What is distance time graph * Your answer?

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Importance of distance-time graphs
Sample Problems
Practice distance time graph questions
Distance time graph GCSE questions

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A distance-time graph is a graphical representation of how far a body has traveled in a specified amount of time. It is used to depict the relationship between distance and time, where distance is plotted on the Y-axis and time is plotted on the X-axis. Let’s first learn the importance of distance-time graphs.

Importance of distance-time graphs

Distance-time graphs help to study the motion of bodies. A distance-time graph shows how far someone or something has traveled and how long it took them/it to travel that distance. A distance-time graph is obtained when the data of distance and time obtained while studying the motion of a body is plotted on a rectangular graph.

Distance-time graphs

The graphs shown above are distance-time graphs for various types of body motion.

When a body is steady or stationary,
When a body is moving at a uniform speed,
When a body is moving non-uniformly with increasing speed, and
When a body is moving non-uniformly with decreasing speed.

Distance-time graph

Conclusions

The following are the points concluded from the distance-time graphs.

When a body is at rest, then the graph is parallel to the axis where time is plotted.
When the motion of a body is uniform, then the distance-time graph is a straight line.
The slope of the distance-time graph is equal to the speed of the body.
The slope of the straight-line graph is constant regardless of the chosen interval. This implies that an object moving uniformly will always move at the same speed.
The speed increases as the graph become steeper.
A negative gradient or slope means the body is returning to the starting point.

Sample Problems

Problem 1: The graph below describes the journey of a man. Determine the speed of the man for each part of the journey.

Solution:

From the given graph, we can conclude that,
Part a: From 10:00 to 11:30, the man traveled 15 km from the starting point in 1.5 hours.
Part b: From 11:30 to 13:30, the man is at rest, so the distance remains the same.
Part c: From 13:00 to 14:30, the man traveled the 15 km back to where they began in 1 hr.
The speed at part a: Speed = 15/1.5 = 10 km/h
The speed at part b: Speed = 0/2 = 0 km/h (not moving)
The speed at part c: Speed = 15/1 = 15 km/hr

Problem 2: Jay is going for a drive in his car. The distance-time graph given below describes his full journey. Calculate his total distance traveled during his journey, as well as his average speed between 4:30 and 5:30.

Solution:

Jay traveled 30 km away from his home and then stopped for a while. He again drove 20 km and stopped briefly. He then traveled 50 km back home.
So, the total distance traveled by Jay = 30 km + 20 km + 50 km = 100km.
From the axis, we can observe that two big squares total 30 minutes. Hence, one big square is worth 15 minutes. From 4:30 to 4:45, Jay is at rest.
So, his speed between 4:30 and 4:45 = 0/0.25 = 0 km/h.
To calculate the average speed of Jay between 4:30 and 5:30, we need to calculate the slope of the graph between 4:45 and 5:00. This period lasted for 15 minutes, which is equivalent to 0.25 hours – this is the “change in x”. During this period, he increased his distance from home from 30 km up to 50 km. That means he traveled 20 km in total – this is the “change in y”.
So, we get, slope = 20/0.25 = 80 km/h.
Hence, the average speed = (0 + 80)/2 = 40 km/h

Problem 3: Construct a distance-time graph from the description of Uma’s journey given below. Uma left her home at 17:00 and, after traveling at a constant speed for an hour, she had covered 28 kilometers before stopping. After half an hour of being stopped, she drove home at a constant speed, and it took her an hour and a half in total to reach home.

Solution:

From the given data, we conclude that,
From 17:00 to 18:00, Uma travels from 0 km away to 28 km away;
From 18:00 to 18:30, she is at rest.
From 18:30 to 20:00, she travels from 28 km away to 0 km away.
The final graph looks like this,

Problem 4: Every morning, Shubham walks from his home to the bus stop nearby, which is 750 meters away from his home. The below graph shows his journey on one particular day. Calculate his average speed from 0 sec to 40 sec and from 50 sec to 110 sec.

