Kinematics Calculator (SUVAT)
Enter any 3 of s, u, v, a and t and get the other 2, with the equation used for each one and every root of the time quadratic, including the ones that get rejected.
Fill in exactly 3 boxes and leave 2 blank. Use SI units and one positive direction for every value.
Both are valid. The object passes the same point twice, for example once going up and once coming down.
Solving for t
- Rearrange s = ut + ½at² so everything is on one side: ½a·t² + u·t − s = 0
- Substitute the known values: −4.9t² + 20t − 15 = 0
- Discriminant, with A = ½a, B and C the coefficients of t², t and 1 above: D = B² − 4AC = 20² − 4 × (−4.9) × (−15) = 106
- Quadratic formula, t = (−B ± √D) / (2A), gives two roots: t₁ ≈ 0.9902 s and t₂ ≈ 3.0914 s
- Kept: t ≈ 0.9902 s and t ≈ 3.0914 s. Both roots are positive, so both are valid answers.
Solution 1 of 2: t ≈ 0.9902 s
- Known: s = 15 m, u = 20 m/s, a = −9.8 m/s². Unknown: v and t.
- Find t with s = ut + ½at²: t is a root of the quadratic above, and this solution uses t ≈ 0.9902 s.
- Find v with v = u + at: v = 20 + (−9.8) × 0.9902 ≈ 10.2956 m/s
- Result: v ≈ 10.2956 m/s, t ≈ 0.9902 s.
Solution 2 of 2: t ≈ 3.0914 s
- Known: s = 15 m, u = 20 m/s, a = −9.8 m/s². Unknown: v and t.
- Find t with s = ut + ½at²: t is a root of the quadratic above, and this solution uses t ≈ 3.0914 s.
- Find v with v = u + at: v = 20 + (−9.8) × 3.0914 ≈ −10.2956 m/s
- Result: v ≈ −10.2956 m/s, t ≈ 3.0914 s.
How it works
For motion in a straight line with constant acceleration, five quantities are linked by five equations. Each equation leaves out one of the letters:
- v = u + at (no s)
- s = ut + ½at² (no v)
- s = vt − ½at² (no u)
- v² = u² + 2as (no t)
- s = ½(u + v)t (no a)
The missing-variable rule: find the letter you neither know nor want, then use the equation that leaves it out. Once one unknown is found, the second follows from any equation that contains it. The calculator does exactly this and names the equation at each step.
Signs matter. Choose one direction as positive and give every value a sign from it. With up as positive, free fall has a = −9.8 m/s², a point below the start has a negative s, and an object moving down has a negative velocity. The equations only apply while the acceleration stays constant, so a motion that speeds up and then brakes has to be split into separate parts.
Worked example
A ball is thrown straight up at u = 20 m/s with a = −9.8 m/s². When is it s = 15 m above the launch point, and how fast is it moving then? The unknowns are v and t, so use s = ut + ½at², the equation with no v.
- Substitute: 15 = 20t − 4.9t², which rearranges to −4.9t² + 20t − 15 = 0.
- Discriminant: D = 20² − 4 × (−4.9) × (−15) = 400 − 294 = 106.
- Quadratic formula: t = (−20 ± √106) / (−9.8), so t ≈ 0.9902 s (going up) and t ≈ 3.0914 s (coming down).
- Then v = u + at gives v ≈ 10.2956 m/s at the first time and v ≈ −10.2956 m/s at the second, the same speed in the opposite direction.
Both times are correct, which is why the calculator lists every root instead of quietly picking one. Try the dropped-stone example above to see a negative root rejected. For more on picking the right equation and on two-part problems, read the kinematic equations guide. To practise the time quadratic on its own, the quadratic formula calculator shows the discriminant and both roots step by step.
Frequently asked questions
What does SUVAT stand for?
SUVAT is a name built from the five letters in the constant-acceleration equations: s for displacement, u for initial velocity, v for final velocity, a for acceleration and t for time. Many US textbooks write v₀ for u and Δx for s, but the equations are the same. Each equation leaves out exactly one of the five letters.
Why does the calculator give two answers for time?
When you know s, u and a (or s, v and a), time comes from a quadratic equation, and a quadratic can have two roots. If both are positive, both are real moments in the motion. A ball thrown up at 20 m/s is 15 m above the launch point twice: once on the way up at about 0.99 s and again on the way down at about 3.09 s.
Why is a negative time rejected?
The equations treat the motion as starting at t = 0, so a root with t below zero describes a moment before the motion began. For a stone dropped from 20 m, the quadratic gives t ≈ 2.02 s and t ≈ −2.02 s. Only the positive root answers the question, so the calculator keeps it and lists the other one as rejected.
What sign should I use for gravity?
Pick one direction as positive and stick to it for every value. With up as positive, gravity pulls down, so a = −9.8 m/s², a point below the start has a negative s, and a falling object has a negative velocity. With down as positive, a = +9.8 m/s² and all those signs flip. Both choices give the same physical answer.
When do the SUVAT equations not apply?
They only hold when the acceleration is constant for the whole stretch of motion and the motion is along one straight line. They do not handle air resistance that changes with speed, a car that speeds up and then brakes (split it into two parts), or circular motion. For a projectile, use them separately on the horizontal and vertical directions.
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