Acceleration Calculator

Most acceleration calculators only divide a change in speed by a time, which answers about a third of the questions people arrive with. Give this one any three of starting speed, final speed, acceleration, time and distance, and the other two follow. Every answer also comes back in g, which is the unit your body actually reports.

Fill in any three of the five boxes. The other two always follow.

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How this acceleration calculator works

Most acceleration calculators do one thing: change in speed divided by time. That answers perhaps a third of the questions people actually arrive with, because the other two thirds involve a distance. This one solves the whole set. Give it any three of starting speed, final speed, acceleration, time and distance, and the remaining two follow, because constant acceleration has exactly that much freedom in it and no more.

Every answer comes back in g as well as in your chosen units, because g is the one people can feel, and it comes with a consistency check: the five values are tested against an equation they were not built from, so a wrong branch cannot slip through looking tidy.

The formula

v = u + at
s = ut + ½at²
v² = u² + 2as
s = (u + v)t ÷ 2

u is the starting speed, v the final speed, a the acceleration, t the time and s the distance. These four are the standard kinematic equations, and each one is missing exactly one of the five quantities: the first has no distance, the second no final speed, the third no time, the fourth no acceleration. That is why knowing three is always enough, and it is also how you pick which equation to use by hand: choose the one that omits the quantity you neither know nor want.

Worked example

A car doing 0 to 60 mph in 5.8 seconds, which is an ordinary quick family car. 60 mph is 26.82 m/s, so a = 26.82 ÷ 5.8 = 4.625 m/s², which is 0.472 g. It covers 77.8 m doing it, about the length of a football pitch.

The 0.472 g is the number worth carrying, because it is what you feel. Hard braking in the same car is about 1 g, roughly twice as strong as its best acceleration, which is why stopping distances are shorter than the equivalent accelerating distances and why brakes are the least appreciated part of a fast car.

And the reverse problem. How long to stop from 60 mph at 1 g? Enter 26.82 as the starting speed, 0 as the final, and 9.80665 m/s2 as a deceleration: 2.74 seconds, over 36.7 m. That 36.7 m is the physics floor, before any thinking time. At 70 mph, with the same braking, it becomes 50 m, because the distance goes with the square of the speed.

Acceleration is not speed, and g is the unit that shows it

The most common confusion here is between going fast and accelerating hard, and they are genuinely unrelated. A passenger jet at cruise is doing 550 mph and accelerating at zero; you can walk down the aisle and pour a drink. The same jet on takeoff roll is much slower and accelerating hard, and you are pinned to the seat. What you feel is never speed, it is always acceleration, which is why a smooth train at 200 mph feels like a room and a lift starting to move does not.

Quoting acceleration in g makes that immediate, because g is the acceleration you are experiencing right now while sitting still. Some anchors: a brisk car pulls around 0.3 g, a hard stop about 1 g, a good rollercoaster 3 to 5 g, a fighter pilot in a sustained turn up to 9 g with a pressure suit, and a car crash at survivable speeds can peak in the tens of g for a few hundredths of a second. That last case is the one that matters most, and it is not about the acceleration at all but about how long it lasts, which is where our momentum calculator and the impulse idea take over.

The word constant is doing all the work

These equations describe motion at constant acceleration and nothing else. Real cars do not manage it: they pull hardest in low gear and taper off as drag builds, so a 0 to 60 figure is an average across a curve rather than a rate that was ever actually held. Real brakes fade as they heat. Real falling objects meet air resistance and stop accelerating altogether once drag balances weight, which is why a skydiver reaches a terminal velocity rather than falling faster forever.

That does not make the equations useless, it makes them a summary. An average acceleration is exactly right for questions about the whole journey, and exactly wrong for questions about any particular moment in it. If you need the moment rather than the journey, the honest tool is calculus rather than algebra, and our derivative calculator is where that starts: acceleration is the derivative of velocity, velocity is the derivative of position, and the four equations above are what you get when you integrate a constant.

Frequently asked questions

How do I calculate acceleration?

Divide the change in velocity by the time it took: a equals v minus u over t. A car going 0 to 60 mph (26.82 m/s) in 5.8 seconds accelerates at 4.625 m/s squared. If you have a distance instead of a time, use v squared equals u squared plus 2as, which is why this page asks for any three of the five quantities rather than insisting on a particular pair.

Which kinematic equation should I use?

The one that leaves out the quantity you neither know nor want. Each of the four standard equations is missing exactly one of the five quantities: v = u + at has no distance, s = ut + half at squared has no final speed, v squared = u squared + 2as has no time, and s = (u+v)t/2 has no acceleration. Pick the equation whose missing quantity is the one you do not care about, and it will be solvable in one step.

What is acceleration in g?

The acceleration divided by 9.80665 m/s squared, the standard acceleration of gravity. It is worth quoting because it is the unit your body reports: a brisk car pulls around 0.3 g, hard braking about 1 g, a good rollercoaster 3 to 5 g, and a fighter pilot in a sustained turn up to 9 g. The sensation of being pressed into a seat is exactly this quantity and never the speed.

Is deceleration different from acceleration?

No, it is the same quantity pointing the other way. A negative acceleration means the change in velocity is opposite to the motion, and there is no separate physics for it. The word deceleration is a convenience rather than a distinct concept, which is why this page reports a signed acceleration and explains the sign rather than switching to a different formula.

How far does a car travel while accelerating?

s equals ut plus half at squared, or equivalently the average speed multiplied by the time. A car doing 0 to 60 mph in 5.8 seconds covers 77.8 m, because its average speed over that stretch is half the final speed. Leave the distance box blank above and it comes out alongside the acceleration.

What is a typical braking deceleration?

About 1 g on dry asphalt with good tyres, which is roughly twice as strong as the acceleration of an ordinary quick car. Stopping from 60 mph at 1 g takes 2.74 seconds and 36.7 m, before any reaction time is added. At 70 mph the same braking needs about 50 m, because stopping distance rises with the square of the speed rather than in step with it.

Why does my answer not match a real car?

Because these equations assume constant acceleration and real cars do not manage it. A car pulls hardest in low gear and tapers off as aerodynamic drag builds, so a published 0 to 60 time is an average across a changing curve rather than a rate that was ever held. The same applies to brakes that fade and to falling objects that meet air resistance and reach terminal velocity. The equations give an honest summary of a whole journey and say nothing reliable about any single moment in it.

Can acceleration happen without a change in speed?

Yes, and it is the case this page cannot handle. Acceleration is a change in velocity, and velocity includes direction, so anything moving in a circle at a steady speed is accelerating constantly toward the centre. That is what keeps a car on a curve and a satellite in orbit. These equations describe straight-line motion only, so for circular motion you want a centripetal acceleration formula instead.

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