Mathematics C (2008) · Mathematics C (2008) Sample assessment instrument and indicativeresponse Extended modelling and problem solving: Topics include Dynamics and Calculus This

Mathematics C (2008) Sample assessment instrument and indicative response

Extended modelling and problem solving: Topics include Dynamics and Calculus This sample is intended to inform the design of assessment instruments in the senior phase of learning. It highlights the qualities of student work and the match to the syllabus standards.

Criteria assessed • Knowledge and procedures (K&P)

• Modelling and problem solving (M&P)

• Communication and justification (C&J)

Assessment instrument The work presented in this sample is in response to an assessment task.

Context Vertical motion under gravity with and without air resistance On 16 August 1960 over New Mexico, USA, Joseph W. Kittinger stepped off the ‘Highest Step in the World’ from a stationary gondola attached to a helium-filled balloon at an altitude of 31333m (which is over three times higher than a commercial airliner typically flies). He descended basically at the mercy of the elements, except for a small stabilizing device) until his main parachute deployed at a height of about 5500 metres above the ground. He then landed about eight minutes later. In addition to his pressure suit, Kittinger carried instruments and safety gear that weighed more than he did. He also wore several layers of clothing to help protect him against the extremes of his high-altitude environment. During his fall, Kittinger experienced air temperatures as low as -70°C. Vision of this amazing accomplishment (which still claims two world records to this day) can be found at the following website. http://www.youtube.com/watch?v=Z8RRkMsHOMU

http://www.youtube.com/watch?v=Z8RRkMsHOMU

2 | Mathematics A (2008): Sample student assessment and responses Topics include Dynamics and Calculus

Task (K&P, M&P) (a) (i) Assuming that there was no air resistance, use integration methods to develop a

velocity model for Kittinger’s motion, and hence show that his speed as the parachute opened would be approximately 2560 km/h.

(ii) If this assumption was changed so that air resistance was considered, briefly discuss the effect of this new assumption on this speed result.

(b) (i) Compare this speed of 2560 km/h against the speed of sound. (ii) Investigate the term Mach number in terms of speed. What Mach number would this

speed of 2560 km/h represent? (iii) What would Kittinger have experienced if he travelled faster than the speed of sound? Linear Relationship between air resistance and velocity You only need to move around on a windy day to realise that air resistance increases with velocity. There is a law of physics called Stokes’ Law, which states that if an object falls through air under certain situations, then the air resistance is proportional to the velocity of the object. The pictures below display the drag effect of an object moving through a resistance force. Note the drag (or turbulence) created behind the object.

Considering resistance to be proportional to velocity provides a reasonable model to represent falling objects in many real-life circumstances. In other words, R = kv where R is the resistance force, v is the velocity and k is a positive constant of proportionality. (c) Indeed, the greatest speed Kittinger reached was a terminal velocity of 1000 km/h. This is

because the forces acting on his body included both gravity and air resistance as represented in the following diagram (assuming R = kv).

Square Flat, edged objects create a large amount of drag. Sphere Spherical objects such as balls create a medium amount of drag

Aerofoil An aircraft wing uses an aerofoil shape to minimise drag and generate lift.

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These measurements were recorded when a large coin was thrown from the top of a building which was 369 metres above the ground. You must discuss/give evidence of at least one (1) strength and one (1) weakness for each of the two models in your solution. Furthermore, while it is expected that you would use a graphical representation of relevant aspects of this data as part of your analysis, it is not required that you use all of this data in determining your conclusion. Data recording variables associated with the vertical fall of a coin from a building Time (seconds) Weight (Newtons) Drag (Newtons) Acceleration (m/s/s) Velocity (m/s) Distance (m)

