Motion graphs

A journey often tells its story better as a picture than as a paragraph. Plot how far a scooter has travelled against the clock, or how its speed rises and falls, and the shape of the line does the explaining for you: a steeper slope means faster motion, a flat stretch means rest or steady speed, and the space trapped under the line even hands you the distance covered, without a single equation of motion in sight. This article builds the two workhorse graphs of straight-line motion — distance-time and velocity-time — and shows exactly what their slopes and areas mean, why the two graphs answer different questions, and how to read a described journey straight off the axes.

The core insight

On a distance-time graph the slope is speed, and on a velocity-time graph the slope is acceleration while the area under the line is distance travelled.

The core idea

Distance-time graphs plot time on the x-axis and distance along the y-axis, and they answer "how far, and how fast is distance piling up?"

A straight line through the origin means uniform speed — equal distances covered in equal time intervals. The steeper the line, the greater the speed.
A horizontal line means the body is at rest: distance is not changing even though time is passing.
A curve means the speed itself is changing from moment to moment.

The single number that describes a straight stretch is its slope:

\text{Speed} = \text{slope of distance-time graph} = \frac{\Delta(\text{distance})}{\Delta(\text{time})}

where: $\Delta(\text{distance})$ is the change in distance (m) over the interval, and $\Delta(\text{time})$ is the corresponding change in time (s), so speed comes out in m/s.

Velocity-time graphs plot time against velocity instead, and they answer a different question: "how is velocity itself changing?"

A horizontal line means uniform velocity — zero acceleration.
A sloped straight line means uniform acceleration: velocity rises steadily if the line climbs, and falls steadily (the body retards) if the line descends. Because only uniformly accelerated motion is in this syllabus, every velocity-time graph you meet here is made of straight segments, never curves.

The slope of this graph is acceleration:

\text{Acceleration} = \text{slope of velocity-time graph} = \frac{v-u}{t}

where: $u$ is the velocity at the start of the interval (m/s), $v$ is the velocity at the end (m/s), and $t$ is the time taken (s), so acceleration comes out in m/s². The velocity-time graph carries one more piece of information that the distance-time graph does not give directly: the region trapped between the line and the time axis is numerically equal to the distance covered in that interval. This is the idea the derivation below makes precise.

Figure

Two graphs side by side, a distance time graph rising in steps and a velocity time graph with a sloped rise followed by a flat top, with the enclosed area shaded under the velocity time line

Derivation ADistance travelled equals the area under a velocity-time graph

The goal is to show, for a body moving with uniform acceleration whose velocity-time graph is a straight line from an initial velocity $u$ to a final velocity $v$ over a time $t$ , that the region enclosed between that line and the time axis works out to exactly the distance the body has actually covered, so that "find the area" and "find the distance" become the same calculation.

Step 1. Picture the shape enclosed. Plot velocity against time with $u$ marked at $t = 0$ and $v$ marked at $t$ , joined by a straight line (since the acceleration is uniform). The region between this line and the time axis, from $0$ to $t$ , is a trapezium: it has two parallel vertical sides of length $u$ and $v$ , and a horizontal width $t$ .

Step 2. Split the trapezium into a rectangle and a triangle. The rectangle sits below, with height $u$ and width $t$ , giving area $u \times t$ . The triangle sits on top of it, with base $t$ and height $(v-u)$ — the rise in velocity — giving area $\tfrac12 \times t \times (v-u)$ .

Step 3. Add the two pieces:

\text{Area} = ut + \tfrac12 t(v-u)

Expand and simplify: $\text{Area} = ut + \tfrac12 vt - \tfrac12 ut = \tfrac12 ut + \tfrac12 vt = \tfrac12(u+v)t$ . This is exactly the standard "average velocity times time" rule, arrived at purely from the geometry of the graph.

Step 4. Confirm this is the actual distance. Over a short slice of time $\delta t$ , the distance covered is (velocity at that instant) $\times \delta t$ — a thin vertical strip under the graph. Adding up all such strips across the whole interval is precisely what "the area under the graph" means, so the total area is the total distance:

\text{Distance} = \tfrac12(u+v)t

The same reasoning applies to any straight-line or piecewise-straight velocity-time graph: break the enclosed region into rectangles and triangles (or trapeziums), find each area with ordinary mensuration, and add them up.

