Skip to content Skip to navigation Skip to collection information

OpenStax_CNX

You are here: Home » Content » Siyavula textbooks: Grade 10 Physical Science » Equations of motion

Navigation

Table of Contents

Lenses

What is a lens?

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

This content is ...

Affiliated with (What does "Affiliated with" mean?)

This content is either by members of the organizations listed or about topics related to the organizations listed. Click each link to see a list of all content affiliated with the organization.
  • Bookshare

    This collection is included inLens: Bookshare's Lens
    By: Bookshare - A Benetech Initiative

    Comments:

    "Accessible versions of this collection are available at Bookshare. DAISY and BRF provided."

    Click the "Bookshare" link to see all content affiliated with them.

  • FETPhysics display tagshide tags

    This module is included inLens: Siyavula: Physics (Gr. 10-12)
    By: Siyavula

    Review Status: In Review

    Click the "FETPhysics" link to see all content affiliated with them.

    Click the tag icon tag icon to display tags associated with this content.

  • Siyavula: Physical Science display tagshide tags

    This collection is included inLens: Siyavula textbooks: Physical Science
    By: Free High School Science Texts Project

    Click the "Siyavula: Physical Science" link to see all content affiliated with them.

    Click the tag icon tag icon to display tags associated with this content.

Recently Viewed

This feature requires Javascript to be enabled.

Tags

(What is a tag?)

These tags come from the endorsement, affiliation, and other lenses that include this content.
 

Equations of Motion

In this chapter we will look at the third way to describe motion. We have looked at describing motion in terms of graphs and words. In this section we examine equations that can be used to describe motion.

This section is about solving problems relating to uniformly accelerated motion. In other words, motion at constant acceleration.

The following are the variables that will be used in this section:

v i = initial velocity ( m · s - 1 ) at t = 0 s v f = final velocity ( m · s - 1 ) at time t Δ x = displacement ( m ) t = time ( s ) Δ t = time interval ( s ) a = acceleration ( m · s - 1 ) v i = initial velocity ( m · s - 1 ) at t = 0 s v f = final velocity ( m · s - 1 ) at time t Δ x = displacement ( m ) t = time ( s ) Δ t = time interval ( s ) a = acceleration ( m · s - 1 )
(1)
v f = v i + a t v f = v i + a t
(2)
Δ x = ( v i + v f ) 2 t Δ x = ( v i + v f ) 2 t
(3)
Δ x = v i t + 1 2 a t 2 Δ x = v i t + 1 2 a t 2
(4)
v f 2 = v i 2 + 2 a Δ x v f 2 = v i 2 + 2 a Δ x
(5)

The questions can vary a lot, but the following method for answering them will always work. Use this when attempting a question that involves motion with constant acceleration. You need any three known quantities (vivi, vfvf, ΔxΔx, tt or aa) to be able to calculate the fourth one.

  1. Read the question carefully to identify the quantities that are given. Write them down.
  2. Identify the equation to use. Write it down!!!
  3. Ensure that all the values are in the correct unit and fill them in your equation.
  4. Calculate the answer and fill in its unit.

Note: Interesting Fact :

Galileo Galilei of Pisa, Italy, was the first to determined the correct mathematical law for acceleration: the total distance covered, starting from rest, is proportional to the square of the time. He also concluded that objects retain their velocity unless a force – often friction – acts upon them, refuting the accepted Aristotelian hypothesis that objects "naturally" slow down and stop unless a force acts upon them. This principle was incorporated into Newton's laws of motion (1st law).

Finding the Equations of Motion

The following does not form part of the syllabus and can be considered additional information.

Derivation of Equation 2

According to the definition of acceleration:

a = Δ v t a = Δ v t
(6)

where ΔvΔv is the change in velocity, i.e. Δv=vfΔv=vf - vivi. Thus we have

a = v f - v i t v f = v i + a t a = v f - v i t v f = v i + a t
(7)

Derivation of Equation 3

We have seen that displacement can be calculated from the area under a velocity vs. time graph. For uniformly accelerated motion the most complicated velocity vs. time graph we can have is a straight line. Look at the graph below - it represents an object with a starting velocity of vivi, accelerating to a final velocity vfvf over a total time tt.

Figure 1
Figure 1 (PG10C2_045.png)

To calculate the final displacement we must calculate the area under the graph - this is just the area of the rectangle added to the area of the triangle. This portion of the graph has been shaded for clarity.