Solution:

To calculate the average speed of Shubam from 0 sec to 40 sec, we need to calculate the slope of the graph from 0 sec to 40 sec. This period lasted for 40 seconds – this is the “change in x”. During this period, he increased his distance from home from 0 m to 300 m. That means he traveled 300m in total – this is the “change in y”.
So, we get, slope = change in y/change in x = 300/40 = 7.5 m/s.
Thus, Shubham’s average speed from 0 to 40 sec is 7.5 m/s.
Now, to calculate the average speed of Shubam from 50 sec to 110 sec, we need to calculate the slope of the graph from 50 sec to 110 sec. This period lasted for 60 seconds – this is the “change in x”. During this period, he increased his distance from home from 200 m to 700 m. That means he traveled 500m in total – this is the “change in y”.
So, slope = 500/60 = 8.34 m/s.
Hence, Shubham’s average speed from 50 to 110 sec is 8.34 m/s.

Problem 5: A tourist bus journey on a day trip to the coast is recorded. Now, calculate the maximum speed of the bus throughout the trip and also how long the bus was stationary.

Solution:

The slope (or) gradient of a distance-time graph is the speed. Therefore, to calculate the fastest average speed, we must find the steepest section of the graph.
The bus traveled at maximum speed from 15:00 to 15:30. It covered 30 km in half an hour.
So, the maximum speed of the bus = 30/(0.5) = 60 km/hr.
The bus stopped for 30 minutes at the 40 km mark.

Here we will learn about distance-time graphs, including how to interpret and construct them.

There are also distance-time graph worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

Distance-time graphs are graphs that show the distance an object or person has travelled against time. They can also be referred to as travel graphs.

A distance-time graph will show the distance (in metres, kilometres, miles etc.) on the vertical axis ( $y$ -axis) and the time (in seconds, minutes, hours etc.) on the horizontal axis ( $x$ -axis).

The distance-time graphs we will look at on this page will all be drawn using straight lines.

Straight lines on a distance-time graph represent constant speeds. To find the speed from a distance-time graph you will need to be able to find the gradient of the straight lines.

The steeper the gradient, the faster the object is travelling.

One way of calculating the speed from a distance-time graph is to think about how far the object has travelled in one time unit.

For example, if an object has travelled $20$ miles in $30$ minutes, it would travel $40$ miles in one hour, therefore the speed is $40 \ mph.$

In order to read distance-time graphs:

Locate any relevant points from the distance-time graph.
Check the information required. For example, distance travelled at a selected time, period of being stationary, speed at a selected point in the journey.
Use the appropriate process from the list.

Distance travelled at a selected time – read from the vertical axis at the given time.
Period of being stationary – find the time when the line is horizontal.
Speed at a selected point in the journey – find the gradient of the line at that time or think how far has the object travelled in one time unit.
Average speed for the whole journey – divide the total distance travelled by the total time.

Get your free distance time graph worksheet of 20+ questions and answers. Includes reasoning and applied questions.

COMING SOON

Get your free distance time graph worksheet of 20+ questions and answers. Includes reasoning and applied questions.

COMING SOON

Distance time graph is part of our series of lessons to support revision on rates of change. You may find it helpful to start with the main rate of change lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

Rate of change
Speed time graph

The distance-time graph shows the journey of a person from their home.

How far away from their home are they after $1$ hour and $30$ minutes?

Locate any relevant points from the distance-time graph.

Locate $1$ hour $30$ minutes on the horizontal axis and go up to the graph line.

2Check the information required. For example, distance travelled at a selected time, period of being stationary, speed at a selected point in the journey.

We require the distance travelled at that time.

3Use the appropriate process from the list. Distance travelled at a selected time – read from the vertical axis at the given time.

Reading from the vertical axis gives a distance of $25 \ km.$

The distance-time graph shows the journey of a person from their home.

At what time did the person first stop and for how long?

Locate any relevant points from the distance-time graph.

Locate the points in the graph when the line changes.

Check the information required. For example, distance travelled at a selected time, period of being stationary, speed at a selected point in the journey.

We required the first period that the person was stationary.

Use the appropriate process from the list. Period of being stationary – find the time when the line is horizontal.

The person was first stationary after $30$ minutes, and remained stationary for $30$ minutes.

The distance-time graph shows the journey of a person from their home.

