Kinematics Calculator
Pick the two quantities you do not know and enter the three you do; the kinematics calculator picks the one SUVAT equation that leaves the other unknown out, then shows the work.
Final velocity v
- Final velocity v From v = u + at: v = 5 + 2 × 4 = 13 m/s.
- 13 m/s
- Displacement s From s = ut + ½at²: s = 5 × 4 + ½ × 2 × 4² = 36 m.
- 36 m
- Knowns used (SI) Everything is converted to meters, seconds and meters per second before the equations run, so mixed units cannot leak into the arithmetic.
- u = 5 m/s, a = 2 m/s², t = 4 s
- Average velocity ½(u + v). Under constant acceleration, and only then, the average velocity is the mean of the start and end velocities, which is why s = ½(u + v)t works.
- 9 m/s
- Acceleration in g Divided by standard gravity, 9.80665 m/s². A useful sense check: a hard road-car launch is a fraction of 1 g, and free fall is exactly 1 g.
- 0.2039 g
- Final velocity in other units
- 13 m/s 46.8 km/h 29.08 mph 42.65 ft/s
- Displacement in other units
- 36 m 0.036 km 118.1 ft 0.02237 mi
| Equation | Leaves out | Left side | Right side | Used here |
|---|---|---|---|---|
| v = u + at | s | 13 | 13 | yes |
| s = ut + ½at² | v | 36 | 36 | yes |
| v² = u² + 2as | t | 169 | 169 | |
| s = ½(u + v)t | a | 36 | 36 | |
| s = vt − ½at² | u | 36 | 36 |
- The v and s boxes are ignored: those are the two being solved for.
- These equations assume the acceleration is constant for the whole interval. If it changes (a car shifting gear, anything with real air resistance), split the motion into pieces that are each roughly constant.
- Results are shown to four significant figures. Round the answer to the precision of your least precise measurement.
How the Kinematics Calculator works
The kinematic formulas, usually called the SUVAT equations, describe motion in a straight line with constant acceleration. There are five quantities: displacement s, initial velocity u, final velocity v, acceleration a and time t. There are five equations, and each one leaves out exactly one of the five:
v = u + at leaves out displacement s.
s = ut + ½at² leaves out final velocity v.
v² = u² + 2as leaves out time t.
s = ½(u + v)t leaves out acceleration a.
s = vt − ½at² leaves out initial velocity u.
Worked example, the one above: a cart starts at 5 m/s and accelerates at 2 m/s² for 4 s. The final velocity is v = u + at = 5 + 2 × 4 = 13 m/s, and the displacement is s = ut + ½at² = 5 × 4 + ½ × 2 × 4² = 20 + 16 = 36 m. As a check, s = ½(u + v)t = ½ × (5 + 13) × 4 = 36 m, and v² = u² + 2as gives 169 = 25 + 144. The table under the result runs that check on all five equations for whatever you enter.
How to choose the equation. Write down the three quantities you know and the one you want. Of the five, exactly one is neither: the one the question does not mention. Use the equation that leaves that one out. Want the stopping distance of a car from its speed and braking deceleration, with no time given? Time is the quantity not involved, so use v² = u² + 2as. Want how long a ball takes to drop a known height from rest, with no final speed asked for? Use s = ut + ½at². This is what the calculator does: the two unknowns you pick decide which equation solves each.
When to use each one. v = u + at is the definition of constant acceleration, for any question about speed after a time. s = ut + ½at² is for distance after a time when the final speed is not known. v² = u² + 2as is the time-free equation: braking distances, take-off runs, how high something thrown upward climbs. s = ½(u + v)t is for when you know both speeds and the time but not the acceleration. s = vt − ½at² is the rarest: it describes the motion looking backward from the end, for when the starting speed is the unknown.
Signs are where most mistakes happen. Pick one direction as positive and keep to it for every quantity. A ball thrown upward at 15 m/s with up positive has u = +15 and a = −9.80665; its displacement on the way back down past the launch point is negative. Mixing a positive displacement with a positive gravitational acceleration on an upward throw is the most common error on these problems, and it usually produces the "no solution" message below.
The quadratic case. Solving for time from s, u and a means solving ½at² + ut − s = 0, which has two roots. Both are shown. When one is negative, it is the moment the same motion, extrapolated backwards, would have passed that point. That root is algebraically valid but not physical. When both are positive the object passes that point twice: a ball thrown upward passes 10 m on the way up and again on the way down. And when u² + 2as is negative there is no real root at all: the acceleration turns the object around before it reaches the displacement you asked about.
What these equations do not cover. They hold only while the acceleration is constant. Air resistance makes acceleration depend on speed, so for anything fast or light the answers are an upper bound. Motion in two dimensions is two of these problems at once. The projectile mode does exactly that, with constant horizontal velocity and free fall vertically. And they are classical: near the speed of light they fail.
Kinematics Calculator FAQ
What are the kinematic equations?
The five SUVAT equations for straight-line motion with constant acceleration: v = u + at, s = ut + ½at², v² = u² + 2as, s = ½(u + v)t and s = vt − ½at². Each leaves out one of displacement s, initial velocity u, final velocity v, acceleration a and time t, so with any three known you can always find the other two.
How do I know which kinematic equation to use?
List the three quantities you know and the one you want. One of the five is not involved at all; use the equation that leaves that one out. No time involved means v² = u² + 2as; no final velocity means s = ut + ½at²; no acceleration means s = ½(u + v)t.
Why does the calculator give two answers for time?
Finding t from s, u and a means solving a quadratic, and a quadratic has two roots. A negative root is the time the motion run backwards would have been there, and is discarded. Two positive roots mean the object passes that point twice, out and back. If there is no real root, the object turns around before it gets there.
What does SUVAT stand for?
The five symbols: s for displacement, u for initial velocity, v for final velocity, a for acceleration and t for time. The name is common in British and Commonwealth teaching; American textbooks more often write x, v0, v, a and t and call them the kinematic equations or the big four or five.
Can I use the kinematic equations when acceleration is not constant?
No. Every one of them is derived assuming a constant a. If the acceleration changes, split the motion into stages that are each roughly constant and chain them, using the final velocity of one stage as the initial velocity of the next. For acceleration that varies continuously, the general method is calculus.