Motion can be shown with three kinds of graph: displacement–time, velocity–time, and acceleration–time. They’re all linked by two simple ideas — the gradient (slope) of one graph and the area under it lead you to the next. Master those two moves and you can read any motion graph.
📘 What you need to know
Displacement–time graph: gradient = velocity. (Area means nothing here.)
Velocity–time graph: gradient = acceleration; area under it = displacement.
Acceleration–time graph:area under it = change in velocity. (Gradient means nothing here.)
A straight diagonal line = constant rate; a curve = a changing rate.
A horizontal line means the quantity isn’t changing (e.g. flat on a d–t graph = at rest).
The y-intercept gives the starting value (initial displacement, velocity, or acceleration).
The two moves: gradient and area
Everything in this topic comes down to going up or down a chain. To go from displacement to velocity to acceleration, take the gradient. To go back the other way, find the area under the graph.
displacement (s–t graph)
⇄gradient → ← area
velocity (v–t graph)
⇄gradient → ← area
acceleration (a–t graph)
Going right (down to acceleration), take the gradient. Going left (back up to displacement), take the area under the line. That single sentence covers almost every motion-graph question.
Displacement–time graphs
On a displacement–time (d–t) graph, the height tells you where the object is, and the gradient tells you the velocity. A steeper line means a faster object.
A straight diagonal line = constant velocity.
A curve = the velocity is changing, so the object is accelerating.
A horizontal line = the object is at rest (not moving).
A negative gradient = moving in the negative direction (back towards the start).
On a d–t graph the gradient is the velocity: steep = fast, shallow = slow, flat = at rest, curved = accelerating.
Velocity–time graphs
This is the most useful graph in kinematics, because it gives you two things at once. The gradient is the acceleration, and the area underneath is the displacement.
A straight diagonal line = uniform (constant) acceleration.
A curve = changing acceleration.
A horizontal line = constant velocity (zero acceleration).
A negative gradient = deceleration (or acceleration in the negative direction).
From a velocity–time graph
acceleration = gradient • displacement = area underneath
On a v–t graph the gradient is the acceleration and the shaded area is the displacement.
Acceleration–time graphs
The simplest of the three. Here the area under the line equals the change in velocity. A horizontal line means the acceleration is constant — for example, an object in free fall has a flat a–t graph at 9.81 m s⁻².
Quick reference: d–t gradient → velocity. v–t gradient → acceleration, v–t area → displacement. a–t area → change in velocity.
The bouncing ball
A bouncing ball is the classic test of whether you really understand these graphs. Take upwards as positive. As the ball rises it has a positive velocity that shrinks to zero at the top; as it falls, the velocity grows negative. But all the way through, gravity pulls down, so the acceleration stays constant at g, directed downwards.
At the highest point: displacement is maximum, velocity is momentarily zero, and the velocity flips from positive to negative. Acceleration is still g, downwards.
At the lowest point (the bounce): the velocity flips instantly from negative to positive — same speed, opposite direction.
For a bouncing ball (up = positive): the d–t graph is a row of arcs; the v–t graph is a sawtooth that jumps from negative to positive at each bounce.
Worked examples
WE 1
Reading acceleration from a v–t graph
A cyclist’s velocity–time graph has five sections. Sections A, C and E are flat; sections B and D slope upward, with D the steepest. (a) In which section is the acceleration largest? (b) Between 5 s and 10 s (a sloped part) the velocity rises by 5 m s⁻¹. Find the acceleration there.
(a) Acceleration = gradient of a v–t graphflat sections (A, C, E) have zero acceleration; the steepest slope winssection D — the steepest slope(b) gradient = rise ÷ runa = 5 ÷ (10 − 5) = 5 ÷ 5a = 1 m s⁻²
WE 2
Displacement from the area under a v–t graph
A vehicle accelerates uniformly from rest, reaching 20 m s⁻¹ after 40 s. Using the velocity–time graph, find the displacement at 40 s.
Displacement = area under the v–t lineThe graph is a triangle (rest → 20 m s⁻¹)area = ½ × base × height= ½ × 40 × 20displacement = 400 m
WE 3
A two-part journey
A car accelerates from rest to 12 m s⁻¹ in 6 s, then travels at a steady 12 m s⁻¹ for another 10 s. Find the total displacement from the v–t graph.
Split the area into a triangle + a rectangleTriangle (0–6 s): ½ × 6 × 12= 36 mRectangle (6–16 s): 12 × 10= 120 mTotal = 36 + 120displacement = 156 m
💡 Top tips
Read the axes first. Always check whether you’re looking at a d–t, v–t, or a–t graph before deciding what gradient and area mean.
Gradient goes “down” the chain, area goes “up”. Slope takes you towards acceleration; area takes you back towards displacement.
For areas, split the shape into triangles and rectangles and add them up.
Draw a big gradient triangle for slopes — bigger triangles give more accurate answers.
A flat line isn’t always “stopped” — on a v–t graph it means constant velocity, not at rest.
⚠ Common mistakes
Confusing the graph types: a flat line means “at rest” on a d–t graph but “constant velocity” on a v–t graph.
Taking the area of a d–t graph — it has no physical meaning. Area only matters for v–t and a–t graphs.
Reading the height instead of the gradient (or vice versa) for the quantity asked.
Forgetting negative areas: below the time axis on a v–t graph, the displacement is negative.
Ignoring direction at a bounce — velocity flips sign even though the speed is unchanged.
Up next: Projectile Motion — what happens when an object moves horizontally and vertically at the same time, and why those two directions can be treated completely separately.
Need help with SL Kinematics?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