The graph shows the person taking a bus to a cafe. They then stopped at the cafe for a drink and snack. They then ran for an hour, before taking a taxi home.

What speed was the person travelling during the taxi journey home?

Locate any relevant points from the distance-time graph.

The journey home began after $2$ hours $30$ minutes and finished at $3$ hours.

Check the information required. For example, distance travelled at a selected time, period of being stationary, speed at a selected point in the journey.

We require the speed between $2.5$ hours and $3$ hours.

Use the appropriate process from the list. Speed at a selected point in the journey – find the gradient of the line at that time or think how far has the object travelled in one time unit.

The person travelled $30 \ km$ in $30$ minutes, this would mean they would travel $60 \ km$ in $1$ hour, therefore, their speed was $60 \ km/h.$

We could also find this using the speed, distance, time formula,

$\text{Speed}=\frac{\text{distance}}{\text{time}}=\frac{30}{0.5}=60\,\text{km/h}$

(it is important to use $0.5$ hours and not $30$ minutes for the time).

Or we could have found the gradient of the line,

$\frac{\text{Change in }y}{\text{Change in }x}=\frac{30}{0.5}=60.$

The distance-time graph shows the journey of a person from their home.

The graph shows the person taking a bus to a cafe. They then stopped at the cafe for a drink and snack. They then ran for an hour, before taking a taxi home.

What was the person’s average speed for their whole journey?

Locate any relevant points from the distance-time graph.

We need to find the total distances travelled.

Check the information required. For example, distance travelled at a selected time, period of being stationary, speed at a selected point in the journey.

We required the average speed for the whole journey.

Use the appropriate process from the list. Average speed for the whole journey – divide the total distance travelled by the total time.

It is important to add all the distances travelled and find the time for the whole journey including any periods when the person was stationary.

$\text{Average speed =}\frac{\text{total distance}}{\text{total time}}=\frac{60}{3}=20 \ \text{km/h.}$

In order to draw distance-time graphs:

Draw/label the horizontal axis for the time and a vertical axis for the distance.
Use the information about the distance or speed of the object to plot points on the graph.
Join the points with straight line segments.

A car starts from home and travels at a constant speed for $30$ minutes until it is $20$ miles from home. It then stops for $1$ hour before returning home, travelling at a constant speed for $45$ minutes.

Draw a distance-time graph to represent the journey.

Draw/label the horizontal axis for the time and a vertical axis for the distance.

The car gets no further than $20$ miles from home and the journey takes a total of $2$ hours $15$ minutes. Draw and label the vertical axis from $0$ to $20$ miles and the horizontal axis from $0$ to $3$ hours.

Use the information about the distance or speed of the object to plot points on the graph.

The car starts at home, so plot the point $(0, 0).$

After $30$ minutes the car has travelled $20$ miles, so plot the point $(0.5,20).$

It remains stationary for $1$ hour, so plot the point $(1.5,20).$

It then returns home after $45$ minutes, so plot the point $(2.25,0).$

Join the points with straight line segments.

A car leaves home at $12$ : $00$ and sets off at a constant speed of $30 \ mph.$

At $12$ : $30$ the car stops for $10$ minutes. It then returns home at a speed of $45 \ mph.$

Draw a distance-time graph to represent the journey.

Draw/label the horizontal axis for the time and a vertical axis for the distance.

Draw the horizontal and vertical axes, but we will finish labelling them once we know more information about the maximum distance and times taken based on the information provided about the speed.

Use the information about the distance or speed of the object to plot points on the graph.

Plot $(12$ : $00,0).$

The car travels at a speed of $30 \ mph$ between $12$ : $00$ and $12$ : $30. \ 30 \ mph$ means it would travel $15$ miles in $30$ minutes, so we will need to plot $(12$ : $30,15).$

The car stops for $10$ minutes, so we will need to plot $(12$ : $40,15).$

The car returns home at a speed of $45 \ mph.$

You can use $\text{time =}\frac{\text{distance}}{\text{speed}}=\frac{15}{45}=\frac{1}{3} \ hour,$ so it will take $20$ mins to travel $15$ miles.

