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Lesson 5 Displacement vs. Distance

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Course: Science 10 [5 cr] - AB Ed copy 1
Book: Lesson 5 Displacement vs. Distance
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Date: Sunday, 7 September 2025, 6:43 PM

Table of contents

  Introduction


How can we measure the motion of an object?

C2.1 Light trails in busy roadway
An object's motion can be measured and described in many ways. Is it speeding up? Is it slowing down? How far has it travelled? Where is it in relation to where it started? These are all questions that we can answer through measurements and calculations, as well as through creating and looking at graphs.

In Section 2, we will explore the ideas of speed and acceleration of an object. We will look at how to create graphs of data, and how to use a graph to describe the motion of an object. This is a continuation from Section 1, where we first looked at the kinetic energy of an object and work done by, or on, an object.

  Targets


By the end of this section, you will be able to


C2.2 Aerial view of a roadway

  1. identify what scalar and vector quantities are, and work with position-time graphs.

C2.3 Speed limit sign on mountain road

  1. use the terms velocity and speed to describe motion, calculate velocity, and work with speed-time and position-time graphs.

C2.4 Toy rocket taking off

  1. describe the motion of an object when velocity is changing over time, calculate acceleration, and work with acceleration-time and speed-time graphs.

  Introduction

Do you know the difference between the displacement of an object and the distance an object travels?



C5.1 A twisty road
How is it that the Earth takes 365.25 days to travel 149.6 million km to make one complete orbit around the sun, yet its displacement is zero? In this lesson, you will learn the difference between the concepts of distance and displacement.  We will investigate how distance and displacement can be calculated. Through calculations and graphs, you will learn how an object’s motion over a specific distance and time, can be communicated.

Targets

By the end of this lesson, you will be able to

  • define, compare, and contrast scalar and vector quantities (define scalar and vector)
  • describe displacement quantitatively
  • investigate scalar motion and work done using calculations and graphs

  Watch This

Distance and Displacement: What Are They and What’s the Difference © YouTube NinetyEast 


This video provides you with a great overview of the difference between distance and displacement. It introduces you to the terms “scalar quantities” and “vector quantities” as well. This will help get you in the right mindset for your first lesson in this section.


  Scalar and Vector Quantities

When asked to explain the motion of an object, it is useful to know not just how far the object travelled, but also where it ended up in relation to where it started.


To describe the motion of an object, there are two main quantities you can use.

Scalar Quantities


C5.2 Measuring tape
Values, that describe the size or amount, but not the direction, are called scalar quantities. These quantities measure only the magnitude because they do not include a direction.

Here are some examples:

  • The length of string I need to tie a bow is 1.2 m.
  • The density of water is 1 g/cm3.
  • The atmospheric pressure outside today is 101.325 kPa.
  • The mass of the box is 4.25 kg.
  • The temperature outside today is 14.2 °C.

Vector Quantities


C5.3 Compass to measure direction
Quantities that describe magnitude, and direction, are called vector quantities. Scientifically, some quantities and processes in our world depend on the direction in which they occur.

Here are some examples:

  • The weight of the box is 41.7 N due to gravity pulling downwards.
  • I need to drive 32.6 km NE to get home from school.
  • The tow truck dragged the car 230 m to the left.
  • I ran 400 m up the hill to catch my dog.
  • The child was lifted upwards 76 cm.

In science, the symbols for a vector quantity are written with an arrow above them, to indicate the quantity is a vector, not scalar, quantity.

For example, a car travelling a velocity of 36 km/h north would be represented by v→=36 km/h N. A car travelling a speed of 36 km/h would be represented by v = 36 km/h.

  Read This

Please read page 137 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on the difference between scalar and vector quantities. Remember, if you have any questions, or do not understand something, ask your teacher!

  Practice Questions

Complete the following practice questions to check your understanding of the concept you just learned. Make sure you write complete answers to the practice questions in your notes. After you have checked your answers, make corrections to your responses (where necessary) to study from.

