Velocity-Time Graph Calculator: Equations of Motion (SUVAT) Explained

Sit in the passenger seat and watch the speedometer for twelve seconds, writing down the needle's reading, in metres per second, once a second. Pulling away from the lights: 0, 2, 4, 6. The driver holds it there: 6, 6, 6, 6. A red light ahead, the brakes: 4, 2, 0 — and if the driver then selects reverse, the next readings need minus signs. That column of numbers is the whole story of the drive. A velocity–time graph is that column drawn as a line, time along the bottom and the needle's reading up the side, and everything on this page is about reading it: what the line's steepness means, what its height means, what the area trapped under it means, and why those are three different things that nearly everybody confuses at least once.

The three records below are ones you make yourself. A cart on a rail with a thruster at each end, fired by holding a button, whose graph draws itself from what your thumb does. A ticker-tape timer and a trolley on a tilted runway — the Leaving Cert's own instrument — whose printed tape you cut into strips that stand up and turn into the graph. And a calculator that takes any three of the five SUVAT quantities and draws the single shape that fixes the other two.

What Does a Velocity-Time Graph Show? Slope, Area and Shape Explained

The line carries three separate pieces of information, and almost every mistake made with it comes from mixing two of them up. The height of the line at any instant is the velocity then: how fast, and by its sign, which way. The steepness of the line is the acceleration: how quickly that velocity is changing. The area between the line and the time axis is the displacement: how far from the start the object has got. Height, slope, area — three questions with three different answers, read off the same drawing in three different ways.

How do you find acceleration from a velocity-time graph?

Acceleration is the slope. A line that climbs from 0 to 6 m/s over 3 s is gaining 2 m/s every second, and that is all "2 m/s²" means: metres per second, per second. Read it as rise over run, the way you would for any straight line:

$a = \frac{\Delta v}{\Delta t} = \frac{v - u}{t}$

where $u$ is the velocity at the start of the stretch you are looking at and $v$ the velocity at its end. A line sloping down has a negative slope, which means the velocity is decreasing — and that is only "slowing down" if the velocity was positive to begin with, a point that gets its own section further on.

Steep is not the same as high. A line can be high and flat: fast, and not accelerating at all. It can be low and steep: slow, and accelerating hard. The exam trap is the point where the line is highest, which is where the object is fastest and where students reach for "maximum acceleration". At a smooth peak the line has just stopped climbing and is about to fall, so its slope there is zero — at the instant of top speed the acceleration is nothing at all. (At a sharp corner, the kind exam graphs are drawn with, it jumps from one value to another instead.) The steepest part of a real graph is usually somewhere unremarkable in the middle of a ramp. The steepest segment you will meet in an exam is a collision on a velocity–time graph: a change of several metres per second in a few milliseconds, drawn as a segment so steep it looks vertical.

What does the area under a velocity-time graph represent?

Drive at a steady 6 m/s for 4 s and you cover 24 m. On the graph that stretch is a flat line at height 6 running for a width of 4, and 6 × 4 is the area of the rectangle under it. This is the reason the graph is drawn the way it is. Velocity times time is distance, so height times width is distance, so the area under the line is how far you went. Any shape works, because any shape is a stack of thin rectangles, each one a short moment of time multiplied by the velocity during it.

For a straight line from $u$ to $v$ the shape is a trapezium, and its area is the average of the two parallel sides times the width:

$s = \tfrac{1}{2}(u + v)\,t$

which is average velocity multiplied by time, as it should be. The same trick — a product of two quantities turning into the area under a graph of one against the other — is what makes work the area under a force–displacement graph, and it is worth seeing that it is the same trick, because it comes back with impulse and with charge.

One subtlety separates the good answers from the full-mark ones. When the line goes below the axis the velocity is negative — the object is moving the other way — and the area down there counts negative towards the displacement, because the object is coming back. It counts positive towards the distance travelled, because an odometer does not care which way you are going. A graph that spends 10 s above the axis and 2 s below has one displacement (above minus below) and one distance (above plus below), and a question can ask for either.

What is the difference between a distance-time graph and a velocity-time graph?

The same journey gives two graphs, and the slope of one is the height of the other. On a distance–time graph (strictly a displacement–time graph, if the object can turn round) the height is where the object is and the steepness is how fast it is going. On the velocity–time graph the height is how fast it is going and the steepness is the acceleration. So a straight sloping line on the distance–time graph — moving at a steady rate — becomes a flat line on the velocity–time graph. A curve that gets steeper and steeper on the distance–time graph, speeding up, becomes a rising line on the velocity–time graph — a straight one if the acceleration is constant. A flat line on the distance–time graph means stopped, and appears on the velocity–time graph as a line lying along the axis. And a flat line above the axis on the velocity–time graph, which looks identical on paper, means the opposite of stopped.