0.00 ‐0.0245 0.0000 ‐9.81 0.00 369.00

1.00 ‐0.0245 0.0126 ‐4.91 ‐7.87 364.54

2.00 ‐0.0245 0.0220 ‐1.04 ‐10.41 355.06

3.00 ‐0.0245 0.0241 ‐0.18 ‐10.89 344.34

4.00 ‐0.0245 0.0245 ‐0.03 ‐10.97 333.40

5.00 ‐0.0245 0.0245 0.00 ‐10.98 322.42

6.00 ‐0.0245 0.0245 0.00 ‐10.99 311.43

7.00 ‐0.0245 0.0245 0.00 ‐10.99 300.45

8.00 ‐0.0245 0.0245 0.00 ‐10.99 289.46

9.00 ‐0.0245 0.0245 0.00 ‐10.99 278.47

10.00 ‐0.0245 0.0245 0.00 ‐10.99 267.49

11.00 ‐0.0245 0.0245 0.00 ‐10.99 256.50

12.00 ‐0.0245 0.0245 0.00 ‐10.99 245.51

13.00 ‐0.0245 0.0245 0.00 ‐10.99 234.53

14.00 ‐0.0245 0.0245 0.00 ‐10.99 223.54

15.00 ‐0.0245 0.0245 0.00 ‐10.99 212.55

16.00 ‐0.0245 0.0245 0.00 ‐10.99 201.57

17.00 ‐0.0245 0.0245 0.00 ‐10.99 190.58

18.00 ‐0.0245 0.0245 0.00 ‐10.99 179.59

19.00 ‐0.0245 0.0245 0.00 ‐10.99 168.61

20.00 ‐0.0245 0.0245 0.00 ‐10.99 157.62

21.00 ‐0.0245 0.0245 0.00 ‐10.99 146.63

22.00 ‐0.0245 0.0245 0.00 ‐10.99 135.65

23.00 ‐0.0245 0.0245 0.00 ‐10.99 124.66

24.00 ‐0.0245 0.0245 0.00 ‐10.99 113.68

25.00 ‐0.0245 0.0245 0.00 ‐10.99 102.69

26.00 ‐0.0245 0.0245 0.00 ‐10.99 91.70

27.00 ‐0.0245 0.0245 0.00 ‐10.99 80.72

28.00 ‐0.0245 0.0245 0.00 ‐10.99 69.73

29.00 ‐0.0245 0.0245 0.00 ‐10.99 58.74

30.00 ‐0.0245 0.0245 0.00 ‐10.99 47.76

31.00 ‐0.0245 0.0245 0.00 ‐10.99 36.77

32.00 ‐0.0245 0.0245 0.00 ‐10.99 25.78

33.00 ‐0.0245 0.0245 0.00 ‐10.99 14.80

34.00 ‐0.0245 0.0245 0.00 ‐10.99 3.81

34.35 ‐0.0245 0.0245 0.00 ‐10.99 0.00

Queensland Studies Authority October 2013 | 5

Instrument-specific criteria and standards Student responses have been matched to instrument-specific criteria and standards; those which best describe the student work in this sample are shown below. For more information about the syllabus dimensions and standards descriptors, see www.qsa.qld.edu.au/1896.html#assessment.

Standard A

Kn

ow

led

ge

and

p

roce

du

res

(K&

P)

The student work has the following characteristics:

application of mathematical definitions, rules and procedures in routine and non-routine simple tasks through to routine complex tasks, in life-related and abstract situations

numerical calculations, spatial sense and algebraic facility in routine and non-routine simple tasks through to routine complex tasks, in life-related and abstract situations

appropriate selection and accurate use of technology

knowledge of the nature of and use of mathematical proof

Mo

del

ling

an

d p

rob

lem

so

lvin

g (

M&

P)


use of problem-solving strategies to interpret, clarify and analyse problems to develop responses from routine simple tasks through to non-routine complex tasks in life-related and abstract situations

identification of assumptions and their associated effects, parameters and/or variables

investigation and evaluation of the validity of mathematical arguments including the analysis of results in the context of problems, the strengths and limitations of models, both given and developed