  1. Describe, in your own words, what the difference is between scalar and vector quantities.

    Your answer should be a variation of the following. A scalar quantity (size or amount) describes only magnitude only. A vector quantity (size or amount) describes only magnitude and direction.

  Distance, Displacement, and Position

When describing the distance that an object travelled, it is often important to use the point where it started, as a reference.



C5.4 Map of Alberta
The distance an object travelled is a scalar quantity—it is the length of the path between two points. If you are talking about the position the object ends up compared to where it started, then you are referring to displacement.

Displacement describes the straight-path distance from one point to another, including its direction.

The symbol for distance is "Δd." The symbol "Δ" is the Greek letter delta, which means change. So Δd means change in the distance of an object from one point to a second point.

The symbol for displacement is "«math»«mo»§#8710;«/mo»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«/math»." Displacement is a vector quantity, so the arrow is written above it.

Let’s look at a practical example of the difference between distance and displacement.

A hockey fan lives in Medicine Hat, Alberta, and wants to travel to Edmonton to watch an NHL hockey game. He checks out the fastest route on a road map.

C5.5 Map of Alberta with route from Medicine Hat to Edmonton
The distance along the highway from Medicine Hat to Edmonton is 560 km.

Δd = 560 km

The blue line in image C5.5 represents this distance.

If a person is able to follow a direct path from Medicine Hat to Edmonton, the distance is 435 km [NW].

«math»«mi»§#916;«/mi»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#160;«/mo»«mn»435«/mn»«mo»§#160;«/mo»«mi»km«/mi»«mo»§#160;«/mo»«mo»[«/mo»«mi»NW«/mi»«mo»]«/mo»«mo»§#160;«/mo»«/math»

The black line in image C5.5 represents this path. It represents displacement, as it describes the straight-path distance from one city to the next, including its direction.

Calculating Distance and Displacement


When calculating the distance travelled by an object, if the object travels along different paths, you can simply add the distances travelled in each path.


When the direction an object is travelling is indicated, there are standards that indicate if the value is positive or negative.

  1. [up] is postive.
  2. [down] is negative.
  3. [right] is positive.
  4. [left] is negative.
  5. [North] is postive.
  6. [South] is negative.
  7. [East] is positive.
  8. [West] is negative.


For example, if a football player back pedals 2.1 m [N], catches the ball, and then runs 7.4 m [S], what distance did the player travel?

To calculate the distance, you would perform the calculation Δd = 21 m + 7.4 m = 9.5 m.

When calculating the displacement of an object, if the object has travelled in more than one path, you need to take into account the direction of each path when adding each length.

In the example of the football player, the player travels 2.1 m [N] and then runs 7.4 m [S]. This is in the opposite direction. So, this needs to be taken into account in the calculation. Also, remember that north is a positive direction and south is a negative direction.

C5.6 Path of football player
To calculate the displacement, you would add the two values, taking into account that north is a positive direction and south is a negative direction.

«math»«mo»§#160;«/mo»«mi»§#916;«/mi»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mo»(«/mo»«mo»+«/mo»«mn»2«/mn»«mo».«/mo»«mn»1«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mo»§#8211;«/mo»«mn»7«/mn»«mo».«/mo»«mn»4«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»)«/mo»«mo»=«/mo»«mn»5«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo».«/mo»«mo»§#160;«/mo»«/math»

A negative answer here means that the player has ended up south of his starting position; south is negative.

Or, you could also look at the calculation for displacement as the operation:

«math»«mi»§#916;«/mi»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mn»2«/mn»«mo».«/mo»«mn»1«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«mo»[«/mo»«mi mathvariant=¨normal¨»N«/mi»«mo»]«/mo»«mo»-«/mo»«mn»7«/mn»«mo».«/mo»«mn»4«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«mo»[«/mo»«mi mathvariant=¨normal¨»S«/mi»«mo»]«/mo»«mo»=«/mo»«mn»5«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«mo»[«/mo»«mi mathvariant=¨normal¨»S«/mi»«mo»]«/mo»«mo»§#160;«/mo»«/math»

Notice that you still find the answer to be that the football player ends up 5.3 m south of his original position.

