Equations of Motion Calculator: Any Three of u, v, a and t
Equations of Motion Calculator: Any Three of u, v, a and t
Give it any three of initial velocity u, final velocity v, acceleration a and time t. It returns the displacement s, fills in the fourth quantity you did not give, and names which of the four equations it used — which is the part a formula sheet never tells you.
Equations of motion: give any three of u, v, a and t
A metro train leaves a station at 4.5 m/s and accelerates uniformly at 1.2 m/s² for 9.0 s. How far does it go, and how fast is it moving at the end?
Four equations, one assumption, and the one with no t in it
- u
- initial velocity, in m/s. NCERT and CBSE write this v0; the letters differ, the quantity does not.
- v
- final velocity at the end of the interval, in m/s.
- a
- acceleration, in m/s², taken as constant. A deceleration is a negative a, not a different formula.
- s
- displacement — the net change in position, in m. NCERT writes this x. It is not the distance travelled when the motion turns round.
- t
- the length of the interval, in s. Always an interval, never a clock reading.
Worked example
A metro train leaves a station at 4.5 m/s and accelerates uniformly at 1.2 m/s² for 9.0 s. How far does it go, and how fast is it moving at the end?
Pick the equation from what you are given. You have u, a and t, and you want s. Only one of the four equations contains exactly those: s = ut + ½at². That choice is the whole problem; the arithmetic after it is bookkeeping.
Substitute, keeping the two terms apart. s = (4.5)(9.0) + ½(1.2)(9.0)² = 40.5 + ½(1.2)(81) = 40.5 + 48.6.
Add. s = 89.1 m. Note that the ½at² term is bigger than the ut term here — dropping the ½ would give 137.7 m, a 55% error, and it is the single commonest arithmetic slip on this equation.
Get the fourth quantity from the first equation. v = u + at = 4.5 + (1.2)(9.0) = 4.5 + 10.8 = 15.3 m/s.
Check it another way. Average velocity = (u + v)/2 = (4.5 + 15.3)/2 = 9.9 m/s, and 9.9 × 9.0 = 89.1 m. The two routes agree, which they only do if the acceleration really was constant — so this check is also a check on the assumption.
Significant figures. The data carry two and three figures, so 89 m is the defensible answer and 89.1 m is what most mark schemes accept. 89.100 m claims a precision the question never gave you.
What the other two equations would have needed. s = (u + v)t/2 would have worked too, but only after finding v first — two steps instead of one. s = (v² − u²)/2a would also have worked, and again only after finding v. Neither is wrong; both are longer, and on a timed paper the equation that uses only what you were given is the right choice every time.
Which equation to use, decided by what the question hands you
| You are given | You want | Equation to use | Why that one |
|---|---|---|---|
| u, a, t | s | s = ut + ½at² | The only form built out of exactly those three. |
| u, v, t | s | s = (u + v)t/2 | No a anywhere in it, so you never need the acceleration. |
| v, a, t | s | s = vt − ½at² | Runs the motion backwards from the finish line. |
| u, v, a | s | s = (v² − u²)/2a | The only form with no t, for questions that never mention time. |
| u, v, t | a | a = (v − u)/t | The definition of acceleration, rearranged. |
| u, a, s | v | v² = u² + 2as | Not on this page: s is given, so use the velocity–displacement form directly. |
| u, s, t | a | a = 2(s − ut)/t² | Not on this page: rearrange s = ut + ½at² for a. |
| a, s, t | u | u = s/t − at/2 | Not on this page: rearrange the same equation for u. |
The same three equations in two lettering conventions
| This page and most UK papers | NCERT and CBSE | What the symbol means |
|---|---|---|
| v = u + at | v = v0 + at | velocity after time t |
| s = ut + ½at² | x = v0t + ½at² | displacement after time t |
| v² = u² + 2as | v² = v0² + 2ax | velocity after a displacement |
| s | x, or x − x0 | displacement; NCERT states the general form with a starting position x0 |
| u | v0 | initial velocity |
| a | a | acceleration — the one letter both conventions agree on |
| t | t | the time interval; also the same in both |
| s = (u + v)t/2 | x = (v0 + v)t/2 | the average-velocity form, which NCERT reads straight off the trapezium under a velocity–time graph |
Why choosing the equation is the hard part, and what the signs are doing
The four equations are not four facts. They are one fact written four ways, and the only skill being tested is picking the writing that matches what you were handed. That is why this page asks which three quantities you have before it asks for any numbers: the choice of equation is the answer, and the arithmetic is what comes after.