Area = 1 2 b × h = 1 2 t × ( v f - v i ) = 1 2 v f t - 1 2 v i t Area = 1 2 b × h = 1 2 t × ( v f - v i ) = 1 2 v f t - 1 2 v i t
(8)
Area = × b = t × v i = v i t Area = × b = t × v i = v i t
(9)
Displacement = Area + Area Δ x = v i t + 1 2 v f t - 1 2 v i t Δ x = ( v i + v f ) 2 t Displacement = Area + Area Δ x = v i t + 1 2 v f t - 1 2 v i t Δ x = ( v i + v f ) 2 t
(10)

Derivation of Equation 4

This equation is simply derived by eliminating the final velocity vfvf in Equation 3. Remembering from Equation 2 that

v f = v i + a t v f = v i + a t
(11)

then Equation 3 becomes

Δ x = v i + v i + a t 2 t = 2 v i t + a t 2 2 Δ x = v i t + 1 2 a t 2 Δ x = v i + v i + a t 2 t = 2 v i t + a t 2 2 Δ x = v i t + 1 2 a t 2
(12)

Derivation of Equation 5

This equation is just derived by eliminating the time variable in the above equation. From Equation 2 we know

t = v f - v i a t = v f - v i a
(13)

Substituting this into Equation 4 gives

Δ x = v i ( v f - v i a ) + 1 2 a ( v f - v i a ) 2 = v i v f a - v i 2 a + 1 2 a ( v f 2 - 2 v i v f + v i 2 a 2 ) = v i v f a - v i 2 a + v f 2 2 a - v i v f a + v i 2 2 a 2 a Δ x = - 2 v i 2 + v f 2 + v i 2 v f 2 = v i 2 + 2 a Δ x Δ x = v i ( v f - v i a ) + 1 2 a ( v f - v i a ) 2 = v i v f a - v i 2 a + 1 2 a ( v f 2 - 2 v i v f + v i 2 a 2 ) = v i v f a - v i 2 a + v f 2 2 a - v i v f a + v i 2 2 a 2 a Δ x = - 2 v i 2 + v f 2 + v i 2 v f 2 = v i 2 + 2 a Δ x
(14)

This gives us the final velocity in terms of the initial velocity, acceleration and displacement and is independent of the time variable.

Exercise 1: Equations of motion

A racing car is travelling north. It accelerates uniformly covering a distance of 725 m in 10 s. If it has an initial velocity of 10 m··s-1-1, find its acceleration.

Solution
  1. Step 1. Identify what information is given and what is asked for :

    We are given:

    v i = 10 m · s - 1 Δ x = 725 m t = 10 s a = ? v i = 10 m · s - 1 Δ x = 725 m t = 10 s a = ?
    (15)
  2. Step 2. Find an equation of motion relating the given information to the acceleration :

    If you struggle to find the correct equation, find the quantity that is not given and then look for an equation that has this quantity in it.

    We can use equation Equation 4

    Δ x = v i t + 1 2 a t 2 Δ x = v i t + 1 2 a t 2
    (16)
  3. Step 3. Substitute your values in and find the answer :
    Δ x = v i t + 1 2 a t 2 725 m = ( 10 m · s - 1 × 10 s ) + 1 2 a × ( 10 s ) 2 725 m - 100 m = ( 50 s 2 ) a a = 12 , 5 m · s - 2 Δ x = v i t + 1 2 a t 2 725 m = ( 10 m · s - 1 × 10 s ) + 1 2 a × ( 10 s ) 2 725 m - 100 m = ( 50 s 2 ) a a = 12 , 5 m · s - 2
    (17)
  4. Step 4. Quote the final answer :

    The racing car is accelerating at 12,5 m··s-2-2 north.

Exercise 2: Equations of motion

A motorcycle, travelling east, starts from rest, moves in a straight line with a constant acceleration and covers a distance of 64 m in 4 s. Calculate

  1. its acceleration
  2. its final velocity
  3. at what time the motorcycle had covered half the total distance
  4. what distance the motorcycle had covered in half the total time.
Solution
  1. Step 1. Identify what information is given and what is asked for :

    We are given:

    v i = 0 m · s - 1 ( because the object starts from rest. ) Δ x = 64 m t = 4 s a = ? v f = ? t = ? at half the distance Δ x = 32 m . Δ x = ? at half the time t = 2 s . v i = 0 m · s - 1 ( because the object starts from rest. ) Δ x = 64 m t = 4 s a = ? v f = ? t = ? at half the distance Δ x = 32 m . Δ x = ? at half the time t = 2 s .
    (18)

    All quantities are in SI units.

