Motion Graphs: What the Slope and the Area Actually Give You

Pick a graph and pick a feature, and this page tells you which quantity it is — including the two cases where the answer is nothing at all. The area under a displacement–time graph is the most commonly computed meaningless number in school physics.

Last updated 11 Nov 2026

Which graph, which feature, which quantity

Gradient, area and intercept on all three graphs

Which graph are you looking at?Check the vertical axis label before anything else. A question that hands you a v–t graph and a student who reads it as an s–t graph will produce a confident, wholly wrong answer.

Which feature do you want to read off?Gradient is a division by time, so it moves you one step down the chain. Area is a multiplication by time, so it moves you one step up.

DisplacementExample

A velocity–time graph rises in a straight line from 0 to 12 m/s over the first 4.0 s, stays flat at 12 m/s for the next 6.0 s, then falls in a straight line to zero over the final 2.0 s. What does the area tell you, and what is it?

One chain, two directions: differentiate going down, integrate coming back

s − gradient → v − gradient → a | a − area → Δv − area → Δs

gradient: a division by time, so it takes you one step along the chain from displacement towards acceleration. On a curve, use a tangent for an instantaneous value and a chord for an average.
area: a multiplication by time, so it takes you one step the other way. Only meaningful when the vertical axis is itself a rate — which is true of v–t and a–t and false of s–t.
signed area: area below the time axis counts negative. This is what makes the area under a v–t graph the displacement rather than the distance.
intercept: the value of whatever is on the vertical axis at t = 0: the starting position, the initial velocity, or the initial acceleration respectively.
straight line: on a v–t graph, uniform acceleration — and therefore permission to use the equations of motion. On a curve, that permission is withdrawn.

Worked example

A velocity–time graph rises in a straight line from 0 to 12 m/s over the first 4.0 s, stays flat at 12 m/s for the next 6.0 s, then falls in a straight line to zero over the final 2.0 s. What does the area tell you, and what is it?
Identify the graph and the feature. The vertical axis is velocity, so the area has units of m/s × s = m, and the area is the displacement. Had the axis been displacement, the area would have had no meaning at all.
Split the shape. A triangle for 0 to 4.0 s, a rectangle for 4.0 to 10.0 s, a triangle for 10.0 to 12.0 s. Exam graphs are drawn to make this possible.
Add the three areas. ½(4.0)(12) = 24 m, then (6.0)(12) = 72 m, then ½(2.0)(12) = 12 m. Total = 108 m.
Note that it is displacement, not distance. Here they agree, because the line never goes below the time axis — the object never reverses. If any part of the line were below the axis, the signed area would give the displacement and you would have to add magnitudes instead for the distance.
Read the gradients while you are there. 12/4.0 = 3.0 m/s² for the first phase, zero for the middle, and 0 − 12 over 2.0 s = −6.0 m/s² for the last. Three straight segments means three separate constant accelerations, so the equations of motion apply to each phase individually and to none of them across a join.
And the intercept. The line starts at the origin, so u = 0: the object began at rest. That one fact is the difference between s = ½at² and the full s = ut + ½at².

All three graphs at once: gradient, area and intercept

The pattern is mechanical once you see it. A gradient divides by time and moves you one step along s → v → a; an area multiplies by time and moves you one step back. Area is meaningless on the s–t graph because displacement is not a rate, so there is nothing for the multiplication by time to undo.
Graph	Vertical axis	Gradient gives	Area gives	Intercept gives
s–t	displacement, m	velocity (m/s)	nothing — units of m·s, no such quantity	starting position s₀
v–t	velocity, m/s	acceleration (m/s²)	displacement (m), signed	initial velocity u
a–t	acceleration, m/s²	rate of change of acceleration (m/s³), off-syllabus	change in velocity v − u (m/s), signed	initial acceleration

Four shapes on a velocity–time graph, and the motion each one describes

The third row is the one examiners like: the object reverses at the crossing point, so the signed area gives a displacement smaller than the distance travelled, and a candidate who adds the two areas as positives has answered a different question. The fourth row is the trap: a curve means the equations of motion are off the table, and substituting into them anyway loses the method marks even if the arithmetic is flawless.
Shape of the line	Acceleration	What the object is doing	Equations of motion?
Straight, through the origin, rising	constant and positive	speeding up from rest, uniformly	Yes
Horizontal, above the axis	zero	constant velocity; the area is simply v × t	Yes, with a = 0
Straight, falling, crossing the axis	constant and negative	slowing, stopping at the crossing, then moving backwards and speeding up	Yes — and note the displacement has areas of both signs
Curved, getting steeper	increasing	a non-uniform acceleration	No — use areas and tangents only
Curved, flattening towards a horizontal line	decreasing towards zero	approaching a terminal or steady speed, as a falling object does against air resistance	No — the area still gives the displacement
Horizontal, below the axis	zero	constant velocity in the negative direction; the area is negative	Yes, with a = 0 — but the signed area matters

The chain, and the two places where there is honestly nothing to find

Graphs are not a separate topic from the equations of motion. They are where the equations come from. NCERT derives the whole set from a velocity–time graph rather than from calculus: the area under a straight v–t line is a trapezium, its area is ½(u + v)t, and that is the average-velocity equation. Read that way, the graph is the primary object and the four equations are consequences of its geometry.

Everything follows from one chain. Displacement, differentiated with respect to time, gives velocity; velocity, differentiated, gives acceleration. Reverse it and acceleration, integrated, gives the change in velocity; velocity, integrated, gives the change in displacement. On a graph, differentiating is taking a gradient and integrating is taking an area. So a gradient moves you one step towards acceleration and an area moves you one step towards displacement, and once you hold that, there are no separate facts left to memorise.