refinement of mathematical models

Co

mm

un

icat

ion

an

d ju

stif

icat

ion

(C

&J)


appropriate interpretation and use of mathematical terminology, symbols and conventions from simple through to complex and from routine to non-routine, in life-related and abstract situations

analysis and translation of information from one representation to another in life-related and abstract situations from simple through to complex and from routine through to non-routine

use of mathematical reasoning to develop coherent, concise and logical sequences within a response from simple through to complex and in life-related and abstract situations using everyday and mathematical language

coherent, concise and logical justification of procedures, decisions and results

justification of the reasonableness of results

provision of supporting arguments in the form of proof


Indicative response — Standard A The annotations show the match to the instrument-specific standards.

(a) (KAP, MAP) (i) Assuming that there was no air resistance, the speed of dare devil Joseph W. Kittinger when he had fallen 25816 metres is to be determined. It is required to show that this approximates to 2560 km/h. Assuming that there was no air resistance, the only force acting upon Joseph W. Kittinger as he jumped from the stationary balloon (and the only acceleration he experienced) was due to gravity.

Assuming that downwards is positive and that the origin is at ground level, where s = 0.

kjgt

dtav

jga

ˆ

ˆ

Finding k as when Joseph W. Kittinger jumped, it was from a stationary balloon, meaning that there was no initial velocity such that when 0, v = 0.

Substituting these in to solve for k,

0

ˆ00

k

kjg

Therefore the expression for the velocity of Joseph W. Kittinger’s descent is given by

jgtv ˆ

Also

cgt

dtjgt

dtvs

2

2

1

ˆ

Finding c given that the stationary balloon is at an altitude of 31333m and assuming that downwards is positive and that the position of this stationary balloon was the initial displacement, then when t = 0, s = -31333. Substituting these in to solve for c,

jc

cjgj

ˆ31333

ˆ02

1ˆ31333 2

Therefore the expression for the displacement of Joseph W. Kittinger is given by:

jjgts ˆ31333ˆ2

1 2

At the point where Joseph W. Kittinger opens his parachute, it is given that he is at an altitude of 5500m. Given that downwards is positive,

js ˆ5500 .

Identification of assumptions and variables Use of mathematical reasoning to develop a coherent, concise and logical sequence within a response in a complex, life-related situation using everyday and mathematical language Application of mathematical definitions and rules in a non-routine, simple, life-related situation


Algebraic facility in a routine simple task in a life-related situation

Coherent, concise and logical justification of procedures and results

Substituting this value into the displacement expression, we obtain:

61.729.4

25833

258339.4

313339.45500

313332

15500

2

2

2

2

t

t

t

t

tg

Rejecting the negative case since t is positive, it takes approximately 72.6 seconds for Joseph Kittinger to open his parachute.

To find the velocity at this time, substitute t = 72.6 into the velocity expression:

6.711

6.728.9

gtv

Thus, as the parachute opened, Kittinger’s speed was approximately 711.6 m/s.

Converting this to km/h, we have

km/h6.25611000

3600711.6m/s9.711

Thus it has been shown that Kittinger’s speed as the parachute opened was approximately 2560 km/h.

(Remember that all motion is in the vertical plane and the j can be removed from the expression without any loss of generality.)

(a)(ii)

If the assumption was changed so that air resistance was considered, the effect of this new assumption would be that the expected maximum speed reached by Kittinger would be largely reduced.

(b)(i)

The speed of sound through air is approximately 343.2 metres per second or 1236 kilometres per hour (or about one kilometre in three seconds). The speed varies according to air temperature and humidity.

(Source: http://en.wikipedia.org/wiki/Speed_of_sound)

(b)(ii)

The term Mach number, M, is a dimensionless quantity representing the speed of an object moving through air or another fluid divided by the local speed of sound. It is commonly used to represent the speed of an object when it is travelling close to or above the speed of sound. This means

that a

vM where M is the Mach number, v is the velocity of the source

relative to the medium, and a is the speed of sound in the medium.