  2. Step 2. Acceleration: Find a suitable equation to calculate the acceleration :

    We can use Equation 4

    Δ x = v i t + 1 2 a t 2 Δ x = v i t + 1 2 a t 2
    (19)
  3. Step 3. Substitute the values and calculate the acceleration :
    Δ x = v i t + 1 2 a t 2 64 m = ( 0 m · s - 1 × 4 s ) + 1 2 a × ( 4 s ) 2 64 m = ( 8 s 2 ) a a = 8 m · s - 2 east Δ x = v i t + 1 2 a t 2 64 m = ( 0 m · s - 1 × 4 s ) + 1 2 a × ( 4 s ) 2 64 m = ( 8 s 2 ) a a = 8 m · s - 2 east
    (20)
  4. Step 4. Final velocity: Find a suitable equation to calculate the final velocity :

    We can use Equation 5 - remember we now also know the acceleration of the object.

    v f = v i + a t v f = v i + a t
    (21)
  5. Step 5. Substitute the values and calculate the final velocity :
    v f = v i + a t v f = 0 m · s - 1 + ( 8 m · s - 2 ) ( 4 s ) = 32 m · s - 1 east v f = v i + a t v f = 0 m · s - 1 + ( 8 m · s - 2 ) ( 4 s ) = 32 m · s - 1 east
    (22)
  6. Step 6. Time at half the distance: Find an equation to calculate the time :

    We can use Equation 4:

    Δ x = v i + 1 2 a t 2 32 m = ( 0 m · s - 1 ) t + 1 2 ( 8 m · s - 2 ) ( t ) 2 32 m = 0 + ( 4 m · s - 2 ) t 2 8 s 2 = t 2 t = 2 , 83 s Δ x = v i + 1 2 a t 2 32 m = ( 0 m · s - 1 ) t + 1 2 ( 8 m · s - 2 ) ( t ) 2 32 m = 0 + ( 4 m · s - 2 ) t 2 8 s 2 = t 2 t = 2 , 83 s
    (23)
  7. Step 7. Distance at half the time: Find an equation to relate the distance and time :

    Half the time is 2 s, thus we have vivi, aa and tt - all in the correct units. We can use Equation 4 to get the distance:

    Δ x = v i t + 1 2 a t 2 = ( 0 ) ( 2 ) + 1 2 ( 8 ) ( 2 ) 2 = 16 m east Δ x = v i t + 1 2 a t 2 = ( 0 ) ( 2 ) + 1 2 ( 8 ) ( 2 ) 2 = 16 m east
    (24)
  8. Step 8. Write down the final results.:
    1. The acceleration is 8 m·s-28m·s-2 east
    2. The velocity is 32 m·s-132m·s-1 east
    3. The time at half the distance is 2,83s2,83s
    4. The distance at half the time is 16m16m east
Equations of motion
  1. A car starts off at 10 m··s-1-1 and accelerates at 1 m··s-2-2 for 10 s. What is its final velocity?
    Click here for the solution.
  2. A train starts from rest, and accelerates at 1 m··s-2-2 for 10 s. How far does it move?
    Click here for the solution.
  3. A bus is going 30 m··s-1-1 and stops in 5 s. What is its stopping distance for this speed?
    Click here for the solution.
  4. A racing car going at 20 m··s-1-1 stops in a distance of 20 m. What is its acceleration?
    Click here for the solution.
  5. A ball has a uniform acceleration of 4 m··s-1-1. Assume the ball starts from rest. Determine the velocity and displacement at the end of 10 s.
    Click here for the solution.
  6. A motorcycle has a uniform acceleration of 4 m··s-1-1. Assume the motorcycle has an initial velocity of 20 m··s-1-1. Determine the velocity and displacement at the end of 12 s.
    Click here for the solution.
  7. An aeroplane accelerates uniformly such that it goes from rest to 144 km··hr-1-1in 8 s. Calculate the acceleration required and the total distance that it has traveled in this time.
    Click here for the solution.

