1. Motion
Physics: Unit 2

2. Describing Movement SCALAR VECTOR Definitions vector scalar
∆ ‘delta’ means ‘change in’
Definitions
A quantity specifies direction as well as magnitude (size).
A quantity specifies the magnitude but not the direction.
Distance is a measure of length of a path taken by an object.
Displacement is a measure of the change in position of an object.
Are distance and displacement scalar or vector quantities?
What is a ?
And what is a ?
vector
scalar
SCALAR
VECTOR
Represented by the symbol ‘𝑥’ and given by the equation
∆𝑥= 𝑥 2 − 𝑥 1

3. Describing Movement Applying these concepts to problems
eg1. Rowan walks from his locker up to his Maths classroom located 50 metres away in a northerly direction, but realises halfway there that he has forgotten his calculator. He walks back to his locker, collects the calculator and goes to class.
a) What is the distance travelled by Rowan after this journey?
b) What is the displacement of Rowan after this journey?
Total Distance = Halfway to maths + Back to Locker + All the way to maths
= 25 m + 25 m + 50 m
= 100 m
∆𝑥= 𝑋 2 − 𝑋 1
=50 𝑚 −0 𝑚
= 50 metres North
Distance is a measure of length of a path taken by an object.
Displacement is a measure of the change in position of an object.

4. Describing Movement SCALAR VECTOR Definitions
Speed is ‘how fast an object moves’, or the rate at which an object covers a distance.
Velocity is a measure of ‘the rate at which an object changes its position’.
Understanding Velocity…… A person moves rapidly - one step forward and one step back - always returning to the original starting position. Because the person always returns to the original position, the motion would never result in a change in position. This would result in a zero velocity, since velocity is defined as the rate at which the position changes.
If a person in motion wishes to maximize their velocity, then that person must make every effort to maximize the amount that they are displaced from their original position.
SCALAR
VECTOR
Are speed and velocity scalars or vectors?

5. Acceleration is the rate of change of velocity.
Describing Movement
Definitions
Acceleration is the rate of change of velocity.
Scalar or Vector?
VECTOR

6. Describing Movement Useful Equations
Average Speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 ( 𝑘𝑚ℎ𝑟 −1 𝑜𝑟 𝑘𝑚/ℎ𝑟 , 𝑚𝑠 −1 𝑜𝑟 𝑚/𝑠 etc.. )
Average Velocity = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒 = ∆𝑥 ∆𝑡 = 𝑋 𝑓𝑖𝑛𝑎𝑙 − 𝑋 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 (𝑎𝑠 𝑝𝑒𝑟 𝑠𝑝𝑒𝑒𝑑 𝑢𝑛𝑖𝑡𝑠)
Converting km/hr to m/s = 𝑘𝑚 ℎ𝑟 𝑎𝑚𝑜𝑢𝑛𝑡 Converting m/s to km/hr = 𝑚 𝑠 𝑎𝑚𝑜𝑢𝑛𝑡 ×3.6
Acceleration = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 = ∆𝑣 ∆𝑡 = 𝑉 𝑓𝑖𝑛𝑎𝑙 − 𝑉 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 ( 𝑚𝑠 −2 𝑜𝑟 𝑚/ 𝑠 2 or 𝑘𝑚/ℎ𝑟/𝑠)

7. ∴ Aimee’s displacement was 2.5 km 53.1° T
Describing Movement
eg2. After school Aimee walks 1.5 km NORTH into town, then turns right and walks a further 2 km to the lake. The journey took her a total of 30 minutes.
a) Sketch Aimee’s journey.
What distance has Aimee travelled?
Total Distance = 1.5 km + 2 km
= 3.5 km
What was the displacement of Aimee?
Find the length of the yellow line using Pythagoras.
𝑐 2 = 𝑎 2 + 𝑏 2
𝑆𝐿 2 =
𝑆𝐿 2 =2.25+4
S𝐿= 6.25
S𝐿=2.5 𝑘𝑚
Town
Start
Lake
2 km
1.5 km
(Vector - Give direction) 𝜽
Direction given by 𝜃. Solve using trigonometry.
tan 𝜃 = 𝑂 𝐴
𝜃= tan −1 ( )
𝜃=53.1°
∴ Aimee’s displacement was 2.5 km 53.1° T

