The Kinematics Calculator solves motion problems under constant acceleration using the four kinematic equations. Enter any three of the five variables, displacement, initial velocity, final velocity, acceleration and time, and the calculator finds the remaining two, states which equation it applied, and explains why that equation was the one that fits.
That last part is the point. Choosing the right equation is the skill being taught in every introductory mechanics course, and a calculator that returns a number without naming the equation teaches nothing.
Solve any kinematics problem from three known values
Constant-acceleration motion is described by five quantities, and knowing any three fixes the other two. That is the whole structure of the topic, and it is why every textbook problem gives you exactly three numbers.
| Symbol | Quantity | SI unit |
|---|---|---|
| s | displacement | metres (m) |
| u | initial velocity | metres per second (m/s) |
| v | final velocity | metres per second (m/s) |
| a | acceleration | metres per second squared (m/s²) |
| t | time | seconds (s) |
Fill in the three you know and leave the others blank. The calculator solves for both unknowns, sometimes in one step and sometimes in two, and shows each step with the equation it used.
Every field has its own unit selector, so a problem stated in kilometres per hour with a time in seconds needs no manual conversion. Scientific notation works in the 3.45e9 form.
If you enter more than three values, the calculator solves from three and checks the rest against the result. When they disagree, it says so, which usually means a typo rather than a physics error.
Choose the right kinematic equation
Four equations describe constant-acceleration motion, and each one leaves out exactly one of the five variables. That single observation is the whole selection rule.
| Equation | Leaves out | Use when you do not have |
|---|---|---|
| v = u + at | s | displacement |
| s = ut + ½at² | v | final velocity |
| s = ½(u + v)t | a | acceleration |
| v² = u² + 2as | t | time |
So the method is: identify the variable you neither know nor need, and use the equation that omits it.
Given initial velocity, acceleration and time, you have no displacement, so v = u + at applies. Given initial velocity, acceleration and displacement with no time available, v² = u² + 2as applies. The equation is not a choice, it follows from what is on the page in front of you.
The calculator states this reasoning for every problem you enter, in the panel above the inputs. Read it a few times and the rule becomes automatic, which is faster than reaching for a calculator at all.
Apply the four kinematic equations
Each equation solves for any variable it contains, and the rearrangements are worth having to hand. v = u + at `` v = u + at u = v − at a = (v − u)/t t = (v − u)/a ``
s = ut + ½at² `` s = ut + ½at² u = (s − ½at²)/t a = 2(s − ut)/t² ``
Solving this one for t needs the quadratic formula, because t appears squared: `` t = [−u ± √(u² + 2as)] / a ``
Two roots come out of that, and both can be physically real. A ball thrown upward passes any given height twice, once going up, once coming down, so two positive times is the correct answer, not an error. The calculator reports both and says which is which. Negative roots are discarded, with a note explaining why.
s = ½(u + v)t `` s = ½(u + v)t u = 2s/t − v v = 2s/t − u t = 2s/(u + v) ``
v² = u² + 2as `` v = √(u² + 2as) u = √(v² − 2as) a = (v² − u²)/(2s) s = (v² − u²)/(2a) ``
When the quantity under the square root goes negative, the object never reaches that displacement. A ball thrown up at 10 m/s cannot reach 20 m, and the calculator says that rather than returning an imaginary number.
Solve for final velocity given u, a and t
A car pulls away at 5 m/s and accelerates at 2 m/s² for 3 seconds. Its final velocity is 11 m/s. 1. List what is known. u = 5 m/s, a = 2 m/s², t = 3 s. Displacement is neither given nor asked for.
2. Choose the equation. No displacement, so the equation that omits s applies: v = u + at.
3. Substitute. v = 5 + (2 × 3)
4. Solve. v = 5 + 6 = 11 m/s
The calculator also returns the fifth variable. With u, a, t and now v all known, displacement follows from s = ½(u + v)t:
s = ½(5 + 11) × 3 = ½ × 16 × 3 = 24 m
So the car covers 24 metres while reaching 11 m/s. Both unknowns solved, two equations, and the second one used the answer from the first.
Solve a two-step problem
Many problems need two equations in sequence, and this is where hand calculation goes wrong, people pick one equation, find they are missing a value, and start over rather than solving for it first. A cyclist decelerates from 27 m/s to 9 m/s at −2 m/s².
How far do they travel, and how long does it take?
Step 1: find time. Known: u = 27, v = 9, a = −2. No displacement, so use v = u + at rearranged for t:
t = (v − u)/a = (9 − 27)/(−2) = (−18)/(−2) = 9 s
Step 2: find displacement. Now u, v, a and t are all known. Use s = ½(u + v)t:
s = ½(27 + 9) × 9 = ½ × 36 × 9 = 162 m
Note the two negatives cancelling in step 1. A negative acceleration with a decreasing velocity gives a positive time, as it must. If a time comes out negative, a sign has gone wrong somewhere in the setup.
The calculator chains these steps automatically and shows both, so a two-equation problem takes no more input than a one-equation problem.
Read the motion graphs
Three graphs describe the same motion, and each one's shape follows from the equations. The calculator draws all three from your solved values. Displacement against time. Under constant acceleration this is a parabola, because s depends on t². Its gradient at any point is the velocity at that moment, steepening curve means speeding up.
Velocity against time. A straight line, because v = u + at is linear in t. Its gradient is the acceleration, and its vertical intercept is the initial velocity.