Applications in the Real-World

What we have learnt in this chapter can be directly applied to road safety. We can analyse the relationship between speed and stopping distance. The following worked example illustrates this application.

Exercise 3: Stopping distance

A truck is travelling at a constant velocity of 10 m··s-1-1when the driver sees a child 50 m in front of him in the road. He hits the brakes to stop the truck. The truck accelerates at a rate of -1.25 m··s-2-2. His reaction time to hit the brakes is 0,5 seconds. Will the truck hit the child?

Solution

  1. Step 1. Analyse the problem and identify what information is given :

    It is useful to draw a timeline like this one:

    Figure 2
    Figure 2 (PG10C2_046.png)

    We need to know the following:

    • What distance the driver covers before hitting the brakes.
    • How long it takes the truck to stop after hitting the brakes.
    • What total distance the truck covers to stop.
  2. Step 2. Calculate the distance AB :

    Before the driver hits the brakes, the truck is travelling at constant velocity. There is no acceleration and therefore the equations of motion are not used. To find the distance traveled, we use:

    v = d t 10 = d 0 , 5 d = 5 m v = d t 10 = d 0 , 5 d = 5 m
    (25)

    The truck covers 5 m before the driver hits the brakes.

  3. Step 3. Calculate the time BC :

    We have the following for the motion between B and C:

    v i = 10 m · s - 1 v f = 0 m · s - 1 a = - 1 , 25 m · s - 2 t = ? v i = 10 m · s - 1 v f = 0 m · s - 1 a = - 1 , 25 m · s - 2 t = ?
    (26)

    We can use Equation 2

    v f = v i + a t 0 = 10 + ( - 1 , 25 ) t - 10 = - 1 , 25 t t = 8 s v f = v i + a t 0 = 10 + ( - 1 , 25 ) t - 10 = - 1 , 25 t t = 8 s
    (27)
  4. Step 4. Calculate the distance BC :

    For the distance we can use Equation 3 or Equation 4. We will use Equation 3:

    Δ x = ( v i + v f ) 2 t Δ x = 10 + 0 s ( 8 ) Δ x = 40 m Δ x = ( v i + v f ) 2 t Δ x = 10 + 0 s ( 8 ) Δ x = 40 m
    (28)
  5. Step 5. Write the final answer :

    The total distance that the truck covers is dABdAB + dBCdBC = 5 + 40 = 45 meters. The child is 50 meters ahead. The truck will not hit the child.

Summary

  • A reference point is a point from where you take your measurements.
  • A frame of reference is a reference point with a set of directions.
  • Your position is where you are located with respect to your reference point.
  • The displacement of an object is how far it is from the reference point. It is the shortest distance between the object and the reference point. It has magnitude and direction because it is a vector.
  • The distance of an object is the length of the path travelled from the starting point to the end point. It has magnitude only because it is a scalar.
  • A vector is a physical quantity with magnitude and direction.
  • A scalar is a physical quantity with magnitude only.
  • Speed (ss) is the distance covered (dd) divided by the time taken (ΔtΔt):
    s=dΔts=dΔt
    (29)
  • Average velocity (vv) is the displacement (ΔxΔx) divided by the time taken (ΔtΔt):
    v=ΔxΔtv=ΔxΔt
    (30)
  • Instantaneous speed is the speed at a specific instant in time.
  • Instantaneous velocity is the velocity at a specific instant in time.
  • Acceleration (aa) is the change in velocity (ΔxΔx) over a time interval (ΔtΔt):
    a=ΔvΔta=ΔvΔt
    (31)
  • The gradient of a position - time graph (xx vs. tt) give the velocity.
  • The gradient of a velocity - time graph (vv vs. tt) give the acceleration.
  • The area under a velocity - time graph (vv vs. tt) give the displacement.
  • The area under an acceleration - time graph (aa vs. tt) gives the velocity.
  • The graphs of motion are summarised in (Reference).
  • The equations of motion are used where constant acceleration takes place:
    vf=vi+atΔx=(vi+vf)2tΔx=vit+12at2vf2=vi2+2aΔxvf=vi+atΔx=(vi+vf)2tΔx=vit+12at2vf2=vi2+2aΔx
    (32)