Worked example

Question: A bus starts from rest and speeds up uniformly to 20 m/s in 10 s, then travels at this constant velocity for the next 15 s, and finally slows down uniformly to rest over the next 5 s. Use the velocity-time graph to find (a) the acceleration during the first 10 s, (b) the retardation during the last 5 s, and (c) the total distance covered over the whole 30 s journey.

Step 1. Describe the graph. It has three straight segments: rising from $(0\text{ s}, 0)$ to $(10\text{ s}, 20\text{ m/s})$ , flat at $20$ m/s from $t = 10$ s to $t = 25$ s, then falling from $(25\text{ s}, 20\text{ m/s})$ to $(30\text{ s}, 0)$ .

Step 2. Acceleration in phase 1, from the slope: $a = \dfrac{v-u}{t} = \dfrac{20-0}{10} = 2\text{ m/s}^2$ . This matches Worked example 2 of the Pass-1 article for the same rising segment.

Step 3. Retardation in phase 3, from the slope: $a = \dfrac{v-u}{t} = \dfrac{0-20}{5} = -4\text{ m/s}^2$ , i.e. a retardation of $4\text{ m/s}^2$ — the bus loses velocity twice as fast as it gained it, because the same speed change (20 m/s) happens over a shorter time.

Step 4. Distance from each part, using area. Phase 1 is a triangle: $\tfrac12 \times 10 \times 20 = 100\text{ m}$ (this reproduces the 100 m of Pass-1's Worked example 2). Phase 2 is a rectangle: $15 \times 20 = 300\text{ m}$ (also matching Pass-1). Phase 3 is a new triangle: $\tfrac12 \times 5 \times 20 = 50\text{ m}$ .

Step 5. Add the three areas: $\text{Total distance} = 100 + 300 + 50 = 450\text{ m}$ . The bus accelerates at 2 m/s², retards at 4 m/s², and covers a total of 450 m over the 30 s journey.

Common traps

Calling the slope of a distance-time graph "acceleration" — it is always speed or velocity; acceleration only ever comes from the slope of a velocity-time graph.
Mixing up what a horizontal line means on the two graphs: flat on a distance-time graph means the body is at rest, but flat on a velocity-time graph means the body is cruising at constant, non-zero velocity.
Treating the area under a sloped part of a velocity-time graph as a rectangle instead of splitting it into a triangle (or trapezium) plus a rectangle, which overstates or understates the distance.
Drawing a curved velocity-time graph at this level — with only uniform acceleration, retardation, and rest in scope, every segment must be a straight line.

Ask the tutor: Want to practise sketching a velocity-time graph from a word problem and splitting its area into triangles and rectangles? Ask the AI tutor to walk through one with you.

Why does the same "slope" trick give speed on one graph and acceleration on the other?

Slope always means "how fast is the y-axis quantity changing with time," so what you get depends entirely on what is plotted on the y-axis. On a distance-time graph, the y-axis is distance, so its rate of change with time is speed. On a velocity-time graph, the y-axis is velocity, so its rate of change with time is acceleration. Same mathematical idea, different physical quantity, because the axes are different.

Can a velocity-time graph ever show a curve in this syllabus?

No, not at this level. This chapter only studies uniform velocity, uniform acceleration, and uniform retardation, and all three produce perfectly straight segments on a velocity-time graph — horizontal for uniform velocity, sloped upward for uniform acceleration, sloped downward for uniform retardation. A curved velocity-time graph would mean the acceleration itself is changing with time, which is a more advanced topic outside this syllabus. If you see a curve in a problem here, re-check the given data rather than trying to read it.

If I only know the area under a velocity-time graph, can I get the equations of motion from it?

Yes — that is exactly how this syllabus derives them. Splitting the trapezium under a uniformly accelerated velocity-time graph into a rectangle and a triangle, the way the derivation above does, gives $s = ut + \tfrac12 at^2$ once you also use $a = (v-u)/t$ to eliminate $v$ . The graph is not just a picture; it is the geometric proof behind the standard equations of motion, covered in full in a separate article.

Motion graphs

The core idea

Worked example

Common traps

Before this

Where this leads