It also explains the two dead ends, which are the most useful thing on this page. The area under a displacement–time graph is a dead end because displacement is not a rate — there is nothing for a multiplication by time to undo, and the result has units of metre-seconds, which measures nothing. The gradient of an acceleration–time graph is a dead end for a different reason: it is a real quantity, the rate of change of acceleration, but it is not on any school syllabus. Students compute both of these regularly, and both are pure lost time.

The sign of the area is where the marks are. On a velocity–time graph, area below the time axis is negative area, because the velocity there is negative and the object is going backwards. The signed total is therefore the displacement. If a question asks for the distance travelled, you must take the magnitude of each piece and add those instead. A line that crosses the time axis is the examiner telling you this matters, and it is the same distinction that separates average speed from average velocity.

Tangent or chord? On a curve, a gradient is only defined at a point, and you get it by drawing a tangent — then reading two widely spaced coordinates off the tangent, not off the curve. That gives the instantaneous value. A chord joining two points on the curve gives the average value over the interval between them. Mark schemes for tangent questions almost always award a mark for the construction and a mark for the coordinates, so showing the triangle on the graph is worth doing even when you can see the answer.

Reading values off a graph is a skill with its own marks attached, separate from knowing what the gradient means. Use as large a triangle as the axes allow, because a small one multiplies your reading error; read the coordinates at points where the line crosses grid intersections if you can; and write the two coordinates down rather than just the answer, because that is what the mark scheme is looking for. For an area, split the shape into triangles and rectangles rather than counting squares when the line is straight, and count squares only when it curves. State the units of whatever you have found: a gradient on a velocity–time graph is m/s divided by s, which is m/s², and showing that division is often itself worth a mark.

One habit to acquire: sketch the other two graphs. Given any one of the three, the other two are determined, and sketching them is often faster than any algebra. A straight-line v–t graph through the origin means a horizontal a–t graph and a parabolic s–t graph. A v–t graph that crosses the time axis means an s–t graph with a maximum at the crossing. An a–t graph made of two rectangles of opposite sign means a v–t graph that rises and then falls in straight lines. Questions that ask you to “sketch the corresponding graph” are testing exactly this, and they mark the shape and the key points rather than any numbers.

The last thing a graph tells you is whether you are allowed to use the equations of motion at all. Those equations assume a constant acceleration, which on a velocity–time graph means a straight line. A graph made of several straight segments, like the worked example above, means the equations apply to each segment separately and to no interval that spans a join. A curved graph means they do not apply anywhere on it, and the only tools left are areas and tangents. Noticing that is often the first mark in the question.

One more thing a graph does that an equation cannot: it shows you a motion made of several phases at a glance. A journey that accelerates, cruises and brakes is three straight segments on a velocity–time graph, and the whole displacement is the sum of three simple areas. The equations of motion would need to be applied three times with the end state of each phase handed to the next, and a single slip propagates through all three. This is why examiners set multi-phase motion as a graph question rather than an algebra one, and why the right first move on such a question is to sketch the graph even when you were not asked to.

Frequently asked questions

What does the area under a displacement–time graph represent?

Nothing. Its units are metres multiplied by seconds, and no quantity in mechanics is measured in metre-seconds. This is worth stating flatly because computing it is one of the most common pieces of wasted work in the topic — it looks like progress and earns no marks. If you are looking at an s–t graph, the useful operation is the gradient, which gives you the velocity.

Does the area under a velocity–time graph give the distance or the displacement?

The displacement, because area below the time axis counts as negative. For a graph that never crosses the axis the two are the same number. For one that does, the signed area gives the displacement and the sum of the magnitudes gives the distance travelled — and the two differ by twice whatever the object retraced. Read the question’s wording carefully: it is choosing between them deliberately.

How do I find the acceleration from a curved velocity–time graph?

Draw a tangent at the point you want and take its gradient, reading two widely separated coordinates off the tangent rather than off the curve. That gives the instantaneous acceleration there. You cannot use the equations of motion anywhere on a curved v–t graph, because the acceleration is not constant — which is exactly why the examiner drew a curve.

What is the difference between a tangent and a chord?

A tangent touches the curve at one point and its gradient is the instantaneous value there. A chord joins two points on the curve and its gradient is the average value over the interval between them. On a straight line they coincide, which is why the distinction never arises for uniform acceleration. Questions say “at t = 3 s” for a tangent and “between t = 2 s and t = 5 s” for a chord.

What does a horizontal line mean on each of the three graphs?

Three different things, which is why the axis label has to be checked first. Horizontal on an s–t graph means the position is not changing: the object is at rest. Horizontal on a v–t graph means the velocity is not changing: constant velocity, not at rest unless the line sits on the axis. Horizontal on an a–t graph means uniform acceleration, which is the licence to use the equations of motion.

Can the gradient of an s–t graph be negative, and what does that mean?

Yes. It means the displacement is decreasing, so the object is moving back towards — and possibly past — the origin. The velocity is negative, which is a statement about direction and not about speed: a gradient of −4 m/s is a speed of 4 m/s in the negative direction. A graph that comes back down to the time axis shows an object that has returned to where it started, with a total displacement of zero and a non-zero distance travelled.

If I am given one graph, how do I sketch the other two?

Go along the chain. From an s–t graph, the gradient at each point is the height of the v–t graph; from a v–t graph, the gradient is the height of the a–t graph and the running area is the height of the s–t graph. So a straight v–t line through the origin gives a horizontal a–t graph and a parabolic s–t graph, and a v–t graph crossing the time axis gives an s–t graph with a maximum at the crossing. Sketch questions mark the shape and the key points, not numbers.

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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.