End of Chapter Exercises: Motion in One Dimension

  1. Give one word/term for the following descriptions.
    1. The shortest path from start to finish.
    2. A physical quantity with magnitude and direction.
    3. The quantity defined as a change in velocity over a time period.
    4. The point from where you take measurements.
    5. The distance covered in a time interval.
    6. The velocity at a specific instant in time.
    Click here for the solution.
  2. Choose an item from column B that match the description in column A. Write down only the letter next to the question number. You may use an item from column B more than once.
    Table 1
    Column AColumn B
    a. The area under a velocity - time graphgradient
    b. The gradient of a velocity - time grapharea
    c. The area under an acceleration - time graphvelocity
    d. The gradient of a displacement - time graphdisplacement
     acceleration
     slope
    Click here for the solution.
  3. Indicate whether the following statements are TRUE or FALSE. Write only 'true' or 'false'. If the statement is false, write down the correct statement.
    1. A scalar is the displacement of an object over a time interval.
    2. The position of an object is where it is located.
    3. The sign of the velocity of an object tells us in which direction it is travelling.
    4. The acceleration of an object is the change of its displacement over a period in time.
    Click here for the solution.
  4. [SC 2003/11] A body accelerates uniformly from rest for t0t0 seconds after which it continues with a constant velocity. Which graph is the correct representation of the body's motion?
    Table 2
    Figure 3
    Figure 3 (PG10C2_047.png)
    Figure 4
    Figure 4 (PG10C2_048.png)
    Figure 5
    Figure 5 (PG10C2_049.png)
    Figure 6
    Figure 6 (PG10C2_050.png)
    (a)(b)(c)(d)
    Click here for the solution.
  5. [SC 2003/11] The velocity-time graphs of two cars are represented by P and Q as shown
    Figure 7
    Figure 7 (PG10C2_051.png)
    The difference in the distance travelled by the two cars (in m) after 4 s is ......
    1. 12
    2. 6
    3. 2
    4. 0
    Click here for the solution.
  6. [IEB 2005/11 HG] The graph that follows shows how the speed of an athlete varies with time as he sprints for 100 m.
    Figure 8
    Figure 8 (PG10C2_052.png)
    Which of the following equations can be used to correctly determine the time tt for which he accelerates?
    1. 100=(10)(11)-12(10)t100=(10)(11)-12(10)t
    2. 100=(10)(11)+12(10)t100=(10)(11)+12(10)t
    3. 100=10t+12(10)t2100=10t+12(10)t2
    4. 100=12(0)t+12(10)t2100=12(0)t+12(10)t2
    Click here for the solution.
  7. [SC 2002/03 HG1] In which one of the following cases will the distance covered and the magnitude of the displacement be the same?
    1. A girl climbs a spiral staircase.
    2. An athlete completes one lap in a race.
    3. A raindrop falls in still air.
    4. A passenger in a train travels from Cape Town to Johannesburg.
    Click here for the solution.
  8. [SC 2003/11] A car, travelling at constant velocity, passes a stationary motor cycle at a traffic light. As the car overtakes the motorcycle, the motorcycle accelerates uniformly from rest for 10 s. The following displacement-time graph represents the motions of both vehicles from the traffic light onwards.
    Figure 9
    Figure 9 (PG10C2_053.png)
    1. Use the graph to find the magnitude of the constant velocity of the car.
    2. Use the information from the graph to show by means of calculation that the magnitude of the acceleration of the motorcycle, for the first 10 s of its motion is 7,5 m··s-2-2.
    3. Calculate how long (in seconds) it will take the motorcycle to catch up with the car (point X on the time axis).
    4. How far behind the motorcycle will the car be after 15 seconds?
    Click here for the solution.
  9. [IEB 2005/11 HG] Which of the following statements is true of a body that accelerates uniformly?
    1. Its rate of change of position with time remains constant.
    2. Its position changes by the same amount in equal time intervals.
    3. Its velocity increases by increasing amounts in equal time intervals.
    4. Its rate of change of velocity with time remains constant.
    Click here for the solution.
  10. [IEB 2003/11 HG1] The velocity-time graph for a car moving along a straight horizontal road is shown below.
    Figure 10
    Figure 10 (PG10C2_054.png)
    Which of the following expressions gives the magnitude of the average velocity of the car?
    1. Area At Area At
    2. Area A+ Area Bt Area A+ Area Bt
    3. Area Bt Area Bt
    4. Area A- Area Bt Area A- Area Bt
    Click here for the solution.
