Example 1
A start from rest, a = 3.5 m/s², t = 8 s. This is a textbook problem of reaching about 62.6 mph.
s = 112.00 m, vₖ = 28.00 m/s (100.8 km/h)
Uniformly Accelerated Motion Calculator
Enter initial speed v₀, constant acceleration a, and time t. The calculator finds distance s = v₀·t + ½·a·t² and final speed vₖ = v₀ + a·t. From rest at 3.5 m/s² for 8 s the distance is 112 m and the final speed is 28 m/s, which is 100.8 km/h. The model does not include air drag or gear changes.
Result
Enter initial speed, acceleration, and a non-negative time. Distance, final speed, and the distance chart appear here.
a = Δv/t alone is the acceleration calculator, a stop under braking is uniformly decelerated motion, and force is F = m·a.
How it works
½ a t²112 m
v₀ t0 m
s112 m
Uniformly accelerated motion is straight-line motion in which speed changes by a constant amount each second. The calculator finds distance and final speed from s = v₀·t + ½·a·t² and vₖ = v₀ + a·t. For v₀ = 0, a = 3.5 m/s², and t = 8 s the shortened formula gives s = ½·3.5·64 = 112 m, and vₖ = 28 m/s. Average speed is (0 + 28)/2 = 14 m/s, and 14 m/s · 8 s = 112 m. The panel shows those values to two decimal places: 112.00 m and 28.00 m/s (100.8 km/h).
The calculator uses the numbers you enter: initial speed v₀ in m/s, acceleration a in m/s², and time t in seconds. A blank field is not zero. A start from rest requires 0 to be entered. For v₀ = 0, a = 1.2 m/s², and t = 10 s the result is 60.00 m and 12.00 m/s (43.2 km/h). When the initial speed is not zero, ½·a·t² alone is not enough: at 5 m/s, 0.8 m/s², and 6 s the distance is 5·6 + ½·0.8·36 = 44.40 m, and the final speed is 9.80 m/s (35.3 km/h).
What the result panel shows. Besides distance and final speed, the calculator reports average speed (v₀ + vₖ)/2, the change in speed a·t, and the entered v₀, a, and t. An acceleration of 3.5 m/s² appears in the panel as 3.50 m/s², and a time of 8 s appears as 8.00 s. The chart draws distance against time: at constant a that curve is part of a parabola. A comma and a period in a field mean the same number.
Limits of the model. The formula assumes constant a along a straight line. It does not include air resistance or gear changes. The acceleration calculator finds a = Δv/t from two speeds. Uniformly decelerated motion takes a positive deceleration and stops the result at zero speed. Here a negative a is allowed, and the motion is computed for the whole time t, including after speed crosses zero. A negative s is then a displacement along the axis, not the length of the path. Force F = m·a is a separate calculator, because this one has no mass field. A negative time produces no result.
The unit switch in the header converts displayed meters to feet and m/s to ft/s, but the formula stays the same. Numbers in the text are in SI as the reference. The number 50 in a field labeled m/s means 180 km/h, about 112 mph, not 50 mph. Sixty miles per hour is entered as about 26.8 m/s. A time of zero gives s = 0 and vₖ = v₀. A blank time field is not zero.
Further examples on the page: v₀ = 0, a = 0.55 m/s², t = 80 s gives 1,760.00 m and 44.00 m/s (158.4 km/h). v₀ = 0, a = 2 m/s², t = 35 s gives 1,225.00 m and 70.00 m/s (252.0 km/h). v₀ = 0, a = 4.5 m/s², t = 2 s gives 9.00 m and 9.00 m/s (32.4 km/h). v₀ = 2 m/s, a = 1.5 m/s², t = 4 s gives 20.00 m and 8.00 m/s (28.8 km/h). The button “Fill this example” enters the same numbers and refreshes the panel and the chart.
Formula
s = v0t + ½a t2
vk = v0 + a t
Average speed on the interval: (v0 + vk)/2. At constant a it equals the speed at half the time.
Model: uniformly accelerated motion in a straight line (constant-acceleration kinematics; Galileo, Discorsi, 1638). The calculator always takes v₀, a, and t. It does not solve for an arbitrary missing quantity from the SUVAT set.
Theory, derivation, and domain
Definition of uniformly accelerated motion
Uniformly accelerated motion is motion in which speed changes linearly with time. Acceleration is constant. Instantaneous speed is v(t) = v₀ + a·t, and displacement from time zero is s(t) = v₀·t + ½·a·t². The SI unit of a is the meter per second squared. In a first course the name often means a positive a, so speed is increasing. This calculator uses the same formulas for a negative a as well, which means speed is falling, and it does not cut the result off at v = 0.
The model describes speeding up on a straight line in a textbook problem, a short start on a track, and a cart pushed by a constant net force, as long as a really does not change with time. It does not describe a launch with gear changes, a climb with a changing grade, or a fall with air resistance. Mass is not in the formula: at a given a the distance does not depend on mass.
History: Galileo and the square of time
Galileo Galilei, in Discorsi e dimostrazioni matematiche intorno a due nuove scienze (1638), described motion at constant acceleration and the inclined-plane experiments. Distance from rest grows with the square of time. That is the family of equations that textbooks call the equations of uniformly accelerated motion, or the SUVAT equations. The relations v = v₀ + a·t and s = v₀·t + ½·a·t² belong to a standard kinematics course (Halliday, Resnick, and Walker; OpenStax University Physics, the section on motion with constant acceleration).
There is no separate discoverer of a “speed-up formula.” It is the integral of a constant acceleration. When v₀ = 0, what remains is s = ½·a·t². When the final speed is zero and the signed acceleration is negative, the same family becomes the braking-distance formula. That second case, with a positive deceleration and a stop, is the uniformly decelerated motion calculator.
Derivation of the formulas this calculator uses
Step 1. Constant a means the speed change over time t is a·t, so vₖ = v₀ + a·t. Step 2. With a linear change in speed, the average on the interval is (v₀ + vₖ)/2 = v₀ + ½·a·t. Step 3. Displacement is that average times the time: s = (v₀ + ½·a·t)·t = v₀·t + ½·a·t². Step 4. Check for v₀ = 0, a = 3.5 m/s², and t = 8 s: vₖ = 28 m/s, the average is 14 m/s, s = 14·8 = 112 m, and ½·3.5·64 = 112 m. Step 5. Check with a nonzero v₀: 5 m/s, 0.8 m/s², and 6 s give vₖ = 9.8 m/s and s = 30 + 14.4 = 44.4 m.
The identity s = v_avg·t is not a second model. At constant a the speed at half the time equals the average, because v(t/2) = v₀ + ½·a·t. For the worked example that is 14 m/s at t = 4 s.
What this calculator does not find
The calculator always reads three fields: v₀, a, and t. It does not pick a missing quantity from the five symbols of the SUVAT set. If you know two speeds and a time, and you do not know a, open the acceleration calculator: a = Δv/t. If you want the distance to a stop at a positive deceleration, open uniformly decelerated motion. When the mass is known and the question is the force, open F = m·a. Kinetic energy at a known vₖ is on its own page.
A negative s means displacement opposite the positive axis, not a “negative path length” along a road. When speed changes sign during t, the path length is longer than the absolute value of s. This calculator reports s from the formula, without splitting the motion into pieces before and after the turnaround.
Orders of magnitude
A start from rest at 3.5 m/s² for 8 s gives 112 m and 28 m/s (100.8 km/h, about 62.6 mph). That is the scale of a textbook car problem, not a claim about one model. A gentler 1.2 m/s² for 10 s gives 60 m and 12 m/s (43.2 km/h). A long interval at 0.55 m/s² and 80 s gives 1,760 m and 44 m/s. A short burst of 4.5 m/s² for 2 s gives 9 m and 9 m/s. Values of 3–5 m/s² are the scale of passenger-car acceleration in textbook problems, not a homologation figure.
Sources
The definition and the derivation come from mechanics textbooks: D. Halliday, R. Resnick, and J. Walker, Fundamentals of Physics, the chapter on motion with constant acceleration; OpenStax, University Physics, Volume 1, “Motion with Constant Acceleration”; H. D. Young and R. A. Freedman, University Physics, “Motion Along a Straight Line”; R. P. Feynman, R. B. Leighton, and M. Sands, The Feynman Lectures on Physics, Volume I, Chapter 8; Galileo Galilei, Discorsi, 1638; BIPM, The International System of Units (SI), 9th edition (meter, second). The page is educational.
How to use
- Enter initial speed v₀ in meters per second. Enter 0 for a start from rest. A blank field does not stand in for zero.
- Enter constant acceleration a, for example 3.5 m/s². A negative number is allowed and means the speed is falling.
- Enter time t in seconds, for example 8. A negative time produces no result. A time of 0 gives s = 0 and vₖ = v₀.
- Click Calculate, or wait for the live result. For 0, 3.5, and 8 the calculator shows 112.00 m and 28.00 m/s.
- Clear removes the inputs and hides the panel. Acceleration from two speeds is the acceleration calculator.
Symbol glossary
The calculator always reads v₀, a, and t, then finds s and vₖ. For v₀ = 0, a = 3.5 m/s², and t = 8 s the result is 112.00 m and 28.00 m/s.
- v₀
- Initial speed [m/s]. An entered 0 is a start from rest. A blank field does not run.
- a
- Constant acceleration [m/s²], with a sign. A positive value raises v, and a negative value lowers it.
- t
- Time of motion [s], greater than or equal to zero. Zero gives s = 0 and vₖ = v₀.
- s
- Displacement [m], s = v₀·t + ½·a·t². If speed changes sign, this is not the path length.
- vₖ
- Final speed [m/s], v₀ + a·t. For the worked example it is 28 m/s.
- v_avg
- Average speed [m/s], (v₀ + vₖ)/2. At constant a it equals the speed at half the time.
Real-life examples
Example 2
A gentle run-up from rest: a = 1.2 m/s² for 10 s.
s = 60.00 m, vₖ = 12.00 m/s (43.2 km/h)
Example 3
Motion begins at 5 m/s on level ground, then a = 0.8 m/s² acts for 6 s.
s = 44.40 m, vₖ = 9.80 m/s (35.3 km/h)
Example 4
A long, gentle interval: v₀ = 0, a = 0.55 m/s², t = 80 s.
s = 1,760.00 m, vₖ = 44.00 m/s (158.4 km/h)
Example 5
A stronger run-up from rest: a = 2 m/s² for 35 s.
s = 1,225.00 m, vₖ = 70.00 m/s (252.0 km/h)
Example 6
A short burst from rest: a = 4.5 m/s² for 2 s.
s = 9.00 m, vₖ = 9.00 m/s (32.4 km/h)
Example 7
A very short, gentle interval: a = 1.2 m/s² for 2 s, from rest.
s = 2.40 m, vₖ = 2.40 m/s (8.6 km/h)
Example 8
Motion is already under way: v₀ = 2 m/s, a = 1.5 m/s², t = 4 s.
s = 20.00 m, vₖ = 8.00 m/s (28.8 km/h)
Ways to use this calculator
- You compute distance and final speed at constant a, for example a start from rest at 3.5 m/s² for 8 s.
- You check how much the v₀·t term adds when the motion does not start from rest.
- You compare vₖ with a speed limit after a known acceleration time, without drag and without gear changes.
Frequently asked questions
What does this calculator find, and from which formula?
The calculator finds distance and final speed in uniformly accelerated motion: s = v₀·t + ½·a·t² and vₖ = v₀ + a·t. Acceleration a stays constant for the whole time t. For v₀ = 0, a = 3.5 m/s², and t = 8 s the result is 112.00 m and 28.00 m/s (100.8 km/h). The calculator does not find acceleration from two speeds.
Which units should I enter for speed, acceleration, and time?
Enter initial speed in meters per second, acceleration in meters per second squared, and time in seconds. About 60 mph is 26.8 m/s. The unit switch in the header converts the result to feet and ft/s, but seconds stay seconds. The number 50 in the m/s field means 50 m/s, about 112 mph, not 50 mph.
What happens when a field is left blank?
The calculator requires three numbers: v₀, a, and t. A blank field is not zero. A start from rest requires 0 to be entered. Missing acceleration or time, and a negative time, produce no result and a message appears.
How is this result different from the acceleration calculator?
The acceleration calculator finds a = Δv/t from two speeds and a time. Here a is given and constant, and the unknowns are distance s and final speed vₖ. One pair of speeds does not replace these two equations.
Can acceleration be negative?
Yes. A negative a means the speed is falling, and the formulas stay the same. The calculator does not stop the motion when speed crosses zero, and it computes the whole interval t. Braking to a stop, with a positive deceleration and a cutoff at v = 0, is on the uniformly decelerated motion page.
Why does distance from rest grow with the square of time?
When v₀ = 0, the formula shortens to s = ½·a·t². Twice the time at the same a makes the distance four times longer. For 3.5 m/s² and 8 s that is ½·3.5·64 = 112 m. When v₀ is not zero, the term v₀·t remains, and ½·a·t² alone is not the whole distance.
What does average speed in the panel mean?
It is the average speed from v₀ to vₖ, equal to (v₀ + vₖ)/2, because at constant a the speed changes in a straight line. From rest to 28 m/s the average is 14 m/s, and 14 m/s times 8 s is 112 m. At constant a the speed at half the time equals that average.
Does the calculator include mass, drag, or gear changes?
No. This is kinematics at constant a. Force F = m·a has its own calculator, and kinetic energy at vₖ has another. Air resistance and gear changes cause the acceleration to vary with time. The formula is then an approximation for one average acceleration.
How do I convert meters per second to miles per hour?
Multiply m/s by 2.2369. 28 m/s is about 62.6 mph. The other way, divide mph by 2.2369: 60 mph is about 26.8 m/s. Multiplying m/s by 3.6 gives kilometers per hour: 28 m/s is 100.8 km/h.
Why does a time of zero produce a result, while a blank time field does not?
A time of 0 is allowed: s = 0 and vₖ = v₀, because there is no displacement on that interval. A blank time field is not zero and produces no result. A comma and a period in a field mean the same number, for example 3,5 and 3.5.
How do I calculate distance with s = v₀t + ½at²?
Multiply v₀ by t, then add half of a·t². From rest with v₀ = 0, a = 3.5 m/s², and t = 8 s: ½·3.5·64 = 112 m.
What is vₖ at v₀ = 0, a = 3.5 m/s², and t = 8 s?
vₖ = v₀ + a·t = 0 + 3.5·8 = 28 m/s, which is 100.8 km/h. Average speed on that interval is (0 + 28)/2 = 14 m/s.
How do I find distance when v₀ is not zero?
Add the v₀·t term to ½·a·t². For 5 m/s, 0.8 m/s², and 6 s: 5·6 + ½·0.8·36 = 44.40 m, and vₖ = 9.80 m/s.
How is this page different from the braking-distance calculator?
Here you enter v₀, a, and t, and the results are s and vₖ. The decelerated-motion page computes s = v₀²/(2a) to a stop with positive deceleration.
Does the calculator solve any three of the five SUVAT quantities?
No. The fields are v₀, a, and t; the unknowns are s and vₖ. There is no mode for v² = v₀² + 2as without time.
Knowledge sources
The definition, the history, and the derivation below come from established physics and metrology sources.
- OpenStax, University Physics Volume 1 — Motion with Constant Acceleration
- Wikipedia — Equations of motion
- Feynman Lectures on Physics, Vol. I, Chapter 8 — Motion
- BIPM — The International System of Units (SI), 9th edition
- NIST SP 330 — The International System of Units
- Galilei, G., Discorsi e dimostrazioni matematiche intorno a due nuove scienze, 1638.
- Halliday, D.; Resnick, R.; Walker, J., Fundamentals of Physics — motion with constant acceleration.
- Young, H. D.; Freedman, R. A., University Physics — Motion Along a Straight Line.
Page updated November 3, 2026.