8. Describing Movement 𝜽 eg2. (continued)
d) What is the average speed for the trip? Express your answer in ms-1
Average Speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 = 𝑘𝑚 ℎ𝑟 = 7 𝑘𝑚ℎ𝑟 −1
In m/s → 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑= = 𝑚𝑠 −1
e) What is the average velocity for the trip? Express your answer in ms-1
Average Velocity = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒 = 2.5 𝑘𝑚 0.5 ℎ𝑟 = 5 𝑘𝑚 ℎ𝑟 −1 at ˚T
In m/s → 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦= = 𝑚𝑠 −1
Total Distance travelled = 3.5 km
c) Aimee’s displacement was 2.5 km, 53.1° T
Town
Start
Lake
2 km
1.5 km 𝜽

9. Now Do
Chapter 5 – Questions 1, 2, 3, 4, 6, 7, 8, 9, 10

10. Instantaneous Velocity and Speed – using graphs
Instantaneous Velocity is the velocity at a particular instant of time.
If we plot a graph of an objects Displacement (Position) versus Time, we can find an objects Instantaneous Velocity by calculating the gradient of the graph at a particular point or interval.
Similarly,
Instantaneous Speed is the speed at a particular instant of time.
If we plot a graph of an objects Distance versus Time, we can find an objects Instantaneous Speed by calculating the gradient of the graph at a particular point or interval.

11. What are they it? What can we use them for?
Graphing Motion
Position v Time Graph
What are they it? What can we use them for?
Plot of an objects position (y-axis) versus time (x-axis)
Shows the journey of an object over a given period of time
Velocity is the rate of change of position, so velocity can be found by calculating the gradient of the graph at a particular instant or over a period of time.
( recall: m = 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 = 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑓𝑖𝑛𝑎𝑙 − 𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑡𝑖𝑚𝑒 𝑓𝑖𝑛𝑎𝑙 − 𝑡𝑖𝑚𝑒 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 )

12. Velocity v Time Graph What are they it? What can we use them for?
Graphing Motion
Velocity v Time Graph
What are they it? What can we use them for?
Plot of an objects velocity (y-axis) versus time (x-axis)
Shows the velocity of an object over a given period of time
Displacement during a time interval is found by calculating the area under the graph
Instantaneous position can only be found if the initial starting position is known
Acceleration is the rate of change of velocity, so acceleration can be found by calculating the gradient of the graph at a particular instant or over a period of time.
( recall: m = 𝑟𝑖𝑠𝑒 𝑟𝑢𝑛 = 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑓𝑖𝑛𝑎𝑙 − 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑡𝑖𝑚𝑒 𝑓𝑖𝑛𝑎𝑙 − 𝑡𝑖𝑚𝑒 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 )

13. Acceleration v Time Graph What are they it? What can we use them for?
Graphing Motion
Acceleration v Time Graph
What are they it? What can we use them for?
Plot of an objects acceleration (y-axis) versus time (x-axis)
Shows the acceleration of an object over a given period of time
Velocity is found by calculating the area under the graph
If the initial velocity is known, we can use the graph to find the velocity at an instant in time.

14. Example 1: Position/Velocity/Acceleration v Time Graph
Graphing Motion
Example 1: Position/Velocity/Acceleration v Time Graph

15. Example 2: Position/Velocity/Acceleration v Time Graph
Graphing Motion
Example 2: Position/Velocity/Acceleration v Time Graph

16. Example 2 (cont’d) : Position/Velocity/Acceleration v Time Graph

17. Graphing Motion – Position vs Time
eg1. During a 100m race between two runners, timekeepers were instructed to record the position of each runner at 3 second intervals. The results can be shown in the table below.
a) The runners ran east, a total distance of 100 m. Plot the data on a Position v Time graph • • • • • • • • • • • •

18. Instantaneous Velocity and Speed – using graphs
eg1 (cont’d)
b) What can be said about the overall velocity of each of the runners?
c) Who wins the race?
d) Calculate the velocity of each of the runners for 3 – 6 second interval.
e) Calculate the average velocity of each of the runners for the entire race.