Investigation and evaluation of the validity of a mathematical argument Analysis and translation of information from one representation to another in a life-related situation in a complex task


Using this definition, the Mach number equivalent of 2560 km/h

would be 07.21236

2560M

(Source: http://en.wikipedia.org/wiki/Mach_number)

A sonic boom is the sound associated with shock waves created by an object travelling through the air faster than the speed of sound. Sonic booms generate enormous amounts of sound energy, sounding much like an explosion. Kittinger himself would have felt changes in pressure and temperature — hence the need for his extra clothing/gloves.

(Source: http://en.wikipedia.org/wiki/Sonic_boom)

(c) (i) Consider the same scenario as in part (a), however air resistance is now present. Let R be the air resistance having the form kvR where k is a positive constant of proportionality and v is the velocity. The net force (the total of the vertical forces) is now reliant on the force of gravity downwards and the resistive force acting upwards. Since the object is moving downwards, gravity must be greater than the air resistance.

dt

dvakvmg

dt

dvm

kvmgma

RFF Gnet

since

Thus Kittinger’s fall can be modelled by the equation kvmgdt

dvm .

(ii)

Solving this differential equation for v:

m

kpdt

pvg

dv

dtv

m

kg

dv

vm

kg

m

kvmg

dt

dv

kvmgdt

dvm

where

Integrating both sides, we obtain

ct

p

pvg

dtpvg

dv

ln

Use of mathematical reasoning to develop a coherent, concise and logical sequence within a response in a complex, life-related situation using mathematical language


Application of mathematical definitions and rules in a complex task in a life-related situation Use of mathematical reasoning to develop a coherent, concise and logical sequence within a complex response in a life-related situation using mathematical language Justification of the reasonableness of results

pt

pt

pt

rpt

rpt

rpt

Aegp

p

g

p

Aev

gAepv

eAAe

ee

epvg

pcrertppvg

pctppvg

1

where

wherln

ln

Therefore the velocity equation for Joseph’s fall when air resistance is

present is ptAegp

v 1

If the object starts from rest, t = 0 and v = 0. Thus

gA

Agp

Aegp

p

so

10

10 0

Substituting this value for A into the velocity equation,

pt

pt

ep

g

gegp

v

1

1

The terminal velocity is 1000k/h or 278m/s, that is when 278, vt .

Thus

0353.0278

01278

1278

gp

p

g

ep

g p

Thus

t

t

e

eg

v

0353.0

0353.0

1278

10353.0

and this is the required equation to describe Kittinger’s velocity when there is air resistance of the form kvR . (d) Graph the equation tv 8.9 when no air resistance is present

against the equation when there is air resistance tev 0353.01278

It can be seen when air resistance is not present the equation is a linear function, whereas when air resistance is present an exponential function is formed. As both graphs pass through the origin, it can be seen that when t = 0, v = 0, which is logical as when t = 0, the parachutist would still be in the stationary balloon, and therefore would have no vertical velocity. Additionally, if the tangent of the velocity function when air resistance is present is developed, it can be seen that its formula would approximately match that when air resistance is not present. Thus the two velocity


Knowledge of the nature of and the use of mathematical proof

graphs begin at the same rate. It can also be seen that as t , the velocity equation with air resistance reaches a terminal velocity of approximately 278 m/s. Some real-life factors which could alter the general shape of the equation,

tev 0353.01278 , that is things that would change the air resistance,

are the outfit used by the parachutist, the surface area that was being covered as the parachutist fell, and the body shape. These would effect the amount of air resistance working against the force of gravity on the object, meaning the velocity and the terminal velocity would be affected. The changes would result in an overall change in the function and hence a corresponding change in the shape of the graph. (e) (MAP)