Examples:

  1. A stray cat is out at night and walks eastward 5.4 m, and then turns and walks westward 8.9 m.
  1. What distance does the cat travel?

    To calculate the distance, you would perform the calculation Δd = 5.4 m + 8.9 m = 14.3 m.
  2. What is the displacement (magnitude and direction) of the cat?

    To calculate the displacement, you would add the two values, taking into account that east is a positive direction and west is a negative direction.

    «math»«mo»§#160;«/mo»«mi»§#916;«/mi»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mo»(«/mo»«mo»+«/mo»«mn»5«/mn»«mo».«/mo»«mn»4«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mo»§#8211;«/mo»«mn»8«/mn»«mo».«/mo»«mn»9«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»)«/mo»«mo»§#160;«/mo»«mo»=«/mo»«mo»§#8211;«/mo»«mn»3«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«/math»

    Because west is negative, then an answer of –3.5 m means that you could express it as «math»«mi»§#916;«/mi»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mo»§#8211;«/mo»«mn»3«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«mi»or«/mi»«mo»§#160;«/mo»«mn»3«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mo»[«/mo»«mi mathvariant=¨normal¨»W«/mi»«mo»]«/mo»«mo».«/mo»«mo»§#160;«/mo»«/math»
  1. A child runs across the playground 12.3 m [N] and then 16.8 m back to the south.
  1. What distance does the child travel?

    To calculate the distance, you would perform the calculation Δd = 12.3 m + 16.8 m = 29.1 m.
  2. What is the displacement (magnitude and direction) of the child?

    To calculate the displacement, you would add the two values, taking into account that north is a positive direction and south is a negative direction.

    «math»«mi»§#916;«/mi»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mo»(«/mo»«mo»+«/mo»«mn»12«/mn»«mo».«/mo»«mn»3«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mo»§#8211;«/mo»«mn»16«/mn»«mo».«/mo»«mn»8«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»)«/mo»«mo»=«/mo»«mo»§#8211;«/mo»«mn»4«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo».«/mo»«mo»§#160;«/mo»«/math»

    Because south is negative, then an answer of –4.5 m means that you could express it as «math»«mi»§#916;«/mi»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mo»§#8211;«/mo»«mn»4«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«mi»or«/mi»«mo»§#160;«/mo»«mn»4«/mn»«mo».«/mo»«mn»5«/mn»«mo»§#160;«/mo»«mo»[«/mo»«mi mathvariant=¨normal¨»S«/mi»«mo»]«/mo»«mo».«/mo»«/math»
  1. A kayaker paddles 32 m [N], then 6 m [S], and then 16 m [N] again.

    1. What distance does the kayaker travel?

      To calculate the distance, you would perform the calculation Δd = 32 m + 6 m + 16 m = 54 m.
    2. What is the displacement (magnitude and direction) of the kayaker?

      To calculate the displacement, you would add the three values, taking into account that north is a positive direction and south is a negative direction.

      «math»«mi»§#916;«/mi»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mo»(«/mo»«mo»+«/mo»«mn»32«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mo»§#8211;«/mo»«mn»6«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»)«/mo»«mo»+«/mo»«mo»(«/mo»«mo»+«/mo»«mn»16«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»)«/mo»«mo»=«/mo»«mo»+«/mo»«mn»42«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo».«/mo»«mo»§#160;«/mo»«/math»

      Because north is positive, then an answer of +42 m means that you could express it as «math»«mi»§#916;«/mi»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mo»+«/mo»«mn»42«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«mi»or«/mi»«mo»§#160;«/mo»«mn»42«/mn»«mo»§#160;«/mo»«mo»[«/mo»«mi mathvariant=¨normal¨»N«/mi»«mo»]«/mo»«mo».«/mo»«mo»§#160;«/mo»«/math»



Do you want a bit more detailed explanation of how to calculate the distance and displacement of an object? Watch this video for more information on how to complete these calculations.

https://adlc.wistia.com/medias/wzrxuitjpd

 












Read This

Please read pages 137 to 138 in your Science 10 textbook. Make sure you take notes on your readings to study from later.  You should focus on how distance and displacement are measured, calculated, and communicated. Remember, if you have any questions, or do not understand something, ask your teacher!