That is the whole "which graph is this?" exam question. The picture means one thing on one graph and another on the other, and the axis label decides which. Read the label first, every time.

Drive Your Own Velocity-Time Graph

The only reliable way to feel the difference between slope and height is to control the slope and nothing else. The cart below has a thruster at each end, and you fire one by pressing and holding its button. While a thruster fires the cart accelerates at the thrust setting; the moment you let go it stops accelerating — and only that. You never set the velocity. You never set the position. Both are consequences, and the graph records them as they happen: the line's slope is whatever your thumb is doing, its height is the velocity that has built up, and the region between the line and the axis fills in teal above the axis and amber below it. A 12 s clock starts when you press Start. If your device has reduced motion switched on, the hold buttons become one-second step buttons, and every number below lands exactly.

Two moments in that run carry the whole lesson. The first is when you let go of the forward thruster. Nothing happened to the cart. It kept going at 6 m/s, arrow unchanged, distance posts still sliding past, and the only thing that changed was that the line stopped climbing. That is Newton's first law drawn for you: with no force there is no change in velocity, and "no change" is a flat line, not a line on the axis. The thrust, while it is on, is Newton's second — the acceleration is the force divided by the mass — and turning the thrust slider is turning that ratio.

The second moment is at 10 s, when the falling line touched the axis. For one instant the cart was stopped: velocity zero, arrow gone, displacement 42 m and not increasing. The back thruster was still firing, the acceleration arrow was still there, unchanged, and a moment later the cart was moving backward with the line continuing down at the same slope as though the axis were not there. As far as the line is concerned it is not. Velocity passed through zero the way a thermometer passes through zero on a cold morning — an instant, with nothing special happening to the rate of change. That is the ball at the top of its throw, and it has its own section below.

How Is a Velocity-Time Graph Measured? The Ticker-Tape Experiment

Where does the graph come from? Not from a textbook. A velocity–time graph is a measurement, and the instrument the Leaving Cert uses to make one is a ticker-tape timer: a small electromagnet vibrating at the mains frequency, 50 times a second, tapping a pin through carbon paper onto a strip of paper tape as the tape is pulled through. Attach the tape to a trolley and let the trolley go. The dots land every 0.02 s regardless of what the trolley is doing, so the spacing of the dots is the trolley's velocity: bunched while it is slow, spread out while it is fast, spreading steadily as it speeds up.

Now cut the tape every five gaps. Each strip is exactly 0.1 s of motion, and its length is the distance the trolley covered in that tenth of a second; divide by 0.1 s and the length is a velocity. Stand the strips side by side in order and you are looking at a velocity–time graph — a bar for every tenth of a second, each bar's height the velocity then. And here is the thing that students who have only ever been handed a graph never quite believe: lay the strips end to end again and you have the tape back, and the tape's length is the distance the trolley travelled. The area under the graph is not like a distance. It is a length of paper.

How do you calculate acceleration from ticker tape?

Each strip is 0.1 s long, so a strip of length $L$ is a velocity of $L \div 0.1$ s — ten times the length, per second, in whatever unit the length is in. If neighbouring strips differ by $\Delta L$, the velocity changed by $\Delta L / 0.1$ during the 0.1 s between them, and dividing that change by the 0.1 s it took gives the acceleration:

$a = \frac{\Delta L / \Delta t}{\Delta t} = \frac{\Delta L}{(\Delta t)^2}$

With $\Delta t = 0.1$ s the denominator is 0.01 s², so the acceleration in m/s² is a hundred times the change in strip length in metres. The default run's strips grow by 11.7 mm each: $0.0117 \div 0.01 = 1.17$ m/s². Better still, take the first and last strips and divide by the number of gaps between them, which averages out the measuring error on any single strip:

$a = \frac{L_{16} - L_{1}}{15\,(\Delta t)^2} = \frac{0.1815 - 0.0059}{15 \times 0.01} = 1.171 \text{ m/s}^2$

There is a small question hiding in "the velocity of a strip". The trolley is speeding up during the tenth of a second, so the strip's length divided by 0.1 s is the average velocity across it — which, when the acceleration is constant, is the velocity at the strip's midpoint. Consecutive midpoints are 0.1 s apart, so dividing the change by 0.1 s is the right arithmetic, and the bar tops fall on a straight line because a straight line is what constant acceleration means.

How do you friction-compensate a runway?