19. Instantaneous Velocity and Speed – using graphs
eg1 (cont’d)
b) What can be said about the overall velocity of each of the runners?
Beryl – Velocity changes throughout the race, slowing down for the last half.
Sam – Velocity is constant throughout the race. We can see this by looking at the gradient of the lines.
c) Who wins the race?
They come a draw. This can be seen as the graph for each runner ends at 15s for both runners.
d) Calculate the velocity of each of the runners for 3 – 6 second interval.
Beryl
velocity= 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑓𝑖𝑛𝑎𝑙 − 𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑡𝑖𝑚𝑒 𝑓𝑖𝑛𝑎𝑙 − 𝑡𝑖𝑚𝑒 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 64−43 6−3 = 21 3 =7 𝑚 𝑠 −1
Sam
velocity= 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑓𝑖𝑛𝑎𝑙 − 𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑡𝑖𝑚𝑒 𝑓𝑖𝑛𝑎𝑙 − 𝑡𝑖𝑚𝑒 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 40−20 6−3 = 20 3 =6.67 𝑚 𝑠 −1

20. Instantaneous Velocity and Speed – using graphs
eg1 (cont’d)
e) Calculate the average velocity of each of the runners for the entire race.
Beryl
velocity= 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑓𝑖𝑛𝑎𝑙 − 𝑃𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑡𝑖𝑚𝑒 𝑓𝑖𝑛𝑎𝑙 − 𝑡𝑖𝑚𝑒 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 100−0 15−0 = =6.67 𝑚 𝑠 −1
Sam

21. What is the acceleration of the hovercraft at 200 sec?
What is the total distance travelled by the hovercraft?

22. What is the acceleration of the hovercraft at 200 sec?
Given by the rate of change of the graph (ie. the gradient)
Choose any two points from the first section of graph.
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 11−0 200−0 = =0.055 𝑚 𝑠 −2
What is the acceleration of the hovercraft at 400 sec?
Choose any two points from the middle section of graph.
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 16−16 600−400 = =0 𝑚 𝑠 −2
What is the acceleration of the hovercraft at 800 sec?
Choose any two points from the third section of graph.
𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡= 𝑦 2 − 𝑦 1 𝑥 2 − 𝑥 1 = 0− −600 = − =0.032 𝑚 𝑠 −2
What is the total distance travelled?
Distance = Area under velocity-time graph
= Triangle + Rectangle + Triangle
= 𝑏ℎ 2 +𝑙𝑤+ 𝑏ℎ 2
= 300× × ×16 2 =
=11200 𝑚

23. Now Do
Simulation Activity

24. Now Do
Chapter 5 – Questions 11, 12, 14, 13

25. Motion Equations – constant acceleration without graphs
For straight line motion with constant acceleration, a number of formulae can be used to represent the motion and calculate unknown quantities.
The terms which we use in the equations are given by:
u = initial velocity (m/s)
v = final velocity (m/s)
a = acceleration (m/s2)
t = time interval (t)
x = displacement (m)

26. Motion Equations – constant acceleration without graphs
The equations which we use to calculate unknown values are given by:
𝑣 = 𝑢 + 𝑎𝑡
𝑥 = 𝑢+𝑣 𝑡
𝑥=𝑢𝑡+ 1 2 𝑎 𝑡 2
𝑥=𝑣𝑡− 1 2 𝑎 𝑡 2
𝑣 2 = 𝑢 2 +2𝑎𝑥

27. Motion Equations – constant acceleration without graphs
𝑣 = 𝑢 + 𝑎𝑡
𝑥 = 𝑢+𝑣 𝑡
𝑥=𝑢𝑡+ 1 2 𝑎 𝑡 2
𝑥=𝑣𝑡− 1 2 𝑎 𝑡 2
𝑣 2 = 𝑢 2 +2𝑎𝑥
eg1. A coin is dropped into a wishing well and takes 2.0 s to reach the water. The coin accelerates at a constant 10 m/s2.
Find: a) The coins velocity as it hits the water
b) How far does the coin fall before hitting the water?