  11. [SC 2002/11 SG] A car is driven at 25 m··s-1-1 in a municipal area. When the driver sees a traffic officer at a speed trap, he realises he is travelling too fast. He immediately applies the brakes of the car while still 100 m away from the speed trap.
    1. Calculate the magnitude of the minimum acceleration which the car must have to avoid exceeding the speed limit, if the municipal speed limit is 16.6 m··s-1-1.
    2. Calculate the time from the instant the driver applied the brakes until he reaches the speed trap. Assume that the car's velocity, when reaching the trap, is 16.6 m··s-1-1.
    Click here for the solution.
  12. A traffic officer is watching his speed trap equipment at the bottom of a valley. He can see cars as they enter the valley 1 km to his left until they leave the valley 1 km to his right. Nelson is recording the times of cars entering and leaving the valley for a school project. Nelson notices a white Toyota enter the valley at 11:01:30 and leave the valley at 11:02:42. Afterwards, Nelson hears that the traffic officer recorded the Toyota doing 140 km··hr-1-1.
    1. What was the time interval (ΔtΔt) for the Toyota to travel through the valley?
    2. What was the average speed of the Toyota?
    3. Convert this speed to km··hr-1-1.
    4. Discuss whether the Toyota could have been travelling at 140km··hr-1-1 at the bottom of the valley.
    5. Discuss the differences between the instantaneous speed (as measured by the speed trap) and average speed (as measured by Nelson).
    Click here for the solution.
  13. [IEB 2003/11HG] A velocity-time graph for a ball rolling along a track is shown below. The graph has been divided up into 3 sections, A, B and C for easy reference. (Disregard any effects of friction.)
    Figure 11
    Figure 11 (PG10C2_055.png)
    1. Use the graph to determine the following:
      1. the speed 5 s after the start
      2. the distance travelled in Section A
      3. the acceleration in Section C
    2. At time t11 the velocity-time graph intersects the time axis. Use an appropriate equation of motion to calculate the value of time t11 (in s).
    3. Sketch a displacement-time graph for the motion of the ball for these 12 s. (You do not need to calculate the actual values of the displacement for each time interval, but do pay attention to the general shape of this graph during each time interval.)
    Click here for the solution.
  14. In towns and cities, the speed limit is 60 km··hr-1-1. The length of the average car is 3.5 m, and the width of the average car is 2 m. In order to cross the road, you need to be able to walk further than the width of a car, before that car reaches you. To cross safely, you should be able to walk at least 2 m further than the width of the car (4 m in total), before the car reaches you.
    1. If your walking speed is 4 km··hr-1-1, what is your walking speed in m··s-1-1?
    2. How long does it take you to walk a distance equal to the width of the average car?
    3. What is the speed in m··s-1-1 of a car travelling at the speed limit in a town?
    4. How many metres does a car travelling at the speed limit travel, in the same time that it takes you to walk a distance equal to the width of car?
    5. Why is the answer to the previous question important?
    6. If you see a car driving toward you, and it is 28 m away (the same as the length of 8 cars), is it safe to walk across the road?
    7. How far away must a car be, before you think it might be safe to cross? How many car-lengths is this distance?
    Click here for the solution.
  15. A bus on a straight road starts from rest at a bus stop and accelerates at 2 m··s-2-2 until it reaches a speed of 20 m··s-1-1. Then the bus travels for 20 s at a constant speed until the driver sees the next bus stop in the distance. The driver applies the brakes, stopping the bus in a uniform manner in 5 s.
    1. How long does the bus take to travel from the first bus stop to the second bus stop?
    2. What is the average velocity of the bus during the trip?
    Click here for the solution.

Collection Navigation

Content actions

Download:

Collection as:

PDF | EPUB (?)

What is an EPUB file?

EPUB is an electronic book format that can be read on a variety of mobile devices.

Downloading to a reading device

For detailed instructions on how to download this content's EPUB to your specific device, click the "(?)" link.

| More downloads ...

Module as:

PDF | EPUB (?)

What is an EPUB file?

EPUB is an electronic book format that can be read on a variety of mobile devices.

Downloading to a reading device

For detailed instructions on how to download this content's EPUB to your specific device, click the "(?)" link.

| More downloads ...

Add:

Collection to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks

Module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks