Kinematics (SUVAT) Calculator
Solve constant-acceleration motion. Give any three of displacement, initial velocity, final velocity, acceleration and time, and the other two are worked out with the right SUVAT equation, a velocity–time graph and every step.
How to Use
- Pick the two quantities you want to find in the Find bar. The other three fields stay open for the values you know.
- Enter the three known values and choose a unit for each: metres or feet, m/s, km/h or mph, m/s² or g, seconds or minutes.
- Choose one direction as positive. A velocity or acceleration pointing the other way is negative, so braking from 30 m/s is a = −6 m/s².
- Read the two answers in the readouts and in the boxes beside their fields; change an answer’s unit at any time to convert it.
- Check the velocity–time graph (its shaded area is the displacement) and the Show Work section for the equation used at each step.
Worked Example
A car goes from 0 to 100 km/h in 8 s. Convert first: 100 km/h = 100 ÷ 3.6 = 27.78 m/s. With u, v and t known, a = (v − u) ÷ t = 27.78 ÷ 8 = 3.47 m/s² (0.354 g), and s = ½(u + v)t = ½ × 27.78 × 8 = 111.1 m.
A ball thrown straight up at 15 m/s. Take up as positive, so a = −9.80665 m/s². At the top v = 0, so t = (0 − 15) ÷ −9.80665 = 1.53 s and s = (0 − 15²) ÷ (2 × −9.80665) = 11.47 m. Asking when it is 10 m up gives a quadratic with two positive roots, 0.98 s and 2.08 s: once rising, once falling.
The common mistake: a positive acceleration for braking. A car braking from 30 m/s at 6 m/s², with a entered as +6, gives s = (0 − 30²) ÷ (2 × 6) = −75 m: a negative stopping distance for a car that went forward. With the sign right, a = −6 m/s², s = (0 − 900) ÷ (−12) = 75 m and t = 5 s.
Show Work
Formulas
From Oxford to Galileo
The equation s = ½(u + v)t is older than Newton. In the 1330s a group of scholars at Merton College, Oxford (William Heytesbury, Richard Swineshead and John Dumbleton) stated the mean speed theorem: a body accelerating uniformly covers the same distance as one moving steadily at its average speed. Around 1350 Nicole Oresme in Paris proved it with a diagram in which the distance is the area of a triangle under a speed line, the same picture as the velocity–time graph above.
Galileo Galilei tested the idea with balls rolling down inclined planes and published the result in 1638 in Two New Sciences: distance from rest grows with the square of time, so the distances in successive equal intervals go 1 : 3 : 5 : 7. Newton’s second law (1687) then explained why: a constant force gives a constant acceleration. The letters s, u, v, a, t are the British school convention, which is where the name SUVAT comes from.
All five equations assume the acceleration is constant. Air resistance, a car changing gear or a rocket burning fuel break that assumption; for those the motion has to be split into short pieces or integrated numerically.
About This Calculator
This calculator solves straight-line motion with constant acceleration for all ten pairs of unknowns. It chooses the SUVAT equation that leaves out the quantity you didn’t give, solves the quadratic for time when it has to and explains which roots are physical, and checks the answer against v² = u² + 2as. Every field takes its own unit, so km/h, feet and g can be mixed; the working shows each conversion to SI.
The velocity–time graph shades the displacement, above the axis for forward motion and below it for backward motion, and the calculator reports the distance travelled separately when the object turns round. Everything runs in your browser; nothing is sent anywhere.
Related tools: Projectile Motion Calculator, Kinetic & Potential Energy Calculator, and Speed, Distance & Time Calculator.
Frequently Asked Questions
Which SUVAT equation should I use?
Use the one that leaves out the quantity you neither know nor need. For a car braking from 30 m/s at 6 m/s² with no time given, v² = u² + 2as leaves out t: s = (0 − 30²) ÷ (2 × −6) = 75 m. This calculator picks the equation for you and shows it in the working.
Why does it sometimes give two times?
Finding t from s = ut + ½at² means solving a quadratic, which has two roots. A ball thrown up at 15 m/s is 10 m above your hand at 0.98 s on the way up and again at 2.08 s on the way down; both are real. A negative root is a time before the start and is rejected.
Is deceleration a negative acceleration?
Yes, when the object moves in the positive direction. Braking from 30 m/s at 6 m/s² is a = −6 m/s². Entering +6 would mean speeding up, and the calculator reports that the velocity moves away from 0 instead of reaching it.
Can I use SUVAT for falling objects?
Yes, with a = g = 9.80665 m/s² as long as air resistance is small. A stone dropped from 20 m lands after 2.02 s at 19.8 m/s. A skydiver is different: drag grows with speed until it balances weight, so the acceleration is not constant.
What is the difference between displacement and distance?
Displacement is the change in position, with a sign; distance is the length of the path. A ball thrown up at 15 m/s and caught 3.06 s later has a displacement of 0 m but has travelled 22.9 m (11.47 m up and 11.47 m down). The calculator shows both.
How do I use the Kinematics (SUVAT) Calculator?
Just type your numbers. The answer shows up right away — there is no button to press. Change anything and it updates by itself.
Do I need to install or sign up for anything?
Not at all — it runs in the browser with nothing to install and no account. After it loads once, it even works without an internet connection.
Is my information private?
Yes. Everything happens in your browser. Nothing you type is sent to a server or saved anywhere.
Common Use Cases
Car acceleration figures
0–60 mph (26.82 m/s) in 6 s is an average of 4.47 m/s², or 0.456 g, covering 80.5 m.
Stopping distances
From 70 mph (31.29 m/s), braking at 7 m/s² takes 4.47 s and 69.9 m, before adding the distance covered while reacting.
Physics homework
A train at 10 m/s accelerating at 0.5 m/s² for 40 s reaches 30 m/s and covers 800 m.
Depth of a drop
A stone that takes 3 s to hit the water fell s = ½gt² = 44.1 m and arrived at 29.4 m/s, ignoring air resistance.
Runway length
An aircraft accelerating at 2.5 m/s² to a take-off speed of 70 m/s needs 980 m of runway and 28 s.
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