28. Motion Equations – constant acceleration without graphs
𝑣 = 𝑢 + 𝑎𝑡
𝑥 = 𝑢+𝑣 𝑡
𝑥=𝑢𝑡+ 1 2 𝑎 𝑡 2
𝑥=𝑣𝑡− 1 2 𝑎 𝑡 2
𝑣 2 = 𝑢 2 +2𝑎𝑥
eg2. A parked car with the handbrake left off rolls down a hill in a straight line with a constant acceleration of 2.0 m/s2.
It stops after colliding with a brick wall at a speed of 12 m/s.
a) For how long was the car rolling?
b) How far did the car roll before colliding with the wall?

29. Now Do
Chapter 5 – Questions 21, 22, 23, 24

30. Force A force is a push or pull applied by one object on another.
Forces can include:
- Actions imposed by people eg. Pushing a door open
- Gravity
- Friction eg. Resistance caused by rough surfaces
- Air Resistance
Force is a vector quantity as it has both a magnitude and direction.
Force is measured in Newtons (N).

31. Force What forces are acting on the falling apple?
We can draw diagrams representing forces acting on an object.
We represent each force with an arrow.
The arrow shows the direction of the force, the length of the arrow indicates its size.
An apple falls from a tree.
What forces are acting on the falling apple?
Air Resistance (upward)
Weight (downward due to gravitational pull)

32. Newtons Laws Newtons First Law
An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
A book sitting on a bench
A book sliding right across a bench

33. Newtons Laws 𝐹 𝑛𝑒𝑡 =𝑚𝑎 Newtons Second Law
The acceleration of an object is dependent upon two variables –
the net force acting upon the object and the mass of the object.
This can be calculated using…..
Net Force = Mass (kg) x Acceleration (m/s2)
𝐹 𝑛𝑒𝑡 =𝑚𝑎
The direction of the net force is in the same direction as the acceleration.
If we are referring to the acceleration due to gravity, then we use a (or g) = 10 m/s2
This gravitational pull is unique to Earth.
Other planets have different gravitational field strength.

34. Newtons Laws 𝐹 𝑛𝑒𝑡 =𝑚𝑎 𝑎= 𝐹 𝑚 Newtons Second Law 𝐹=𝑚𝑎 =𝑚𝑔
eg1. Calculate the Force of a 100 g apple falling from a tree.
eg2. Calculate the acceleration that results when a 12 N force is applied to a 3 kg object.
𝐹=𝑚𝑎
=𝑚𝑔
=0.100 𝑘𝑔 ×10 𝑚/ 𝑠 2
=1 𝑁 ↓
𝐹=𝑚𝑎
𝑎= 𝐹 𝑚
= 12 𝑁 3 𝑘𝑔
=4 𝑚/ 𝑠 2

35. Newtons Laws 𝐹 𝑛𝑒𝑡 =𝑚𝑎 𝐹=𝑚𝑎 𝑊=𝑚𝑔 𝑊=70𝑘𝑔 × 10 𝑚/𝑠 2 𝑊=700 𝑁
Newtons Second Law
𝐹 𝑛𝑒𝑡 =𝑚𝑎
Weight is a force which can be calculated using this formula.
eg3. Calculate the weight of a 70kg person.
We use acceleration due to gravity = 10 m/s2
𝐹=𝑚𝑎
𝑊=𝑚𝑔
𝑊=70𝑘𝑔 × 10 𝑚/𝑠 2
𝑊=700 𝑁

36. For every action there is an equal and opposite reaction.
Newtons Laws
Newtons Third Law
For every action there is an equal and opposite reaction.
If two objects interact with each other, they exert forces upon each other.
Forces always come in pairs - equal and opposite action-reaction force pairs. 
The size of the forces on the first object equals the size of the force on the second object.
The direction of the force on the first object is opposite to the direction of the force on the second object.
These two forces are called action and reaction force.

