quickly velocity changes, in metres per second per second. A car going from rest to 27.8 m/s in five seconds accelerates at 5.56 m/s² — every second, its speed rises by 5.56 m/s.
Because velocity is a vector, acceleration includes changes of direction as well as speed. An object moving in a circle at constant speed is accelerating continuously, because its direction is changing even though its speed is not.
Choosing the right equation
When you know the time, a = (v − u) ÷ t follows directly from the definition. When you know the distance instead, use a = (v² − u²) ÷ 2s, which comes from the same relationships with time eliminated.
The distance form is the more useful one for braking problems, because stopping distance is what you actually care about and the time is rarely known. It also makes the squared dependence explicit: at twice the speed, stopping at the same rate takes four times the distance.
The constant-acceleration assumption
Every equation here assumes the acceleration does not change. That is a good approximation for free fall over short distances, for a braking car with locked-in friction, and for many textbook problems.
It is a poor one for a car accelerating through its gears, where the rate falls off markedly at higher speeds, or for anything where drag is significant. In those cases the result is an average over the interval, which is still useful — provided you know that is what it is.
Formula & methodology
What this calculator assumes
- Assumes constant acceleration throughout. For a car changing gear or anything with significant drag, the result is an average over the interval rather than an instantaneous value.
- Straight-line motion. Circular motion involves acceleration even at constant speed, which these equations do not describe.
- Ignores air resistance, so free-fall figures are upper bounds.
Sources
- Young & Freedman, University Physics, 15th edition, Pearson, 2019 — motion with constant acceleration.
The constant-acceleration equations
| Known | Equation | Gives |
|---|---|---|
| u, v, t | a = (v − u) ÷ t | Acceleration |
| u, v, s | a = (v² − u²) ÷ 2s | Acceleration |
| u, a, t | v = u + at | Final speed |
| u, v, t | s = (u + v)t ÷ 2 | Distance |
| u, a, t | s = ut + ½at² | Distance |
Known as the SUVAT equations, after the five symbols. Each omits one variable, so the one to use is whichever leaves out the quantity you neither know nor want.
Frequently asked questions
What does a negative acceleration mean?
That the object is slowing down, if it is moving in the positive direction. It is the same quantity as acceleration, just opposite in sign, and is labelled deceleration above.
What is 1 g?
The acceleration of free fall at the Earth’s surface, 9.80665 m/s². Expressing an acceleration in g is the most intuitive way to judge how it would feel.
Why does the distance form use squares?
Because it comes from eliminating time between the velocity and distance equations. The practical consequence is that stopping distance grows with the square of speed.
Can I use this for free fall?
Yes. Enter a starting speed of zero and the acceleration works out at about 9.81 m/s², provided air resistance is negligible.