From rest at 3 m/s² for 8 s
- Find
- s and v (know u, a, t)
- Initial velocity u
- 0 m/s
- Acceleration a
- 3 m/s²
- Time t
- 8 s
- Final velocity v
- 24 m/s
- Displacement s
- 96 m
Checked against: v = 0 + 3·8 = 24; s = ½·3·8² = 96
Kinematics (SUVAT) calculator
Kinematics calculator for the SUVAT equations: enter three of displacement, initial and final velocity, acceleration and time to get the other two.
By AdminDrafted with AI writing toolsUpdated Checked examples: 6
Final velocity v
m/s
Final velocity v: 20 m/s
- Displacement s
- 100m
- Initial velocity u
- 0m/s
- Acceleration a
- 2m/s²
- Time t
- 10s
Displacement is 100 m and final velocity is 20 m/s. The shaded area under the velocity line equals the displacement.
Velocity against time
Displacement against time
How it's calculated S
Knowns in SI units
Final velocity
Displacement
About the kinematics (SUVAT) calculator
Motion with constant acceleration is described by five quantities: displacement s, initial velocity u, final velocity v, acceleration a and time t. Each of the five SUVAT equations leaves one of them out, so any three known values fix the other two. When time is unknown, s = ut + ½at² is a quadratic and can have two valid roots.
Physics students use it for braking, free-fall and launch problems. With the defaults, an object starting from rest and accelerating at 2 m/s² for 10 s reaches 20 m/s and covers 100 m. A ball thrown straight up at 20 m/s passes 15 m twice, at 0.99 s and 3.09 s, and both times are listed.
Acceleration must stay constant and the motion must lie along one line. Values are signed: choose a positive direction and give opposing quantities a minus sign, so free fall with up as positive uses a = −9.80665 m/s².
SUVAT equations for constant acceleration
Under constant acceleration, five equations link displacement s, initial velocity u, final velocity v, acceleration a and time t: v = u + at, s = ut + ½at², v² = u² + 2as, s = ½(u + v)t and s = vt − ½at². A car at 10 m/s accelerating at 2 m/s² for 5 s reaches 20 m/s in 75 m.
The motion must also stay on one straight line. Know any three of the five quantities and the other two follow. OpenStax's constant-acceleration lesson derives all five.
Which SUVAT equation to use
Each equation leaves out one quantity. Pick the one that omits the value you neither know nor need:
- v = u + at has no s.
- s = ut + ½at² has no v.
- v² = u² + 2as has no t.
- s = ½(u + v)t has no a.
- s = vt − ½at² has no u.
For the car above, the distance without the final speed comes from s = ut + ½at² = 10 × 5 + ½ × 2 × 5² = 50 + 25 = 75 m.
Every row below describes one motion from rest: u = 0, a = 2 m/s², t = 10 s, v = 20 m/s and s = 100 m. The first column names the two unknowns, so sv means "find s and v from u, a and t".
| Find (unknowns) | Known SI values | Calculated SI values |
|---|---|---|
| sv | u=0, a=2, t=10 | s=100, v=20 |
| vt | u=0, a=2, s=100 | v=20, t=10 |
| st | u=0, a=2, v=20 | s=100, t=10 |
| sa | u=0, t=10, v=20 | s=100, a=2 |
| va | u=0, t=10, s=100 | v=20, a=2 |
| at | u=0, s=100, v=20 | a=2, t=10 |
| ua | t=10, s=100, v=20 | u=0, a=2 |
| ut | a=2, s=100, v=20 | u=0, t=10 |
| uv | a=2, t=10, s=100 | u=0, v=20 |
| su | a=2, t=10, v=20 | s=100, u=0 |
Speed, distance and time formula
At constant speed, distance = speed × time, so speed = distance ÷ time and time = distance ÷ speed. Covering 100 m at 5 m/s takes 100 ÷ 5 = 20 s. This is the SUVAT case with a = 0.
Average speed over a trip is total distance divided by total time, not the average of the speeds. Driving 60 km at 30 km/h and coming back at 60 km/h takes 2 h + 1 h = 3 h for 120 km: an average of 40 km/h, not 45.
Velocity formulas
Velocity is speed with a direction. Under constant acceleration the final velocity is v = u + at, or v² = u² + 2as when the time is unknown. A cyclist at 4 m/s who accelerates at 0.5 m/s² for 6 s reaches 4 + 0.5 × 6 = 7 m/s.
Average velocity is displacement divided by time, and with constant acceleration it equals (u + v)/2. A runner who completes a 400 m lap in 80 s has an average speed of 5 m/s but an average velocity of zero, because the lap ends where it started.
Signs and direction
Choose a positive direction and keep it for every quantity. With up as positive, a stone dropped from rest has a = −9.80665 m/s², the standard value of g. After falling 20 m it is at s = −20 m, moving at v = −19.81 m/s, 2.02 s after release.
