Distance-Time Graphs
A distance-time graph is the most-read graph in school mathematics, and every feature of it means something specific. Steepness is speed, flat is stationary, falling is returning, and curved is a speed that will not stay still.
Two people walk the same route. One walks steadily. The other sprints, stops for a rest, then walks slowly. Sketch what each graph of distance against time would look like, and say which single feature of the second graph tells you when the rest happened.
On a distance-time graph the gradient is the speed. Steeper means faster, horizontal means stationary, and a section that falls means the distance from the reference point is decreasing: the object is returning.
$$\text{speed} = \text{gradient} = \frac{\text{distance}}{\text{time}}$$
A straight section means a constant speed and a curved section means the speed is changing. A curve getting steeper is speeding up; a curve flattening out is slowing down. That single distinction carries most of the marks in this topic.
Know
- That the gradient of a distance-time graph is the speed
- That a horizontal section means stationary and a falling section means returning
- That a straight section has constant speed and a curved section does not
Understand
- Why a falling section does not mean a negative distance
- Why a curved section has no single speed
Can Do
- Calculate speeds from the straight sections of a distance-time graph
- Describe a journey in words from its graph, including the initial and final points
- Describe a curved section qualitatively as speeding up or slowing down
On a distance-time graph, each visual feature has a fixed meaning.
Steepness is speed. The gradient is distance divided by time, which is a speed. A steeper section is a faster one.
Horizontal means stationary. Zero gradient is zero speed. The object is not moving, though time continues, which is why the section has a length.
Rising means moving away from the reference point, and falling means returning towards it.
Straight means a constant speed, and curved means the speed is changing.
The starting and finishing points matter too. The height at the left edge is where the object began, which need not be zero: a graph starting at $(0, 15)$ describes something that was already $15$ km from the reference point when timing started. The height at the right edge is where it finished.
A falling section never means a negative distance. The vertical axis usually measures distance from a particular place, and that measurement genuinely gets smaller as the object comes home.
For a straight section, the speed is the gradient, calculated from the two endpoints and quoted with units.
A section from $(0,0)$ to $(3, 60)$, with kilometres up and hours across:
$$\text{speed} = \frac{60 - 0}{3 - 0} = 20 \ \text{km/h}$$
For a falling section from $(4.5, 60)$ to $(7, 0)$, the gradient is
$$\frac{0 - 60}{7 - 4.5} = \frac{-60}{2.5} = -24$$
and the natural description is "returning at $24$ km/h". The negative sign records the direction of travel; quoting a speed as a negative number is unusual, so let the word do the work.
Average speed is a different quantity and is calculated differently:
$$\text{average speed} = \frac{\text{total distance travelled}}{\text{total time}}$$
For the whole journey above, the object travelled $60$ km out and $60$ km back, so $120$ km in $7$ hours, giving about $17.1$ km/h. That is not the average of $20$ and $24$, because the two phases lasted different lengths of time, and the stationary hour and a half counts towards the total time as well.
A curved section means the speed is not constant, and the shape of the curve says which way it is changing.
A curve getting steeper is speeding up. Each successive time interval covers more distance than the last, so the graph rises more sharply as it goes.
A curve flattening out is slowing down. Each interval covers less distance than the one before.
The same applies to a falling section: a fall that steepens is a return that is speeding up, which is what the diagram above shows.
What you cannot do is quote a single speed for a curved section, because there is no single speed to quote. Two honest responses are available:
describe it qualitatively, as "speeding up throughout" or "slowing down towards the end";
calculate the average speed over the section, using its two endpoints, and say that is what it is.
The average over a curved section is the gradient of the straight line joining its endpoints, which is a real and useful number. Calling it "the speed" is the error; calling it "the average speed over that interval" is correct.
Two different graphs describe the same journey, and reading the axis label is the only way to tell them apart.
Distance from a fixed point. This is the common one. It rises on the way out, is flat during a stop, and falls on the way home, finishing at zero if the object returns to where it started.
Total distance travelled. This can never fall, because distance already travelled cannot be undone. It rises on the way out, is flat during a stop, and rises again on the way home. It finishes at the total journey length.
So for the journey in the diagram, the first version ends at $0$ km and the second ends at $120$ km. Same journey, entirely different graph.
A quick check when reading an unfamiliar graph: if any section falls, the axis must be measuring distance from somewhere, not distance travelled. If nothing ever falls, either could apply and the label decides.
Speed-time graphs are a third kind again, with speed rather than distance on the vertical axis, and their gradients mean something different once more. Those belong to senior courses; the point here is simply to read the label before interpreting anything.
Watch Me Solve It · 3 examples
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1(a) Describe each phase from its shapeThe traveller leaves home and drives steadily away for two hours, reaching $90$ km. They then stop for one hour. They then drive steadily home, arriving after a total of five hours.