We will need to plot $(13$ : $00,0).$

Choose appropriate scales for the axes and plot the points.

Join the points with straight line segments.

A person walks from home for $10$ minutes to a distance of $1 \ km.$ They stop for $5$ minutes, and then run at a speed of $8 \ km/h$ away from home. After $45$ minutes they immediately change direction and run towards home at a speed of $7 \ km/h.$

Draw a distance-time graph to represent their journey.

Draw/label the horizontal axis for the time and a vertical axis for the distance.

Draw the horizontal and vertical axes, but we will finish labelling them once we know more information about the maximum distance and times taken based on the information provided about the speed.

Use the information about the distance or speed of the object to plot points on the graph.

The person starts from home, so we will need to plot $(0, 0).$

The person walks from home for $10$ minutes to a distance of $1 \ km,$ so we will need to plot $(10,1).$

They stop for $5$ minutes, so we will need to plot $(15,1).$

Then run at a speed of $8 \ km/h$ away from home for $45$ minutes. $45$ minutes $= 0.75$ hours. Using $\text{distance = speed}\times \text{time = }8\times 0.75=6 \ km.$

We will need to plot $(60,7).$

They immediately change direction and run towards home at a speed of $7 \ km/h,$ they have $7 \ km$ to run which will take $60$ minutes, so we will need to plot $(120,0)$

Choose appropriate scales for the axes and plot the points.

Join the points with straight line segments.

Confusing distance-time graphs with speed-time graphs

It is common for students to confuse distance-time graphs with speed-time graphs. This may result in students just reading a value from the vertical axis instead of calculating the gradient of a distance-time graph to find a speed.

Calculating average speed incorrectly

The average speed could be incorrectly calculated by finding the mean of the different speeds during a journey. Another incorrect method is to only use the periods when the object is moving and to forget to include the periods when it is stationary.

Drawing impossible distance-time graphs

There are some graphs that would be impossible for a distance-time graph.

For example, vertical lines would represent an infinite speed.

Objects travelling back in time.

Practice distance time graph questions

The object is accelerating.

The object is stationary.

The object is moving at a constant speed.

The object is decelerating.

A horizontal line shows the object is covering zero distance for a period of time.

Greater speeds have a steeper gradient.

The object travels $40 \ km$ in $2$ hours.

The object travels $120 \ km$ in $1.5$ hours.

$60 \ km/h$ for $30$ minutes gives a distance of $30 \ km. \ 0 \ km$ at $40 \ km/h$ will take $45$ minutes.

$40 \ km$ at $80 \ km/h$ will take $30$ minutes. The journey ends after $90$ minutes.

Distance time graph GCSE questions

1. A salesperson was driving on a motorway from London to York. He stopped at a motorway service station halfway into his journey.

Which of the distance-time graphs could represent his journey?

(1 mark)

2. The distance-time graph shows part of a journey Sarah took on a bike ride.

(a) What did Sarah do between $09$ : $45$ and $10$ : $00?$

(b) What was Sarah’s speed between $10$ : $00$ and $10$ : $30,$ in $km/h?$

(3 marks)

(a)

She stopped moving/was stationary.

(1)

(b)

Process of dividing $6 \ km$ by $0.5$ hours or equivalent method.

(1)

12 \ km/h

(1)

3. The distance-time graph shows the first part of a journey of a person on a shopping trip.

The person walked from their home to the shop and arrived at $12$ : $00.$

The person stays in the shop for $15$ minutes and then catches a bus home.

The bus stopped outside their door and travelled at an average speed of $20$ miles per hour.

(a) At what speed did they walk to the shops? Give your answer in miles per hour.

(b) Complete the distance-time graph showing the remaining parts of their journey.

(4 marks)

(a)

5 \ mph

(1)

(b)

Horizontal line to $(12$ : $15, 5).$

(1)

Process to find the time for the bus ride ( $15$ minutes).

(1)

Line from $(12$ : $15, 5)$ to $(12$ : $30,0).$

(1)

You have now learned how to:

Read and interpret distance-time graphs
Draw distance-time graphs
Calculate speed from distance-time graphs

Conversion graphs
Units of measurement
Scale maths

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