Distance-time graphs

A distance-time graph turns a journey into a picture. Instead of describing in words how far something travelled and when, you plot distance on the vertical axis against time on the horizontal axis, and the shape of the resulting line tells the whole story of the motion at a glance. A straight line means the object is covering equal distances in equal times — moving at a steady speed. A flat, horizontal stretch means the object is not moving at all. A curve means the speed itself is changing from moment to moment. Once you know how to read a distance-time graph, you can look at any such graph — for a cyclist, a train, a falling ball's shadow — and immediately say whether it sped up, slowed down, paused, or kept a constant pace, and calculate exactly how fast it was going in each stage, without ever being handed a table of numbers.

The core insight

The slope of a distance-time graph at any point equals the speed of the object at that point, so straight segments mean uniform speed, flat segments mean rest, and curved segments mean changing speed.

The core idea

A distance-time graph plots the total distance travelled by an object (y-axis) against the time elapsed (x-axis). The single rule that connects the picture to the physics is that slope equals speed:

\text{Speed} = \text{slope of the graph} = \frac{\text{change in distance}}{\text{change in time}}

where: the change in distance is read off the y-axis between two points on the line, and the change in time is read off the x-axis between the same two points. Three shapes cover every case a Class 9 student needs to recognise. A straight line through any two points has a constant slope everywhere along it, so it represents uniform motion — equal distances covered in equal time intervals, at one fixed speed. The steeper the line, the greater that constant speed; a nearly flat-but-rising line is a slow uniform motion, and a steeply-rising line is a fast one. A horizontal line has zero slope: the distance value does not change while time keeps advancing, so the object is at rest for that stretch — parked, paused, or waiting at a signal. A curved line has a slope that keeps changing from point to point, so it represents non-uniform motion — the object's speed is different at different instants, either increasing (the curve bends upward more steeply as time goes on) or decreasing (the curve flattens out).

Figure

A distance-time graph showing three stages of a journey: a rising straight line for uniform motion, a flat horizontal line for rest, and a steeper rising straight line for a faster uniform motion

Since distance is a scalar equal to the total path length covered, it can never decrease while an object keeps moving — a distance-time graph therefore only rises or stays flat, and it never comes back down. This is a key difference from a velocity-time graph, which can dip below the axis when an object reverses direction. Also, the area enclosed under a distance-time graph has no standard physical meaning at this level; that trick belongs to velocity-time graphs, where the area under the graph gives distance or displacement. It is worth practising sketching a distance-time graph purely from a word description, without any numbers, since board questions often ask exactly that. For example: "a scooter rides at a steady pace for 5 s, halts for 3 s, then rides off again faster than before" should immediately translate into a straight-flat-straight shape, the second straight portion drawn steeper than the first, because both are uniform-motion segments with the second at higher speed.

Derivation ASpeed from two points on a distance-time graph

The goal is to turn any two marked points on a straight-line stretch of a distance-time graph into a single number for the speed during that stretch, using nothing beyond the basic definition of slope as rise over run, so that the same method works whether the two points are given as coordinates, read off a printed graph, or described in a word problem.

Step 1. Label the two points. Call the earlier point $(t_1, d_1)$ and the later point $(t_2, d_2)$ , where $d$ is the distance already covered and $t$ is the time elapsed since the start of observation. Both points must lie on the same straight segment of the graph, because the formula only gives a single, constant speed for a stretch where the slope does not change.

Step 2. Find the change in distance and the change in time separately. The change in distance is $\Delta d = d_2 - d_1$ , the vertical rise between the two points; the change in time is $\Delta t = t_2 - t_1$ , the horizontal run between them. Both differences are taken in the same order — later point minus earlier point — so that both come out positive for a rising line.

Step 3. Divide rise by run to get the slope, which is the speed for that segment.

v = \frac{\Delta d}{\Delta t} = \frac{d_2 - d_1}{t_2 - t_1}

where: $v$ is the speed of the object during that segment, $d_1$ and $d_2$ are the distances at the two chosen instants, and $t_1$ and $t_2$ are those two instants of time. The units of $v$ come out automatically from the axis labels — metres divided by seconds gives metres per second.