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2(b) Find the outward speed$\frac{90 - 0}{2 - 0} = 45 \ \text{km/h}$
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3(b) Find the return speed$\frac{0 - 90}{5 - 3} = -45 \ \text{km/h}$So the return is also at $45$ km/h, with the negative sign recording that the distance from home is decreasing. The two phases happen to be at the same speed, which the equal steepness of the two segments shows directly.
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4(c) Use total distance over total time$\frac{90 + 90}{5} = \frac{180}{5} = 36 \ \text{km/h}$Note this is not $45$, even though both moving phases were at $45$ km/h, because the stationary hour counts towards the total time while adding no distance.
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1(a) Read the curvatureThe graph rises throughout, so the object is moving away from the start the whole time. It gets steeper, so each hour covers more distance than the one before: the object is speeding up.
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2(b) Explain the difficultyThe gradient is different at every point of a curve, and the speed is the gradient, so the speed is different at every moment. There is no single number to quote.
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3(c) Average over the interval$\frac{80 - 0}{4 - 0} = 20 \ \text{km/h}$This is the gradient of the straight line joining the two endpoints, and it is a genuine quantity: the average speed over the four hours.
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4(d) Compare the end speed with the averageGreater. The object is speeding up throughout, so it was slower than average early on and must be faster than average later to compensate. The curve at $t = 4$ is steeper than the straight line joining the endpoints, and steeper means faster.
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1(a) Build the first graph phase by phaseRises in a straight line from $(0,0)$ to $(40, 12)$; horizontal from $(40,12)$ to $(60,12)$; falls in a straight line from $(60,12)$ to $(90, 0)$. Time is in minutes and distance in kilometres.
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2(a) State the final valueIt finishes at $0$ km, because the cyclist is back home and so is zero distance from it.
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3(b) Build the second graphRises from $(0,0)$ to $(40,12)$, identical to the first for this phase. Horizontal from $(40,12)$ to $(60,12)$, since no distance is covered while resting. Then rises again from $(60,12)$ to $(90, 24)$, because riding home still adds to the distance travelled.
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4(b) State the final value and compareIt finishes at $24$ km, the total length of the ride. The two graphs agree for the first hour and diverge completely afterwards: one falls to zero and the other rises to $24$. Reading the axis label is the only way to tell which is being shown.
Brain Trainer · 5 problems
Five items on distance-time graphs. Work each one, then reveal the answer.
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1 What does a horizontal section of a distance-time graph mean?
The gradient, and so the speed, is zero.The object is stationary -
2 A section runs from $(0,0)$ to $(4, 100)$, in km and hours. Find the speed.
Gradient is $\tfrac{100}{4}$.$25$ km/h -
3 What does a falling section mean, on a graph of distance from home?
The measured distance is getting smaller.Returning towards home -
4 A section curves and gets steeper. What is happening?
Steeper means faster.Speeding up -
5 Someone travels $30$ km in $1$ h, rests $1$ h, then returns in $1$ h. Find the average speed.
Total distance over total time: $\tfrac{60}{3}$.$20$ km/h
Multiple Choice · 5 questions
On a distance-time graph, the gradient of a section represents:
On a graph of distance from home against time, a section that falls means:
A section of a distance-time graph curves upward, becoming steeper. This means:
A journey covers $40$ km in $1$ hour, then rests for $1$ hour. The average speed for the two hours is:
A distance-time graph never falls at any point. This tells you:
Short Answer · 3 questions
(a) Describe the journey in words.
(b) Find the speed in each moving phase, in km/h.
(c) State which phase was faster, and by how much.
(d) Find the average speed for the whole $45$ minutes.
(a) Describe the motion, referring to the shape of the curve.
(b) Explain why you cannot state a single speed for the journey.
(c) Find the average speed over the six hours.
(d) State whether the speed at $t = 1$ was greater or less than the speed at $t = 5$, and whether the speed at $t = 1$ was greater or less than the average, justifying both.
(a) Describe the graph of distance from the park against time, giving the coordinates of every corner.
(b) Describe the graph of total distance run against time, giving the coordinates of every corner.
(c) State the final value on each graph and explain why they differ.
(d) Find the average speed for the whole outing, and explain why it would be the same whichever of the two graphs you used.
(b) Explain why a distance-time graph can never be vertical, and what a nearly vertical section would describe.
(c) A graph of distance from home rises, falls back to zero, then rises again. Describe a journey it could represent, and explain why the average speed cannot be found from the final height alone.
Gradient is speed
Steeper means faster
Flat and falling
Flat is stationary; falling is returning
Curves
Steepening is speeding up; flattening is slowing down
Average speed
Total distance over total time, never an average of speeds
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