  Digging Deeper

There are two methods that can be used to measure and communicate an object’s direction. They both use the mathematical method of a coordinate system grid.


©Wikimedia Commons
C5.7 Cartesian Method

Cartesian Method: uses a coordinate system grid with an x-axis and y-axis similar to a graph. Directions are started from the x-axis, which is the starting point at 0˚, and directions are determined in a counterclockwise direction.


C5.8 Navigator Method


Navigator Method: uses a coordinate system grid with the directions of north [N], south [S], east [E], and west [W]. Directions are started from north, which is the starting point at 0˚, and directions are determined in a clockwise direction.


  Practice Questions

Complete the following practice questions to check your understanding of the concept you just learned. Make sure you write complete answers to the practice questions in your notes. After you have checked your answers, make corrections to your responses (where necessary) to study from.

  1. A soccer player runs down the field for 2.14 m [S] and then turns and runs 7.45 m back [N].
      1. What distance does the player travel?

        To calculate the distance, you would perform the calculation
        «math»«mo»§#160;«/mo»«mi»§#916;«/mi»«mover»«mi»d«/mi»«mrow»«/mrow»«/mover»«mo»=«/mo»«mn»2«/mn»«mo».«/mo»«mn»14«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«mo»+«/mo»«mo»§#160;«/mo»«mn»7«/mn»«mo».«/mo»«mn»45«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»=«/mo»«mn»9«/mn»«mo».«/mo»«mn»59«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo».«/mo»«mo»§#160;«/mo»«/math»
      2. What is the displacement (magnitude and direction of the player?

        3. To calculate the displacement, you would perform the calculation
        «math»«mo»§#160;«/mo»«mi»§#916;«/mi»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mn»2«/mn»«mo».«/mo»«mn»14«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»-«/mo»«mn»7«/mn»«mo».«/mo»«mn»45«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»=«/mo»«mo»§#8208;«/mo»«mn»5«/mn»«mo».«/mo»«mn»31«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«mi»or«/mi»«mo»§#160;«/mo»«mn»5«/mn»«mo».«/mo»«mn»31«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«mo»[«/mo»«mi mathvariant=¨normal¨»N«/mi»«mo»]«/mo»«mo».«/mo»«mo»§#160;«/mo»«/math»

    1. A cross-country skier skies 24 m [W], then 57 m [E], and then 68 m [W] again.
        1. What distance does the skier travel?

          To calculate the distance, you would perform the calculation
          «math»«mo»§#160;«/mo»«mi»§#916;«/mi»«mi»d«/mi»«mo»=«/mo»«mn»24«/mn»«mo»§#160;«/mo»«mtext»m«/mtext»«mo»+«/mo»«mn»57«/mn»«mo»§#160;«/mo»«mtext»m«/mtext»«mo»+«/mo»«mn»68«/mn»«mo»§#160;«/mo»«mtext»m=149§#160;m«/mtext»«mo».«/mo»«mo»§#160;«/mo»«/math»

        2. What is the displacement (magnitude and direction) of the skier?

          To calculate the displacement, you would perform the calculation
          «math»«mo»§#160;«/mo»«mi»§#916;«/mi»«mover»«mi»d«/mi»«mo»§#8594;«/mo»«/mover»«mo»=«/mo»«mn»24«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»-«/mo»«mn»57«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»+«/mo»«mn»68«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»=«/mo»«mn»35«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«mi»or«/mi»«mo»§#160;«/mo»«mn»35«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»§#160;«/mo»«mo»[«/mo»«mi mathvariant=¨normal¨»W«/mi»«mo»]«/mo»«mo».«/mo»«mo»§#160;«/mo»«/math»

Watch This

Distance and Displacement ©ADLC  https://adlc.wistia.com/medias/9ubslvsli4


This video will provides you with a great wrap-up of the difference between distance and displacement, and this section of Lesson 4.
 


  Creating Graphs in Physics

Why do we use graphs to show an object’s motion?


