Kinematics Calculator
Motion in a straight line at a constant acceleration is completely described by five quantities — initial velocity u, final velocity v, acceleration a, time t and displacement s — and any three of them fix the other two. That is what the four SUVAT equations say: v = u + a·t, s = u·t + ½·a·t², s = ((u + v) ÷ 2)·t, and the timeless v² = u² + 2·a·s. Pick the quantity you want, enter the three you know, and this returns it along with the other four, the arithmetic that produced it, and the position and velocity plotted over the whole interval. Solving for time is the interesting case: it is a quadratic, so there are usually two answers, and the calculator reports the first arrival while telling you what the second root means.
- Accurate
- Real-time
- Easy to use
- 100% free
Displacement
150.0000 m
Final velocity (v)
25.0000 m/s
Details
Updates as you typeEach mode reads three of the five boxes and works out the rest. Boxes a mode does not use are ignored, so whatever they hold cannot stop it answering.
Speed at the moment the clock starts, in metres per second. Negative means it is already moving the other way. Read by every mode.
Rate of change of velocity, assumed constant throughout. Negative is deceleration. For an object falling freely near the Earth, use 9.8 measured downwards.
How long the motion lasts, in seconds. Read when solving for displacement, final velocity or acceleration.
Speed at the end of the interval. Read only when solving for acceleration; every other mode works it out for you.
How far the object travels along its line of motion. Read only when solving for time; every other mode works it out for you.
Summary
Displacement
150.0000 m
- Initial velocity (u)
- 5.0000 m/s
- Final velocity (v)
- 25.0000 m/s
- Acceleration (a)
- 2.0000 m/s²
- Time (t)
- 10.0000 s
- Displacement (s)
- 150.0000 m
- Average velocity
- 15.0000 m/s
- Every figure here assumes the acceleration is constant for the whole interval and the motion is along a straight line. That covers free fall in a vacuum, a car braking at a steady rate, and a trolley on a ramp. It does not cover air resistance, a changing thrust, a spring, or a curved path — for those the acceleration is a function of time or position and these five numbers no longer determine each other.
- One direction is positive and the opposite one is negative; which is which is your choice, but it has to be the same choice for all five quantities. A ball thrown upward at 20 m/s with "up" positive has u = 20 and a = −9.81. Displacement is not distance travelled: an object that goes out and comes back has a displacement of zero while its odometer says otherwise.
How this is calculated
- Initial velocity u (known)
- 5.0000 m/s
- Acceleration a (known)
- 2.0000 m/s²
- Time t (known)
- 10.0000 s
- Distance at constant u, u × t
- 50.0000 m
- Extra from accelerating, ½ × a × t²
- 100.0000 m
- s = u·t + ½·a·t²
- 150.0000 m
- v = u + a·t
- 25.0000 m/s
- Check: v² = u² + 2·a·s, left side
- 625.0000
- Check: right side
- 625.0000
Seconds
Compare scenarios
See how one change moves the result
- CurrentYour inputs as they stand150.0000 mCurrent
- Initial velocity (u)m/s 6.3163.0000 m
- Acceleration (a)m/s² 2.5175.0000 m
- Time (t)s 12.5218.7500 m
Frequently asked questions
What are the four SUVAT equations?
They are v = u + a·t; s = u·t + ½·a·t²; s = ((u + v) ÷ 2)·t; and v² = u² + 2·a·s. Here u is the velocity at the start, v the velocity at the end, a the constant acceleration, t the elapsed time and s the displacement. Each equation deliberately omits one of the five quantities — the third has no acceleration in it and the fourth has no time — which is what lets you pick the one that uses only the numbers you actually have.
Why does solving for time give two answers?
Because s = u·t + ½·a·t² is a quadratic in t, and a quadratic has two roots. When both are positive the object genuinely passes that displacement twice: a ball thrown upward is 10 m above your hand on the way up and again on the way down. When the second root is negative it describes where the object would have been before the clock started had it been moving this way all along, which is not part of the journey you asked about. This calculator reports the smallest root that is not negative and tells you what the other one is.
What happens if the acceleration is zero?
The quadratic collapses. With a = 0 the velocity never changes, the ½·a·t² term vanishes, and s = u·t solves in a single division. The calculator branches to that case explicitly rather than dividing by zero or letting a limit take care of it, so setting the acceleration slider to 0 gives the plain constant-speed answer with the working to match.
What does "never reached" mean?
It means no time at all satisfies the equation. Under a deceleration an object stops after −u ÷ a seconds, having covered −u² ÷ (2a) metres, and then comes back the way it went; ask for a displacement beyond that turning point and there is no answer to give. Rather than refuse, the calculator reports the furthest it actually gets and when. Algebraically this is the case where the discriminant u² + 2·a·s is negative, so the quadratic has no real root.
How do I use this for free fall or a projectile?
Set the acceleration to the acceleration due to gravity, 9.81 m/s² at the Earth's surface, and be consistent about which direction you are calling positive. Dropping something from rest with "down" positive means u = 0 and a = 9.81. Throwing it upward with "up" positive means a = −9.81, and the time it takes to come back to your hand is the larger root of the quadratic. This page handles the vertical component of a projectile; the horizontal component is separate motion at a constant velocity, because gravity does not act sideways.
Is displacement the same as distance travelled?
No, and the difference matters here. Displacement is the straight-line change in position from start to finish, and it carries a sign. Distance travelled is how far the object actually went, which is never negative and never less than the size of the displacement. An object that goes out 10 m and comes back has travelled 20 m and been displaced 0 m. Everything on this page is displacement.
Does this work if the acceleration changes?
No. Constant acceleration is the assumption the whole method rests on, and it is what makes the average velocity equal to (u + v) ÷ 2. It holds well for free fall in a vacuum, a car braking at a steady rate, or a trolley rolling down a ramp. It fails for anything with air resistance, a changing thrust, a spring, or a curved path — those need calculus with the acceleration written as a function of time or position.