(i) Prove that

cxa

xa

adx

xaln

2

1122

dxxaxa

dxxa

LHS

1

122

Using partial fractions

1so

1

xaBxaA

xaxa

xaBxaA

xa

B

xa

A

xaxa

When x = a,

aB

aB

aaB

2

1

12

1

When x = -a,

aA

aA

aaA

2

1

12

1

Thus

RHS

cxa

xa

a

cxaxaa

dxxa

axa

adxxaxa

xaa

xaa

xaxa

ln2

1

lnln2

1

2

1

2

11

and

2

1

2

11

Provision of supporting arguments in the form of proof


Refinement of a mathematical model

(ii) Consider the same scenario as in part (a) except there is air resistance present in the form of quadratic drag, ie, the air resistance is of

the form 2kvR where R is the air resistance, k is a positive proportionality constant and v is the velocity.

The net force (the total of the vertical forces) is now reliant on the force of gravity downwards and the resistive force acting upwards. Since the object is moving downwards, gravity must be greater than the air resistance.

dt

dvakvmg

dt

dvm

kvmgma

RFF Gnet

since2

2

Thus Kittinger’s fall can be modelled by the equation 2kvmgdt

dvm .

Solving this differential equation, we obtain:

C

gavaC

vC

gC

m

kCCvg

m

kvg

m

kvmg

dt

dv

kvmgdt

dvm

where

where

22

2

2

2

2

2

We express the differential equation in this form so that we can use the integration developed in the previous section. We can now solve the required differential equation.

dtva

dv

C

dtvaC

dv

dtvaC

dv

vaCdt

dv

22

22

22

22

1

Using the results generated from the previous section, we obtain for this integral:

Use of problem-solving strategies to interpret, clarify and analyse problems to develop a response to a complex task in a life-related situation Appropriate interpretation and use of mathematical terminology, symbols and conventions in a complex, non-routine, abstract situation Application of mathematical definitions and rules in a complex, non-routine, abstract situation

dt

va

va

aC

ln2

11


Algebraic facility in a complex task in a life-related situation Appropriate interpretation and use of mathematical terminology, symbols and conventions in a complex, non-routine task in a life-related situation

Simplifying this expression in term of v:

adatCva

va

adatva

va

C

adtva

va

C

22ln

22ln1

2ln1

Substituting C

ga we obtain

dCgtCg eeva

va

dCgtCg

dC

gCt

C

gC

dtC

gC

va

va

22and

22

22

2ln

Given that C and d are constants, we can let dCgeA 2 and

1

1

1

2

2

2

22

2

2

tCg

tCg

tCg

tCgtCg

tCg

tCg

Ae

Aeav

Aea

aaAevAev

Aevava

Aeva

va

Kittinger dropped from a stationary balloon, so that when t = 0, v = 0, and hence we obtain

1

0

0 02

A

Aa

a

Aea

a Cg

Substituting A = 1 into the velocity equation we obtain

1

12

2

tCg

tCg

e

eav

Divide both numerator and denominator by tCge2 to obtain

tCg

tCg

e

eav

2

2

1

1

To obtain the terminal velocity, we consider the situation as t .

Also as t , .0,0 xe x Thus

a

av t

01

01


Coherent, concise and logical justification of procedures, decisions and results

Therefore in order to determine Kittinger’s particular velocity equation, since the terminal velocity is 278 m/s, then a = 278. So

tCg

tCg

e

ev

2

2

1

1278

Since av and C

ga , then

4

2

2

1027.1

278

8.9

2788.9

278

C

C

C

g

Substituting this value for C and g into the velocity equation, we obtain

t

t

tCg

tCg

e

e

e

ev

071.0

071.0

2

2

1

1278

1

1278

Therefore Kittinger’s particular velocity function when a quadratic relationship exists between air resistance and velocity has been determined. (f) M&P Consider the strengths and weaknesses of the models (based on the linear and quadratic relationships between air resistance and velocity) in order to determine the preferred air resistance model for the data set of a falling object as given. Compare and contrast the real-life effectiveness of both of the two velocity models — one based on the linear relationship between air resistance and velocity and the other based on a quadratic relationship between these 2 factors. By generating graphs of both velocity functions, it can be seen that both similarities and differences exist between the velocity function based on a quadratic relationship between air resistance and velocity and the velocity function based on a linear relationship for Joseph W. Kittinger skydiving.