Step 4. Repeat separately for each straight segment of the graph if the journey has more than one uniform-motion stage, since each straight portion has its own constant slope and hence its own speed; a horizontal segment gives $\Delta d = 0$ so $v = 0$ , consistent with the object being at rest.

Worked example

Question: A distance-time graph for a car shows three stages. From $t = 0$ to $t = 5\ \text{s}$ , the distance rises from $0$ to $50\ \text{m}$ along a straight line. From $t = 5\ \text{s}$ to $t = 10\ \text{s}$ , the distance stays fixed at $50\ \text{m}$ . From $t = 10\ \text{s}$ to $t = 15\ \text{s}$ , the distance rises from $50\ \text{m}$ to $125\ \text{m}$ along a straight line. Describe the motion in each stage and find the speed in the first and third stages.

Step 1. Identify stage 1 as a rising straight line, so it is uniform motion, and apply the slope formula with $(t_1, d_1) = (0, 0)$ and $(t_2, d_2) = (5, 50)$ .

v_1 = \frac{50 - 0}{5 - 0} = 10\ \text{m/s}

Step 2. Identify stage 2 as horizontal, since the distance stays at $50\ \text{m}$ for the entire $5$ s interval, so $\Delta d = 0$ and the car is at rest — perhaps stopped at a traffic signal.

Step 3. Identify stage 3 as another rising straight line, so it is uniform motion again, and apply the slope formula with $(t_1, d_1) = (10, 50)$ and $(t_2, d_2) = (15, 125)$ .

v_3 = \frac{125 - 50}{15 - 10} = \frac{75}{5} = 15\ \text{m/s}

Final answer: the car travels at a uniform $\mathbf{10\ m/s}$ for the first $5$ s, halts for $5$ s, then travels again at a higher uniform $\mathbf{15\ m/s}$ for the last $5$ s — the steeper slope of stage 3 on the graph is the visual sign of the greater speed. (As a quick single-line check: a straight line through the origin reaching $60\ \text{m}$ at $t = 10\ \text{s}$ would give $v = 60/10 = 6\ \text{m/s}$ , the same formula applied to a single uniform stage.)

Common traps

Calling a steeper straight line "acceleration" — a straight line, however steep, is always uniform (non-accelerating) motion; only a curve represents a changing speed.
Reading a horizontal line as the object moving backward or vanishing — it only means the distance is not increasing, i.e., the object is at rest during that interval.
Trying to compute the "area under the graph" on a distance-time graph and expecting a meaningful answer — that technique belongs to velocity-time graphs, not distance-time graphs.
Mixing up $\Delta d$ and $\Delta t$ in the slope formula, or subtracting the points in opposite orders in the numerator and denominator, which flips the sign or the units of the answer.

Ask the tutor: If you have a distance-time graph from your textbook that mixes straight and curved stretches, ask the AI tutor to walk through each stage with you and find the speed wherever it is constant.

Why can a distance-time graph never slope downward?

Because distance is the total length of the path travelled so far, and that total can only stay the same or grow as time passes — it can never shrink, even if the object turns around and heads back toward its starting point, because the ground already covered still counts. This is different from a position or displacement graph, which can decrease if the object moves back toward its starting point. So on a genuine distance-time graph you will only ever see the line rise or run flat, never dip.

How do I know from the graph whether a curve means the object is speeding up or slowing down?

Look at how steep the curve is getting as time increases. If the curve bends upward more and more steeply — the slope is increasing — the object is covering more distance per second than before, so its speed is increasing. If the curve is flattening out — getting less steep as time goes on — the slope is decreasing, so the speed is dropping. At the Class 9 level it is enough to describe this qualitatively as "speed is increasing" or "speed is decreasing"; formally defining acceleration is done later using velocity-time graphs, not distance directly.

If two straight segments on the same graph have the same slope, does that mean the same speed?

Yes. Slope is speed, and slope depends only on how steeply the line rises, not on where it sits on the graph. Two separated straight segments with the same steepness represent two stages of motion at the same constant speed, even if there was a pause or a different speed in between them. You can check this by picking any two points on each segment and confirming the ratio $\Delta d / \Delta t$ comes out equal.

Distance-time graphs

The core idea

Worked example

Common traps

Before this

Where this leads