All four follow from two definitions and one assumption. The definitions are that acceleration is the rate of change of velocity and that velocity is the rate of change of displacement. The assumption is that a is constant over the whole interval. Drop that assumption and every equation here fails: a car that accelerates hard and then coasts has no single a, and the honest move is to split the motion into intervals that each do have one, solve them separately, and add. Examiners set that deliberately, and a candidate who applies one equation across the join loses the marks for the method rather than for the arithmetic.
Signs are where most marks go. There is no separate formula for deceleration. You choose a positive direction once — rightwards, forwards, upwards, whichever the question makes natural — and then u, v, a and s all carry signs relative to that choice. A car slowing down while moving forwards has positive u, positive v, and negative a. A ball thrown upwards and caught again has positive u, negative v, negative a, and a displacement of zero. If your answer comes out negative, read it as a direction before you look for a mistake.
The one that catches people is displacement against distance. Displacement is the net change in position and is what every equation on this page returns. Distance is the length of the path actually travelled. They are the same number only while the motion does not turn round. The figures under the answer give both, and the note tells you when the velocity has passed through zero inside your interval, because that is exactly when they come apart. A ball thrown up and caught has a displacement of zero and a distance of twice the maximum height, and a question that asks for “the distance travelled” wants the second number. That distinction is worked through separately on the average speed and average velocity page.
Units before numbers. These equations are dimensionally consistent only in a consistent set of units. A speed in km/h with a time in seconds produces nonsense that looks like an answer. Convert to metres and seconds first — 72 km/h is 20 m/s, dividing by 3.6 — and convert back at the end if the question asked for km/h. A mass in grams does the same damage a line later, once forces arrive.
One practical habit is worth more than the four equations put together: write down what you have before you write down an equation. List u, v, a, s and t down the side of the page, fill in the three the question gives, mark the one it asks for, and the equation that contains exactly those four and not the fifth is the one to use. Examiners can tell the difference between a candidate who did that and one who tried s = ut + ½at² first and hoped. The habit also catches the questions that give you only two quantities, which are the ones that need a second piece of information from elsewhere in the problem — a conservation law, a geometric constraint, or a second phase of the motion.
It is worth knowing where the equations come from, because it tells you when to distrust them. v = u + at is the definition of constant acceleration, rearranged. s = (u + v)t/2 is the statement that for constant acceleration the average velocity is the mean of the two end velocities — true for a straight line on a velocity–time graph and false for anything else. The other two follow by substituting one into the other, and the fifth, v² = u² + 2as, is what you get when you eliminate t. So the whole set rests on the straight-line graph, and the moment the graph curves the entire set is unavailable. NCERT derives them exactly this way, from the area and the gradient of a velocity–time graph rather than from calculus.
A last word on units of time and on the word ‘interval’. Every t in these equations is a length of time, never a reading on a clock. A question that says “between the fourth and the seventh second” is giving you an interval of three seconds, and the initial velocity for that interval is the velocity at t = 4 s, not the velocity at the start of the whole motion. Candidates who substitute 7 into an equation whose u belongs to t = 0 get an answer for a different question. Work out the state of the object at the start of the interval you care about, then treat that as a fresh problem with its own u and its own t. The same discipline handles a motion made of several phases, and it is the only way to handle one.
Which equations a board calls “allowed” varies. Some specifications print all four and expect you to quote the one you used; others print three and expect s = vt − ½at² to be derived on the spot. NCERT Chapter 2 derives the set from a velocity–time graph rather than from calculus, which is worth knowing because it means the graph is the primary object in that treatment and the equations are a consequence — the subject of the motion-graph page.
