Give feedback
Chapter Overview
Learning Goals
Learning Goals
Maths
Summary
Distance-time graphs give information about journeys. Not only do they tell you about the distance travelled in different lengths of time, they can also be used to calculate speed by finding the gradient of line segment.
For example, the following graph is a distance-time graph:
It has distance from the start position on the $y$-axis and time taken on the $x$-axis. This example shows different sections of a journey. In the first hour, four kilometres are covered. In the following hour, no distance is covered, so the graph stays at four kilometres.
Between hour two and hour three, the graph slopes down. On distance-time graphs this means that the direction is backwards - or in other words, travelling back towards the start position. Between hours three and four, motion has stopped again and for the final hour, travelling back to the start position resumes. Hence by hour five, the traveller is back to the start.
By taking any point on a distance-time graph, you will find the distance from the start position ($y$-coordinate) and the time since the start of the journey ($x$-coordinate). Note: Sometimes the $x$-axis is simply time of day rather than time since the beginning of the journey.
a. How far from the starting position is the traveller after three hours?
Reading off the graph at $x=3$ you find that the point on the graph has coordinates $(3,400)$, so the traveller has travelled $\underline{400}$ kilometres in the first three hours.
b. Does the traveller return to their start position?
Since the graph does not return to $y=0$ after the start, the traveller has not returned to the start. They stopped twice, but didn't turn around.
The gradient of a distance time graph represents the speed of that segment's motion, if the axes' units are compatible. Recall that
$\text{gradient}=\frac{\text{change in }y}{\text{change in }x}$
Note that distance is on the $y$-axis and time is on the $x$-axis, so it follows from
$\text{speed}=\frac{\text{change in distance}}{\text{change in time}}$
that gradient gives speed. Note: A negative gradient represents that the motion is in the direction back towards the beginning. Technically, the word velocity should be used rather than speed since velocity has a direction while speed does not.
Find the speed and direction of the three segments of the journey.
Firstly, the different segments are signified by the different line segments. So the first segment is between hour zero and hour two, the second segment is between hour two and hour three and the third segment is from hour three to hour four.
Recall that speed is given by the size of the gradient and direction is given by the positivity or negativity of the gradient. The gradient of the first line segment is found by using the coordinates of the start and end of the segment: start $(0,0)$, end $(2,5)$. Thus the gradient is
$\text{gradient}=\frac{5-0}{2-0}=\frac52$
So the speed is $\underline{2.5}$ kilometres per hour and is heading away from the start position. Note: The unit here is taken from the units on the axes.
The second line segment is horizontal so has a gradient of zero, hence speed is zero kilometres per hour.
The last line segment has a gradient of
$\text{gradient}=\frac{0-5}{4-3}=-5$
So the speed is $\underline{5}$ kilometres per hour, but back in the opposite direction.
Distance-time graphs give information about journeys. Not only do they tell you about the distance travelled in different lengths of time, they can also be used to calculate speed by finding the gradient of line segment.
For example, the following graph is a distance-time graph:
It has distance from the start position on the $y$-axis and time taken on the $x$-axis. This example shows different sections of a journey. In the first hour, four kilometres are covered. In the following hour, no distance is covered, so the graph stays at four kilometres.
Between hour two and hour three, the graph slopes down. On distance-time graphs this means that the direction is backwards - or in other words, travelling back towards the start position. Between hours three and four, motion has stopped again and for the final hour, travelling back to the start position resumes. Hence by hour five, the traveller is back to the start.
By taking any point on a distance-time graph, you will find the distance from the start position ($y$-coordinate) and the time since the start of the journey ($x$-coordinate). Note: Sometimes the $x$-axis is simply time of day rather than time since the beginning of the journey.
a. How far from the starting position is the traveller after three hours?
Reading off the graph at $x=3$ you find that the point on the graph has coordinates $(3,400)$, so the traveller has travelled $\underline{400}$ kilometres in the first three hours.
b. Does the traveller return to their start position?
Since the graph does not return to $y=0$ after the start, the traveller has not returned to the start. They stopped twice, but didn't turn around.
The gradient of a distance time graph represents the speed of that segment's motion, if the axes' units are compatible. Recall that
$\text{gradient}=\frac{\text{change in }y}{\text{change in }x}$
Note that distance is on the $y$-axis and time is on the $x$-axis, so it follows from
$\text{speed}=\frac{\text{change in distance}}{\text{change in time}}$
that gradient gives speed. Note: A negative gradient represents that the motion is in the direction back towards the beginning. Technically, the word velocity should be used rather than speed since velocity has a direction while speed does not.
Find the speed and direction of the three segments of the journey.
Firstly, the different segments are signified by the different line segments. So the first segment is between hour zero and hour two, the second segment is between hour two and hour three and the third segment is from hour three to hour four.
Recall that speed is given by the size of the gradient and direction is given by the positivity or negativity of the gradient. The gradient of the first line segment is found by using the coordinates of the start and end of the segment: start $(0,0)$, end $(2,5)$. Thus the gradient is
$\text{gradient}=\frac{5-0}{2-0}=\frac52$
So the speed is $\underline{2.5}$ kilometres per hour and is heading away from the start position. Note: The unit here is taken from the units on the axes.
The second line segment is horizontal so has a gradient of zero, hence speed is zero kilometres per hour.
The last line segment has a gradient of
$\text{gradient}=\frac{0-5}{4-3}=-5$
So the speed is $\underline{5}$ kilometres per hour, but back in the opposite direction.
Travel graphs: Distance-time graphs
FAQs
Question: What do distance-time graphs tell you?
Answer: By taking any point on a distance-time graph, you will find the distance from the start position (y-coordinate) and the time since the start of the journey (x-coordinate).
Question: What does a negative gradient mean on a distance-time graph?
Answer: A negative gradient represents that the motion is in the direction back towards the beginning.
Question: What does the gradient represent in a distance-time graph?
Answer: The gradient of a distance-time graph represents the speed.
Theory
Exercises
© 2020 – 2023 evulpo AG
Your data protection
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners who may combine it with other information that you’ve provided them or that they’ve collected from your use of their services. By clicking on either "Accept cookies" or "Necessary cookies only", you agree to this (read more in our Privacy Policy). Privacy Policy