Acceleration against time. A horizontal line, because the acceleration is constant. That is the assumption the whole topic rests on.
The relationship worth knowing: the area under the velocity-time graph equals the displacement. For constant acceleration that area is a trapezium with parallel sides u and v and width t, and its area is ½(u + v)t, which is one of the four equations. The graph and the equation are the same statement.
The calculator shades that area on the velocity-time plot. It is a standard exam question and easier to see once than to memorise.
Understand uniformly accelerated motion
These four equations apply only when acceleration is constant. That is what "uniformly accelerated" means, and it is a real restriction rather than a technicality.
Constant acceleration covers a large share of introductory problems: an object in free fall near the Earth's surface with air resistance neglected, a vehicle braking at a steady rate, a puck sliding on ice with constant friction.
In each case the force producing the acceleration does not change during the motion.
It does not cover motion where the acceleration varies. A car whose driver eases off the throttle, a spring-driven oscillation where force depends on displacement, a falling object where drag grows with speed, none of these can be solved with SUVAT. Those need calculus, or the problem has to be broken into segments over which acceleration is roughly constant.
A useful check: if the problem describes a force that changes during the motion, the four equations do not apply. If the force is steady, they do.
Use standard gravity for free fall
Free fall problems set the acceleration to g, the acceleration due to Earth's gravity, which is defined as exactly 9.80665 m/s². Most textbooks round to 9.81 or 9.8, and any of the three is fine for coursework as long as you are consistent.
The sign depends on your convention. Taking up as positive makes g negative at −9.80665 m/s²; taking down as positive makes it positive. Either works. What matters is that every quantity in the problem uses the same convention, and that you state which one you chose.
An object dropped from rest falls with u = 0. After 3 seconds, taking down as positive:
s = ut + ½at² = 0 + ½ × 9.80665 × 3² = ½ × 9.80665 × 9 = 44.13 m
Two things this ignores. Air resistance, which matters for anything light or fast, a feather and a hammer fall together only in a vacuum, and a skydiver reaches terminal velocity precisely because drag grows until it balances weight. And local variation in g, which ranges from about 9.764 to 9.834 m/s² across the Earth's surface with latitude and altitude. Neither matters for a textbook problem; both matter for real measurement.
Frequently asked questions
What are the SUVAT equations?
SUVAT is a mnemonic for the five variables in constant-acceleration motion: displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (t). Four equations link them, each omitting one variable. Knowing any three lets you find the other two, which is why every textbook problem supplies exactly three.
Which kinematic equation should I use?
Identify the variable you neither know nor need, then use the equation that leaves it out. Without displacement, use v = u + at. Without final velocity, use s = ut + ½at². Without acceleration, use s = ½(u + v)t. Without time, use v² = u² + 2as. The Kinematics Calculator names the equation and the reason for every problem entered.
What is uniformly accelerated motion?
Uniformly accelerated motion is motion in which acceleration stays constant throughout. Free fall with air resistance neglected, steady braking and constant-force motion all qualify. Motion where the force changes during the interval, a spring oscillating, an object reaching terminal velocity, does not, and the four kinematic equations cannot be applied to it.
Do the kinematic equations work for free fall?
Yes. Free fall near the Earth's surface has a constant acceleration of 9.80665 m/s², so all four equations apply with a set to g. Choose a sign convention and hold it: taking up as positive makes g negative. The equations neglect air resistance, so they are accurate for dense compact objects over short falls and inaccurate for light or fast ones.
Why do I need three values to solve a problem?
The four equations each contain four of the five variables, so any single equation needs three knowns to solve for its fourth. With three knowns the calculator can always find one more, and with four it can find the fifth. Two values leave more than one unknown in every equation and the system has no unique solution.
Can acceleration be negative?
Yes, and it usually means deceleration. Acceleration is a vector, so its sign records direction relative to your chosen positive axis. An object slowing while moving in the positive direction has negative acceleration. An object speeding up in the negative direction also has negative acceleration. The sign describes direction, not whether something is speeding up.
What happens if the object changes direction?
The equations handle it correctly as long as acceleration stays constant and the signs are consistent. A ball thrown upward has positive velocity, then zero at the top, then negative on the way down, with g negative throughout. Its displacement when caught is zero, while the distance travelled is twice the peak height, which is why the two quantities are not interchangeable.
Do these equations account for air resistance?
No. The kinematic equations assume constant acceleration, and air resistance grows with speed, so an object experiencing it does not accelerate uniformly. Results are accurate for dense compact objects over short distances and increasingly wrong for light objects, high speeds or long falls. A skydiver reaching terminal velocity is the clearest case where they fail entirely.
Why does solving for time sometimes give two answers?
Because s = ut + ½at² is quadratic in t, and a projectile passes a given height twice, once rising, once falling. Both positive roots are physically real and the calculator reports both, labelling which corresponds to the upward pass. Negative roots describe times before the motion began and are discarded.
Summary
The Kinematics Calculator solves constant-acceleration motion from any three of displacement, initial velocity, final velocity, acceleration and time, returning the remaining two along with the equation it used and the reason that equation applies.
Four equations describe this motion and each omits exactly one variable, so the selection rule is simply to use the equation that leaves out the quantity you neither know nor need.
Problems needing two equations are chained automatically. The equations hold only while acceleration is constant, which covers free fall with air resistance neglected and steady braking but excludes springs, drag-limited falls and any motion where the force changes during the interval.