37. For every action there is an equal and opposite reaction.
Newtons Laws
Newtons Third Law
For every action there is an equal and opposite reaction.
When you sit in your chair, your body exerts a downward force on the chair and the chair exerts an upward force on your body.
There are two forces resulting from this interaction - a force on the chair and a force on your body – the action-reaction forces.
We call the reaction forces the ‘Normal’ reaction force.

38. Forces are balanced, the apple stays in the tree
Forces in Action
Apple in a tree vs Apple falling from a tree
Forces are balanced, the apple stays in the tree
Forces are not balanced, the Net force is downward, the Apple falls from the tree

39. Now Do
Chapter 6 – Questions 1, 2, 3, 5, 7

40. First lets recall Newtons First Law….
The Net Force
Using what we know about Forces and Newtons Laws, we can start to draw diagrams to indicate the forces acting on objects.
First lets recall Newtons First Law….
An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
This means that if an object is stationary or moving with a constant velocity
(ie. Acceleration = 0 m/s2), then forces acting on the object are balanced and we say that the Net Force, 𝐹 𝑁𝑒𝑡 =0

41. The Net Force Recall: Newtons First Law
An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
A book sitting on a bench
𝐹 𝑁𝐸𝑇 =0
A book sliding right across a bench
𝐹 𝑁𝐸𝑇 >0

42. The Net Force
Using this we can start to draw diagrams to indicate the forces acting on objects and calculate Forces acting on them.
eg1. What is the Net force acting on this purple box?
eg2. What is the Net Force acting on the red box?
eg3. If the Net Force = 0, calculate the missing force on the system.
100 N WEST
20 N WEST vs
220 – 170 = 50
Missing Force = 50 N WEST

43. The Net Force
eg4. Calculate the Net Force on the system of Forces shown in the diagram.
Right Angled triangle: 𝑈𝑠𝑒 𝑃𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑎𝑠…..
𝑐 2 = 𝑎 2 + 𝑏 2
𝑐 2 =
𝑐 2 =
𝑐 2 =80 000 𝑐=
NET FORCE = 𝑁 𝐸𝐴𝑆𝑇
If not right angled consider separately: 𝑈𝑠𝑒 𝑇𝑟𝑖𝑔….
cos 𝜃 = 𝐴 𝐻
𝑥=200 cos 45
𝑥= 𝑁 𝐸𝐴𝑆𝑇
200 𝑁
200 𝑁
90 °
𝑁𝑒𝑡 𝐹𝑜𝑟𝑐𝑒
45°
200 𝑁

44. The Net Force
Eg4 continued... Calculate the Net Force on the system of Forces shown in the diagram.
Otherwise consider separately: 𝑈𝑠𝑒 𝑇𝑟𝑖𝑔….
cos 𝜃 = 𝐴 𝐻
𝑥=200 cos 45
𝑥= 𝑁 𝐸𝐴𝑆𝑇
NET FORCE = N EAST
45°
200 𝑁
45°
200 𝑁

45. The Net Force
eg5. a) Calculate the Net Force on the system of forces shown in the diagram.
b) If the Net Force of the system = 0,
what additional force needs to be applied?
173.21𝑁 𝐸𝐴𝑆𝑇
200 N
100 N
50°
𝑎 2 = 𝑐 2 − 𝑏 2
𝑎 2 = − 100 2
𝑎 2 = −10 000
𝑎 2 =30 000 𝑎=
𝑎= 𝑁 WEST

46. The Net Force
eg6. If the Net Force = 200 N EAST, find the missing force in the system below.
cos 𝜃 = 𝐴 𝐻
𝑥=100 cos 30
𝑥=86.6
𝐹𝑜𝑟𝑐𝑒=86.6𝑁 𝐸𝐴𝑆𝑇
Two triangles ∴𝐹𝑜𝑟𝑐𝑒=2×86.6=173.2 𝑁 𝐸𝐴𝑆𝑇
Net = 73.2 N EAST
100 𝑁
30° 𝑥
173.2 𝑁
100 𝑁
But if Net Force = 200 N EAST, Missing force is:
Another Force of 200 – 73.2 = N EAST is required

47. Now Do
Chapter 6 – Questions 9, 10, 11