C5.9 Graph paper and pencil
Graphing is an easy and visual way to show the relationship between two variables. In Unit A of this course, you reviewed manipulated and responding variables. These are the variables whose information are presented in data tables and graphs.

Graphs are useful because they can present a lot of information into one single picture. When you collect data, you sometimes don’t know the relationship between the different variables that are being measured. A graph can give you an idea about how these variables change compared to one another.

  Watch This

Working with Data © ADLC  https://adlc.wistia.com/medias/rajbs9ceky
This video provides you with a great overview of how to collect information during a science experiment, variables in an experiment, and then how to use the data collected to create graphs. It then discusses how graphs can be used to determine
 

  What makes a good graph?

Let’s look at an example and use it to go over all the components of creating a good graph.


A child rides a bicycle on a flat road for 60 s, and the position of the child and her bike are, relative to her initial position, is measured every 10 s. The data table presents the information collected.


  Time (s) 
  Position (m) 
0.0

 0.00
10

 5.00
20

 10.00
30
 15.00
40
 20.00
50
 25.00
60
 30.00

 

https://adlc.wistia.com/medias/efiasqqsov



The manipulated variable is plotted on the x-axis, and the responding variable is plotted on the y-axis. Correct labels, with units, must be included.

In the example, the manipulated variable is time (s) and the responding variable is position (m).

Note: You learned about manipulated and responding variables in Unit A. If you need to review what these variables are, click here.  https://adlc.wistia.com/medias/rajbs9ceky
C5.10 Step 1: label axis
Each axis must be clearly marked with a scale.

To create a good scale,

  • look at the smallest and largest values for the variable
  • spread out the scale values so that at least half of the graph paper is used
  • the scale does not have to start at 0 unless there is a value of 0 in the data collected
  • each box on the graph paper must represent the same amount as the next box

In the example, time (s) is on the x-axis. The largest value is 60 s, and smallest value is 0.0 s. The scale must be spread out from 0 to 60.

Position (m) is on the y-axis. The largest value is 30.0 m, and the smallest value is 0.00 m. The scale must be spread out from 0.00 to 30.0.
C5.11 Step 2: scale for x-axis and y-axis
Each piece of data in the table is plotted in the correct position on the graph.

In the example, seven data points will be plotted on the graph.
C5.12 Step 3: plot data points

A line of best fit may be drawn for linear graphs. A linear graph is a graph where the data points appear to follow a direct straight path. It is important to know that a line of best fit does not have to touch all the dots. It should show the general trend of the data, not the exact path the data points follows.

For non-linear graphs, you many need to connect the plotted points to show the trend of the data. A non-linear graph is a graph where the data appear to follow a direct curved path.

In the example, the data points appear to follow a straight path.
C5.13 Step 4: line of best fit

The graph is given a title that describes the information presented in the data.

In the example, the data represents the position of the child and her bike, relative to her initial position, every 10 s for 60 s.
C5.14 Step 5: title




  Read This

Please read pages 473 and 474 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on how to step-by-step, correctly create a scatterplot graph step by step and drawing a line of best fit. Remember, if you have any questions, or do not understand something, ask your teacher!

  Practice Questions

Complete the following practice questions to check your understanding of the concept you just learned. Make sure you write complete answers to the practice questions in your notes. After you have checked your answers, make corrections to your responses (where necessary) to study from.

  1. In your own words, describe the five steps of creating a graph in science.

    Your answer should be a variation of the following.

    1. Plot the manipulated variable is on the x-axis and the responding variable on the y-axis. Correct labels, with units, must be included.
    2. Each axis must be clearly marked with a correct scale.
    3. Each piece of data is plotted in the correct position on the graph.
    4. Draw a line of best fit (for linear graphs), or connect the plotted points to show the trend of the data (for non-linear graphs).
    5. Give a title to the graph that describes the information presented in the data.

  Position-Time Graphs

What can a position-time graph tell us about an object’s motion?