Push a trolley along a level runway and it slows down, because the wheels and bearings resist rolling. On the tape the dots bunch and the strips shrink — the 0° run above lost 1.96 mm a strip and stopped after 2.55 s. Tilt the runway and gravity pulls the trolley down the slope; the net acceleration along it is

$a = g\,(\sin\theta - \mu\cos\theta)$

with $\mu$ the rolling-resistance coefficient, 0.020 in the model, and $g = 9.81$ m/s² (the model's value; the worked examples below use the exam papers' 9.8). There is one angle at which the slope's pull exactly pays for the friction: $\sin\theta = \mu\cos\theta$, so $\tan\theta = \mu$, and for $\mu = 0.020$ that is $\theta = 1.15°$. On a runway tilted to that angle a trolley given a push keeps its velocity — every strip the same length, a flat line, the 1.15° run above. That is a friction-compensated runway, and it is why the mandatory experiment tilts the runway first and checks the tape for equal spacing before it does anything else. Once the runway is compensated, any extra force you apply — a mass hanging on a string over a pulley — produces the full acceleration $F/m$ (with $m$ the whole moving mass, trolley and hanging mass together), with nothing subtracted for friction, and the tape measures Newton's second law rather than Newton's second law minus a nuisance.

The modern kit swaps the tape for two light gates and a card of known length on the trolley. Each gate records how long the card takes to pass, which gives a velocity at that gate; the acceleration is the change in velocity divided by the time between the gates. It is the same arithmetic as the tape, with two strips instead of sixteen and a computer reading the ruler.

What Are the SUVAT Equations and Where Do They Come From?

Five letters: $s$ for displacement, $u$ for the initial velocity, $v$ for the final velocity, $a$ for acceleration and $t$ for time. Under constant acceleration the velocity–time graph is a straight line from $(0, u)$ to $(t, v)$, and the region under it is a trapezium. Its two vertical sides are $u$ and $v$. Its base is $t$. Its slope is $a$. Its area is $s$. All five quantities are dimensions of one shape, and each of the SUVAT equations is a statement about that shape. Fix any three of the five and the trapezium is fixed, which forces the other two — that is why every question gives you three and asks for a fourth, and why there is no equation for the case where you only know two.

v = u + at: the slope

The slope of the line is the rise, $v - u$, over the run, $t$. That slope is the acceleration:

$a = \frac{v - u}{t} \qquad\Longrightarrow\qquad v = u + at$

Two lines, and it is the definition of acceleration rearranged. Everything that follows is this line plus one area.

s = ½(u + v)t: the trapezium

A trapezium's area is the average of its parallel sides times the distance between them. The parallel sides are $u$ and $v$ and the distance between them is $t$:

$s = \tfrac{1}{2}(u + v)\,t$

In words: displacement is average velocity times time — and only under constant acceleration is the average velocity the plain mean of the first and last, because only then is the line straight.

s = ut + ½at²: the rectangle plus the triangle

Draw a horizontal line across the trapezium at height $u$. Below it is a rectangle, $u$ high and $t$ wide. Above it is a triangle with base $t$ and height $v - u$, and from the slope equation $v - u = at$:

$s = ut + \tfrac{1}{2}\,t\,(at) = ut + \tfrac{1}{2}at^2$

The rectangle is how far the object would have gone with no acceleration; the triangle is the extra the acceleration added. Starting from rest, $u = 0$, only the triangle is left — a fact that turns up in every falling-body calculation as $s = \tfrac{1}{2}gt^2$.

v² = u² + 2as: eliminating time

Some questions never mention time — a braking distance, a height reached — and there is an equation with no $t$ in it. Take the trapezium's area and substitute the slope equation's $t = (v - u)/a$:

$s = \tfrac{1}{2}(u + v)\cdot\frac{v - u}{a} = \frac{v^2 - u^2}{2a} \qquad\Longrightarrow\qquad v^2 = u^2 + 2as$

Nothing new went in. It is the same trapezium, with its base written in terms of its sides and its slope. (A fifth equation, $s = vt - \tfrac{1}{2}at^2$, is the rectangle-plus-triangle measured from the right-hand side instead of the left. It exists, and you will almost never need it.)

When can you not use the SUVAT equations?

Only when the line is straight. Every one of those equations is a fact about a trapezium, and the region under a curve is not a trapezium. Air resistance curves the line: drag grows with speed, so a skydiver's velocity–time graph bends over and flattens towards terminal velocity, and no SUVAT equation describes any part of it. A car with a real engine curves it the same way for a different reason — the engine's pull falls off as the speed rises, so the slope decays. A rocket burning fuel curves it upward, because the same thrust on a lighter rocket gives more acceleration every second.