Investigation and evaluation of the validity of mathematical arguments including the strengths and limitations of models


However, both graphs generally have the same shape, with the greatest similarities occurring initially and after much time, t, has elapsed. It is apparent that both velocity functions share approximately the same terminal velocity, v, of 277.77ms-1 and this is observed as they both plateau out at around the same dependent-axis coordinate. The greatest contrast between these two models occurs in the intermediate area (from the transition from the initial to the terminal velocity).

It can be observed that the quadratic relationship’s model has a steeper slope than the linear relationship’s, with a velocity increasing at a greater rate initially than the other model. This notion is supported by the time, t, it takes both models to actually reach the terminal velocity. The quadratic relationship function takes approximately 161 seconds to reach terminal velocity. The velocity function based on the linear relationship requires at least 297 seconds to reach its terminal velocity, which is nearly 140 seconds longer. It is clear that Joseph W. Kittinger reaches terminal velocity faster under the model with the quadratic relationship.

In terms of the real-life effectiveness of these two velocity models, it is clear that the model based on the quadratic relationship will be more realistic. This is because it is clearly stated that no law of nature says that the relationship between air resistance and velocity is linear. Furthermore, the task stipulates that if an object were heavy enough to produce turbulence, the relationship between air resistance and velocity would not be linear; and thus quadratic. It can be logically assumed that the body of Joseph W. Kittinger would cause more disturbance (and thus turbulence) to the air than a soap bubble as it is a heavier object and thus falls at a faster speed. Therefore the second model would be more effective. This is also reinforced by the fact that terminal velocity would be reached sooner. The difference between the air resistance (drag force) and force due to gravity would be greatest initially and would thus the air resistance provided would work rapidly and balance out the person’s weight faster initially than that suggested by the first model. This is because this is the first ‘opportunity’ the drag force has to work on the force due to gravity and thus the greatest effects would be seen initially. This is where the air resistance acts to balance out g, thus reaching terminal velocity faster (when drag force and gravitational force equate). This initial balancing out of the forces is not reflected in the first velocity function and is therefore unrealistic.

Brief research on air friction has uncovered that for higher velocities and larger objects, the frictional drag produced (that is, air resistance) is approximately proportional to the square of the velocity. It can be logically deduced that a 90kg man is a large object and speeds up to 1000km/h would certainly be considered ‘high velocity.’

Thus within the context of a real-life situation, the velocity function modelled by a quadratic relationship between air resistance and velocity would be more suitable.

Justification of the reasonableness of results Use of problem-solving strategies to interpret, clarify and analyse problems to develop a response to a complex task in a life-related situation


Refinement of mathematical models

Numerical calculations in a complex task in a life-related situation

Investigation and evaluation of the validity of mathematical arguments including strengths and limitations of models

Using the data for a coin falling from a building, we now consider what is the preferred model for an object falling under air resistance.

Method 1

For the first model with air resistance: ctec

gv 1

Need to find c.

From given data, when 99.10,0 va (both have been changed to

positive values as using downwards as positive)

This model was made given that cvga

Substituting in data given to find c:

892.0

99.108.90

c

c

Substituting in this value for c into the model:

t

t

e

ev

892.0

892.0

199.10

1892.0

8.9

For the second model with air resistance: ctgc

ctgc

e

gcegcv

1

1

2

121

1

Need to find c.