C5.15 runner on running track
Sometimes in science, you are asked to plot data on a graph yourself, and sometimes a graph is provided to you. Either way, a graph can tell us a lot about an object’s motion. Distance-time graphs, also known as position-time graphs, can be used to interpret the direction an object is travelling, whether it is moving or stationary, and the distance travelled over a specific time period.

  Virtual Activity


Distance-Time Graphs © Explore Learning

Work through this activity to discover how the motion of a runner can be plotted on a distance-time graph. You will learn what information can be interpreted from the shape of the line on a distance-time graph.

The activity shows a graph and a runner on a track. You can control the motion of the runner by manipulating the graph (drag the red dots).

Please note: if you scroll down while in the Gizmo you will see a list of questions. You DO NOT need to complete these questions. You are able to complete them for extra practice if you would like.

  1. Click on the play button to open the activity. This activity can also be accessed in the Online Resources for Print Students section of your online course.
  1. Check that "Number of points" is 2, and that under "Runner 1" both "Show graph" and "Show animation" are turned on.

    The graph should look like the one shown in image C5.16 – one point at (0, 0) and the other point at (4, 40).
  1. Click the green button stopwatch to start the runner. What happens?

    The runner runs from left to right for 4 seconds, stopping at the 40-metre line.
  2. Click the red button on the stopwatch to reset the runner. The vertical green probe on the graph allows you to see a snapshot of the runner at any point in time. Drag it back and forth. As you do, watch the runner and the stopwatch.

    1. What was the position of the runner at 1 second?

      10 metres
    2. What are the coordinates of the point on the graph that tells you this?

    3. When was the runner on the 30-metre line?

      At 3 seconds
    4. What are the coordinates of the point on the graph that tells you this?

    3. Part A:
    Runner Position
    		Get the simulation ready:
    • Click the red button on the stopwatch to reset the runner.
    • Be sure the "Number of points" is 2.
    ©Explore Learning
    C5.17 appearance of runner position

©Explore Learning
C5.16 Starting graph appearance
In the simulation, run the “race” many times with a variety of different graphs. To run the race you must click on the green button on the stop watch. (You can change the graph by dragging the two red points on the graph up and down). Pay attention to what the graph tells you about the runner.

  1. If a distance-time graph contains the point (4, 15), what does that tell you about the runner? (Be specific, and answer in a complete sentence.)

    This tells you that after 4 seconds of running, the runner was 15 metres from the starting line.
  2. Look at the graph in image C5.18. Notice where the green probe is. If you could see the runner and the stopwatch at this moment, what would you see?

    The runner would be on the 20-metre line, facing left. The stopwatch would read approximately 1.5 seconds (or 1:50).
©Explore Learning
C5.18 green probe position
  1. Look at image C5.19, from the simulation. What must be true about this runner’s graph?


    ©Explore Learning
    C5.19 runner at 3.25 s


    The graph of the runner must include the point (3.25, 25). Also, the graph has a positive, or “uphill,” slope at that point. This is because the runner is facing forward in the image, so he must be running from left to right.

    If the runner had been running from right to left, the graph would have a negative, or “downhill,” slope at that point.
  1. The point on the graph that lies on the y-axis (vertical axis) is called the y-intercept. What does the y-intercept tell you about the runner?

    The y-intercept indicates the starting position of the runner.
  2.  In the simulation, set the "Number of points" to 3. Then create a graph of a runner who starts at the 20-metre line, runs to the 40-metre line, and finishes at the 30-metre line.

    1. Sketch what your graph would look like.


      ©Explore Learning
      C5.20 graph of a runner with described points


    2. What is the y-intercept of your graph?

      (0, 20) or 20 m
      Part B:
      Runner direction and speed
      		Get the simulation ready:
      • Click the red reset button on the simulation and make sure you have three points still selected.
      ©Explore Learning
      C5.21 appearance of graph

      Run the simulation several times with different types of graphs. (Remember, the red points on the graph can be dragged vertically.) Pay attention to the speed and direction of the runner.

  1. Create a graph of a runner that is running forward (from left to right) in the simulation.

    1. Sketch what your graph would look like.

      There are several possible correct answers. This is an example:


      ©Explore Learning
      C5.22 graph of a runner with described motion


    2. If the runner is moving from left to right in the simulation, how will the graph always look?

      The graph will have a positive slope. (The line will go from the lower left to the upper right.)
  1. Click the red reset button. Create a graph of a runner that is running from right to left.

    1. Sketch what your graph would look like.

      There are several possible correct answers. This is an example:


      ©Explore Learning
      C5.23 graph of a runner with described motion


    2. How does the graph always look if the runner is moving from right to left in the simulation?

      The graph will have a negative slope. (The line will go from upper left to lower right.)
  1. Change the "Number of points" to 5. Create a graph of a runner that runs left-to-right for 1 second, rests for 2 seconds, and then continues running in the same direction.

    1. Sketch what your graph would look like.

      There are several possible correct answers. This is an example:


      ©Explore Learning
      C5.24 graph of a runner with described motion


    2. How does a graph show a runner at rest?

      The line will be horizontal when the runner is at rest.

  1. With "Number of points" set to 3, create the graph in image C5.25.
    Your graph should include (0, 0), (2, 10), and (4, 40).

    1. Where does the runner start?

      0-metre line
    2. Where will he be after 2 seconds?

      10-metre line
    3. Where will he be after 4 seconds?

      40-metre line
    4. In which time interval do you think the runner will be moving most quickly?

      2 to 4 seconds

©Explore Learning
C5.25 number of points set to 3
  1. Click the green start button and watch the animation. What changed about the runner after 2 seconds of running?

    The speed of the runner increased.
  2. Speed is a measure of how fast something is moving. To calculate speed, divide the distance by the time. In the simulation, the units of speed is metres per second (m/s).

    1. In the first 2 seconds, how far did the runner go?

      10 metres
    2. In this time interval, how far did the runner go each second?

      5 metres
    3. In this time interval, what was the runner’s speed?

      «math» «mi»v«/mi» «mo»=«/mo» «mfrac» «mrow» «mo»§#8710;«/mo» «mi»d«/mi» «/mrow» «mrow» «mo»§#8710;«/mo» «mi»t«/mi» «/mrow» «/mfrac» «mspace linebreak=¨newline¨»«/mspace» «mi»v«/mi» «mo»=«/mo» «mfrac» «mrow» «mn»30«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»m«/mi» «/mrow» «mrow» «mn»2«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»s«/mi» «/mrow» «/mfrac» «/math»

      5 metres/second (5 m/s)
  1. Now look at the last 2 seconds represented on the graph.

    1. In the last 2 seconds, how far did the runner go?

      30 metres
    2. In this time interval, how far did the runner go each second?

      15 metres
    3. In this time interval, what was the runner’s speed?

      «math» «mi»v«/mi» «mo»=«/mo» «mfrac» «mrow» «mo»§#8710;«/mo» «mi»d«/mi» «/mrow» «mrow» «mo»§#8710;«/mo» «mi»t«/mi» «/mrow» «/mfrac» «mspace linebreak=¨newline¨»«/mspace» «mi»v«/mi» «mo»=«/mo» «mfrac» «mrow» «mn»30«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»m«/mi» «/mrow» «mrow» «mn»2«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»s«/mi» «/mrow» «/mfrac» «/math»

      15 metres/second (15 m/s)
  2. Please return to the top of this page and click on analysis to complete the analysis questions.

  Did you know?

C5.26 Canada goose

Canada geese migrate in order to return to the area where they were born for mating and nesting.

Not all Canada geese migrate, but many do who make their summer homes in Canada do migrate.

Migrations can be as long as 3 200 km to 4 800 km, and the geese are capable of flying up to 2 400 km in a single day if the weather is good.

This measurement is distance flown by the geese, not displacement. The geese do not fly in a directly straight path.
  1. In general, how does a distance-time graph show you which direction the runner is moving?

    The runner’s direction is given by the slope of the graph. A positive slope ( / ) indicates the runner is moving forward, or from left to right. A negative slope ( \ ) indicates the runner is moving backward, or from right to left.

  2. How can you estimate the speed of the runner by looking at a graph?

    The steeper the line on the graph, the faster the runner.

  Read This

Please read pages 137 to 138, and 142 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on how to create a position-time graph, and the information that can be interpreted from a position-time (distance-time) graph. Remember, if you have any questions, or do not understand something, ask your teacher!

  Practice Questions

Complete the following practice questions to check your understanding of the concept you just learned. Make sure you write complete answers to the practice questions in your notes. After you have checked your answers, make corrections to your responses (where necessary) to study from.

  1. Explain how distance-time graphs are useful in explaining an object’s motion.

    Your answer should be a variation of the following.
    Distance-time graphs show the starting position, direction, and speed of a moving object.
  2. Use the graph in image C5.27 to answer the following questions.

    C5.27 Position-Time Graph



    1. Describe the motion of the object from point A to point B.

      A positive slope indicates the object is moving forward.
    2. Describe the motion of the object from point B to point C.

      The flat portion of the graph indicates the object is not moving.
    3. Describe the motion of the object from point C to point D.

      A negative slope indicates the object is moving backward.
    4. What is the total distance travelled by the object?

      Read from the graph that point A is at 1 m and point B is at 3 m. The distance from point A to point B is 2 m.
      Read from the graph that point B is at 3 m and point C is at 3 m. The distance from point B to point C is 0 m.
      Read from the graph that point C is at 3 m and point D is at 0 m. The distance from point C to point D is 3 m.
      To calculate the distance, you would perform the calculation Δd = 2.0 m + 0.0 m + 3.0 m = 5.0 m.
    5. What is the total displacement of the object?

      To calculate the displacement, you would add the three values, taking into account that positive slope is a positive direction and negative slope is a negative direction.

      «math» «mi»§#916;«/mi» «mover» «mi»d«/mi» «mo»§#8594;«/mo» «/mover» «mo»=«/mo» «mo»(«/mo» «mo»+«/mo» «mn»2«/mn» «mo».«/mo» «mn»0«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»m«/mi» «mo»)«/mo» «mo»+«/mo» «mo»(«/mo» «mn»0«/mn» «mo».«/mo» «mn»0«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»m«/mi» «mo»)«/mo» «mo»+«/mo» «mo»(«/mo» «mo»§#8211;«/mo» «mn»3«/mn» «mo».«/mo» «mn»0«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»m«/mi» «mo»)«/mo» «mo»§#160;«/mo» «mo»=«/mo» «mo»§#8211;«/mo» «mn»1«/mn» «mo».«/mo» «mn»0«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»m«/mi» «/math»
       
      Because negative slope is negative, then an answer of –1.0 m means that you could express it as «math» «mo»§#160;«/mo» «mi»§#916;«/mi» «mover» «mi»d«/mi» «mo»§#8594;«/mo» «/mover» «mo»=«/mo» «mo»-«/mo» «mn»1«/mn» «mo».«/mo» «mn»0«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»m«/mi» «mo»§#160;«/mo» «mi»or«/mi» «mo»§#160;«/mo» «mn»1«/mn» «mo».«/mo» «mn»0«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»m«/mi» «mo»§#160;«/mo» «mo»[«/mo» «mi»backward«/mi» «mo»]«/mo» «mo».«/mo» «/math»

  Displacement vs. Distance


C5.28 Professional cyclists on racing track
Quantities in physics can be classified as either scalar or vector quantities. In this lesson, you explored the difference between these types of quantities through looking at the displacement of an object compared to the distance an object travels.

You also learned how distance and displacement can be calculated, both through direct information provided about an object’s motion and by interpreting a position-time graph. The steps of creating a graph in physics, and how to interpret graphs, are skills that we will use through the rest of this section of Unit C.

  Watch This

Physics – Motion – Distance and Displacement © YouTube Don’t Memorise 


Watch this video for a great review of the difference between distance and displacement. It provides examples to reinforce this concept, as well as the difference between scalar and vector quantities.


3.3 Assignment

Unit 3 Section 2 Formative Assessment


It is now time to complete 3.3 Assignment. Click on the button below to go to the assignment page.

3.3 Assignment