A car braking from 10 m/s to rest in 50 m has a = (0² − 10²) ÷ (2 × 50) = −1 m/s² and stops after t = 2 × 50 ÷ (10 + 0) = 10 s. The minus sign means the acceleration points against the motion.
Negative acceleration slows an object only while it moves in the positive direction. Starting at −5 m/s with a = −2 m/s², the velocity after 3 s is −11 m/s: the object is speeding up.
Two answers for the time
When time is the unknown, s = ut + ½at² is a quadratic in t. A ball thrown up at 20 m/s with a = −10 m/s² is 15 m above the start at t = 1 s, rising at 10 m/s, and again at t = 3 s, falling at −10 m/s.
Both roots are real events, so keep the one the question asks about. A negative root lies before the start and is discarded. A negative discriminant means the object never reaches that displacement.
Common mistakes
- Mixing km/h and m/s. Divide km/h by 3.6 before using SI equations: 36 km/h is 10 m/s and 72 km/h is 20 m/s.
- Using SUVAT when acceleration changes. A skydiver approaching terminal velocity, or a car whose acceleration fades with speed, needs a different model.
- Treating displacement as distance. After a round trip s is zero, but the distance traveled is not.
- Switching values of g. Use the value the problem gives (9.8, 9.81 or 9.80665 m/s²) and keep it throughout.
Input formats and units
- Choose the two unknowns and enter the other three quantities.
- Each quantity takes its own unit, such as km/h, mph, ft/s, minutes or g₀ for standard gravity. Answers are in SI units: m, m/s, m/s² and s.
- Displacement and velocities are signed. Time cannot be negative, and it must be above zero when an equation divides by it.
For launches at an angle, use the projectile motion calculator. To find the acceleration from forces and friction, use the force, friction and incline calculator; for energy-based answers, the work, energy and power calculator.
Worked examples
Ball thrown up at 20 m/s passes 15 m twice
- Find
- v and t (know s, u, a)
- Displacement s
- 15 m
- Initial velocity u
- 20 m/s
- Acceleration a
- -9.81 m/s²
- Time t
- 0.990719 s
- Second time (other root)
- 3.08675 s
- Final velocity v
- 10.2811 m/s
- Final velocity at the second time
- -10.2811 m/s
Checked against: t = (20 ∓ √(400 − 2·9.81·15))/9.81 = 0.9907186…, 3.0867534…; v = 20 − 9.81t = ±10.2810505…
Car braking from 10 m/s to rest in 50 m
- Find
- a and t (know s, u, v)
- Displacement s
- 50 m
- Initial velocity u
- 10 m/s
- Final velocity v
- 0 m/s
- Time t
- 10 s
- Acceleration a
- -1 m/s²
Checked against: t = 2s/(u+v) = 10; a = (0 − 100)/(2·50) = −1
Zero acceleration (uniform motion)
- Find
- v and t (know s, u, a)
- Displacement s
- 100 m
- Initial velocity u
- 5 m/s
- Acceleration a
- 0 m/s²
- Time t
- 20 s
- Final velocity v
- 5 m/s
Checked against: t = s/u = 20 (linear case of ½at² + ut − s = 0)
Questions
What does SUVAT stand for?
SUVAT names the five quantities in the constant-acceleration equations: s for displacement, u for initial velocity, v for final velocity, a for acceleration and t for time. The equations are v = u + at, s = ut + ½at², v² = u² + 2as, s = ½(u + v)t and s = vt − ½at². Each omits a different variable, so you pick the one that leaves out the quantity you neither know nor need.
When can you use the SUVAT equations?
Only when acceleration is constant and the motion is in a straight line. Free fall near the Earth's surface qualifies while air resistance is small; standard gravity is 9.80665 m/s², the value adopted by the 3rd General Conference on Weights and Measures in 1901. A skydiver nearing terminal velocity, or a car whose acceleration fades as it gains speed, needs calculus or a numerical model instead.
How do you calculate stopping distance with SUVAT?
Set the final velocity to zero in v² = u² + 2as, which gives s = u² ÷ (2 × deceleration). A car braking from 10 m/s (36 km/h) at 1 m/s² stops in 50 m after 10 s. Doubling the starting speed quadruples the braking distance. The distance covered during the driver's reaction time comes on top and is uniform motion, s = ut.
Why are there two answers for time?
When time is the unknown, s = ut + ½at² is a quadratic in t and can have two positive roots. A ball thrown upward at 20 m/s with a = −9.81 m/s² is 15 m above the start at 0.99 s on the way up and again at 3.09 s on the way down, moving at 10.28 m/s each time but in opposite directions. Negative roots are dropped because time starts at zero.
What value of g should I use?
Use 9.80665 m/s², the defined standard gravity, unless the problem states another value; many textbooks round it to 9.81 or 9.8 m/s². Real sea-level gravity varies with latitude, from about 9.780 m/s² at the equator to 9.832 m/s² at the poles in the WGS 84 model. That spread changes answers by about 0.5%.
About this calculator
Sources
Checked against references
This calculator includes 6 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.