From given data when a = 0, v = 10.99 (both have been changed to positive values as downwards is assumed to be positive)

This model was made given that 2cvga

Substituting in data given to find c:

081.0

99.108.90 2

c

c

Substituting in this value for c into the model:

t

t

t

t

e

e

e

ev

782.1

782.1

081.08.9081.02

1081.08.9081.021

1

78.12078.120

1

8.9081.08.9081.01

1

The first model strengths:

it produces a graph that has an increasing velocity with a decreasing rate of change as expected by the data

that it reaches the same terminal velocity as recorded in the data.

The first model limitations:

the velocity does not increase at as quick a rate as recorded by the actual data and thus the terminal velocity is not reached as soon as recorded in the actual data

the model predicts negative velocity values for negative time, both of which are impossible in this situation.


The second model strengths:

the model matches the data almost perfectly. It produces a graph that has an increasing velocity with a decreasing rate of change as expected by the data.

the terminal velocity predicted is equal to the terminal velocity reached in the actual data. Furthermore, terminal velocity is reached at approximately the same point in time.

the slope of the curve of the model matches that of the slope formed by the data points.

The second model weakness:

it predicts negative velocity values for negative time which are impossible in this situation.

Research indicates that the first model applies more to situations involving small particles falling at low velocities, and the second model applies to situations involving larger objects at faster speeds (HyperPhysics 2000).The object that was falling to produce the data given is described as a ‘large coin’. A coin is many times larger than a particle, so in theory the second model should fit the data better. As explained above, this is the case; hence overall the second model provides the best fit for the data.

Alternative forms for the two models are:

linear model: tev 892.0199.10

quadratic model: t

t

e

ev

782.1

782.1

1

199.10

Method 2:

The models are based on kvR (linear) verses 2kvR (quadratic) where R = Resistance and v = velocity. Since the resistance force is the same as the drag force, we will use the data for the coin to determine whether the relationship between drag and its corresponding velocity is linear or quadratic. The absolute value of the velocity has been used to simplify the calculations, and the data is limited to that until terminal velocity is reached. The data has been reversed so that the velocity becomes the independent variable.

Velocity (m/s)

Resistance (Newtons)

0 0

7.87 0.0126

10.41 0.022

10.89 0.0241

10.97 0.0245

10.98 0.0245

10.99 0.0245

10.99 0.0245

Investigation and evaluation of the validity of mathematical arguments including strengths and limitations of models


Appropriate selection and accurate use of technology

Graphing this data, the following relationship of Resistance as a function of velocity is demonstrated.

This relationship appears to be non-linear, so a quadratic model appears more reasonable.

Using technology to determine the regression coefficients for both linear and quadratic nature, the following results were calculated.

Appropriate selection and accurate use of technology

0

0.005

0.01

0.015

0.02

0.025

0.03

0 2 4 6 8 10 12

Res

ista

nce

(D

rag

) (N

)

Velocity (m/s)

Relationship between R and v

y = 0.0023x - 0.0012R² = 0.9625

-0.005

0

0.005

0.01

0.015

0.02

0.025

0.03

0 2 4 6 8 10 12Res

ista

nce

(D

rag

) (N

)

Velocity (m/s)

Line of Best Fit -Linear Relationship between

R and v


Coherent, concise and logical justification of procedures and results

In conclusion, the most suitable model for this data would be the

quadratic model, that is 2kvR .

Acknowledgments The QSA acknowledges the contribution of Moreton Bay College in the preparation of this document.

y = 0.0002x2 + 7E-06x + 2E-07R² = 1

0

0.005

0.01

0.015

0.02

0.025

0.03

0 2 4 6 8 10 12

Res

ista

nce

(D

rag

) (N

)

Velocity (m/s)

Quadratic Relationship between R and v

Documents

Mathematics C (2008) · Mathematics C (2008) Sample assessment instrument and indicativeresponse Extended modelling and problem solving: Topics include Dynamics and Calculus This