Two of these equations are worth deriving once, by hand, because the derivation is shorter than the memorisation. Start from a straight line on a velocity–time graph running from u at t = 0 to v at time t. Its gradient is (v − u)/t and that is the acceleration, which rearranges to v = u + at. The area beneath it is a trapezium of parallel sides u and v and width t, so the displacement is (u + v)t/2. Substituting the first into the second eliminates v and gives s = ut + ½at²; eliminating t instead gives v² = u² + 2as. Four equations, two pieces of graph geometry, and no memorising — and as a bonus the derivation makes it obvious why a curved velocity–time graph destroys all four at once: the gradient is no longer a single number and the area is no longer a trapezium.
Frequently asked questions
Why does this page ask which three quantities I have instead of letting me leave boxes empty?
Because the calculator engine refuses to produce a figure while any input box is empty, and that refusal is deliberate — a blank box would otherwise be read as zero and the page would answer confidently from numbers you never gave it. So instead of blank boxes you pick the combination, and the box you did not pick locks itself and fills in from the other three. The practical benefit is that you can see all four numbers agreeing at once, which is the fastest check there is on an exam answer.
What is the difference between s and the distance travelled?
s is the displacement: the net change in position, with a sign. The distance travelled is the length of the path, always positive. They are equal while the motion keeps going the same way. The moment the velocity passes through zero and reverses, they differ: a ball thrown straight up and caught at the same height has s = 0 and a distance of twice the maximum height. The figures under the answer give both, and the note tells you when the reversal has happened inside your interval.
Which equation do I use if the question gives me s and asks for t?
That combination is not on this page, and the reason is worth knowing: s = ut + ½at² is a quadratic in t, so it has two roots and you have to decide which one the question means. Both can be positive and physically real — a ball passes a given height once on the way up and once on the way down. The free fall and vertical motion page solves that quadratic and shows both roots side by side.
Is a deceleration a negative acceleration, or do I use a different formula?
A negative acceleration, and there is no different formula. Once you have chosen a positive direction, deceleration simply means a has the opposite sign to the velocity. The common error is to make a negative and change a plus to a minus in the equation, which cancels the sign and gives the answer for an acceleration. Put the sign in the number and leave the equation alone.
Why do I get a strange answer when I mix km/h with seconds?
Because the equations have no way to know. They are relationships between numbers, and they are only true when every quantity is in a consistent set of units. Work in metres and seconds throughout: divide km/h by 3.6 to get m/s, multiply m/s by 3.6 to go back. A speed of 54 km/h is 15 m/s; using 54 with a time in seconds overstates every distance by a factor of 3.6.
My textbook writes these with v-nought and x instead of u and s. Are they the same equations?
Yes, letter for letter. NCERT and CBSE write v0 where UK papers write u, and x where UK papers write s, and NCERT also states the general form with a starting position x0 so that x − x0 is the displacement. The nickname SUVAT comes from the UK lettering and means nothing in an Indian textbook. The second table on this page sets the two conventions side by side.
The question gives me a time like “between the 4th and the 7th second”. What is t?
Three seconds, the length of the interval — never 7. Every t in these equations is an interval, so you must first work out the state of the object at the start of it. Find the velocity at t = 4 s from the original motion, call that your u, set t = 3 s, and treat the interval as a fresh problem. Substituting 7 into an equation whose u belongs to t = 0 answers a different question and is one of the most reliably set traps in the topic.
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References
- NCERT, Physics: Textbook for Class XI, Part I, Chapter 2 “Motion in a Straight Line”, §2.4 “Kinematic Equations for Uniformly Accelerated Motion”. Derives the three equations writing v0 for the initial velocity and x for the displacement, and takes g as 9.8 m s−2 (10 m s−2 in Example 2.3).
- OCR, Data, Formulae and Relationships Booklet for AS and A Level GCE Physics A (H156 and H556), 8 pages. Lists the acceleration of free fall as g = 9.81 m s−2 alongside the thirteen other physical constants the papers assume.
- BIPM, The International System of Units (SI), 9th edition (2019). Definitions of the metre and the second, and the convention that the radian is the coherent SI unit of plane angle with dimension one — which is why every trigonometric formula in mechanics is written in radians.
- NIST, CODATA Internationally Recommended Values of the Fundamental Physical Constants — complete listing. The standard acceleration of gravity is tabulated as 9.806 65 m s−2, marked exact rather than carrying an uncertainty.
physics.nist.gov/cuu/Constants/