When that happens the graph is still right. Its slope at any instant — the slope of the tangent there — is still the acceleration at that instant, and the area under it is still the displacement, found by counting squares if there is nothing better. What fails is the shortcut. SUVAT is the special case where the graph is so simple that its area and slope can be written as formulas, and it is worth learning as that: the geometry of a straight line, and no more. One more condition — the equations are for motion along a single line. A projectile uses them twice, once per axis, with the two axes sharing nothing but the clock.

SUVAT Calculator: Solve Any Equation of Motion From the Graph

Below is the trapezium with the five quantities on it. Mark three as known and type them; the calculator draws the shape those three fix and reads the other two off it. The sides you gave are drawn solid with filled labels, the sides the shape forced are dashed with outlined labels, and the equations it used are printed at the top of the canvas. Exactly three quantities are known at any moment — marking a fourth releases whichever has been known the longest. There are no example buttons, on purpose: the steps below are the examples, and typing them is the point.

Why Is Acceleration Not Zero at the Top of a Throw?

Throw a ball straight up at 20 m/s and its velocity–time graph is one straight line. It starts at +20, falls at a slope of −9.8 m/s² the whole way, crosses the axis at $20 / 9.8 = 2.04$ s, and reaches −20 at 4.08 s, when the ball is back in your hand. At 2.04 s the velocity is zero — for an instant. The acceleration at that instant is −9.8 m/s², exactly as it was a second before and exactly as it will be a second after. The line has no kink and no flat bit. If the acceleration were zero at the top the line would go horizontal there, and a horizontal line means constant velocity — the ball would hang in the air. Gravity does not switch off because the ball has paused. "Stopped" is a statement about $v$; it says nothing at all about $a$.

The two areas say the rest. The teal triangle from 0 to 2.04 s has area $\tfrac{1}{2} \times 20 \times 2.04 = 20.4$ m, the height the ball reached. The amber triangle from 2.04 s to 4.08 s subtracts the same 20.4 m, so the displacement at landing is zero and the distance travelled is 40.8 m. And the vertical component of a projectile's velocity is exactly this line — any ball thrown at any angle has this graph for its up-and-down motion, paired with a flat line for its sideways motion, and the two together trace the parabola. The cart's axis crossing at 10 s in the first interactive is the same moment with a thruster standing in for gravity: the reversal happens with nothing changing but the sign of the line's height.

How Are Velocity-Time Graphs Examined in the Leaving Cert?

This is Strand 1.1 of the new Physics specification, Particle motion in a straight line — the first thing in the course, first examined in 2027. Its three learning outcomes, quoted from the specification, are: model motion of a particle in a straight line; investigate constant and varying linear motion using primary and secondary data; derive the kinematic equations. The content list beneath them names "graphical representation and interpretation: displacement-time graphs, velocity-time graphs" and the three equations $v = u + at$, $s = ut + \tfrac{1}{2}at^2$ and $v^2 = u^2 + 2as$. The three interactives on this page are those three outcomes in order: drive a motion and watch it modelled; measure one with primary data; derive the equations from the shape.

The question shapes are old and stable. Describe the motion shown on a printed velocity–time graph, segment by segment, in words — accelerating, constant velocity, decelerating, and whether it turned round. Find the acceleration from one segment, which is a slope read off the grid. Find the displacement or the distance from the area, which means splitting the shape into rectangles and triangles and watching the sign of anything below the axis. And a SUVAT calculation, often one that turns on a sign: a ball thrown upward, a car braking, a stone dropped down a well. The one thing the "derive" outcome adds that the old syllabus did not stress is that you can be asked to show where an equation comes from, and the answer is the trapezium — two lines for the slope, two for the area, as in the section above.

The old syllabus's mandatory experiment on measuring velocity and acceleration is the ticker-tape (or light-gate) procedure in the second interactive, and the new specification's "primary data" outcome keeps it alive. Plot velocity against time; the slope of the best-fit line is the acceleration, and the scatter of the points about it is your uncertainty. Say in the write-up why the runway was tilted before the run — friction compensation is the part of the method that examiners ask about, because it is the part that shows you understood what was being measured.

The same graph-reading is the opening unit almost everywhere else. GCSE Physics has a required practical on acceleration that uses exactly this kit — a trolley, a runway, light gates or a ticker timer — to measure how the acceleration depends on the force and the mass. A-Level mechanics opens with SUVAT and the graphs, and hands you printed grids to read gradients and areas from. AP Physics 1 makes the graphs its language: a question may give a velocity–time graph and ask you to sketch the position–time and acceleration–time graphs that go with it. JEE's one-dimensional kinematics runs the same ideas with calculus alongside, and its multiple-choice papers lean hard on "which graph" questions.

Worked Examples for Physics Exams