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v1 Algebra, STD2 A4 2009 HSC 28c

The brightness of a lamp \((L)\) is measured in lumens and varies directly with the square of the voltage \((V)\) applied, which is measured in volts.

When the lamp runs at 7 volts, it produces 735 lumens.

What voltage is required for the lamp to produce 1820 lumens? Give your answer correct to one decimal place.   (3 marks)

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 `11.2\ \text(volts)`

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♦♦ Mean mark 22%
TIP: Establishing `L=k V^2` is the key part of solving this question.

`L prop V^2\ \ => \ \ L=kV^2`

`text(Find)\ k\ \text{given}\ L = 735\ \text{when}\ V = 7:`

`735` `= k xx 7^2`
`:. k` `= 735/49=15`

 
`text(Find)\ V\ text(when)\ L = 1820:`

`1820` `= 15 xx V^2`
`V^2` `= 1820/15=121.33…`
`V` `= sqrt{121.33} = 11.2\ text(volts)\ \ text{(to 1 d.p.)}`

Filed Under: Non-Linear: Exponential/Quadratics (Std 2-X) Tagged With: Band 5, num-title-ct-patha, num-title-qs-hsc, smc-4239-40-a prop other, smc-830-20-Quadratics, smc-830-60-Proportional

v1 Algebra, STD2 A4 EO-Bank 1

Taylor discovers that for a Spotted stingray, its mass is directly proportional to the square of its wingspan.

One Spotted stingray has a wingspan of 60 cm and a mass of 5400 grams.

What is the expected wingspan of a Spotted stingray with a mass of 9.6 kg?   (3 marks)

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`80.0\ text{cm}`

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`text(Mass) prop text(wingspan)^2\ \ =>\ \ m = kw^2`

`text(Find)\ k:`

`5400` `= k xx 60^2`
`k` `= 5400/60^2= 1.5`

 
`text(Find)\ w\ text(when)\ \ m = 9600:`

`9600` `= 1.5 xx w^2`
`w^2` `= 9600/1.5=6400`
`:. w` `= 80\ text{cm}`

Filed Under: Non-Linear: Exponential/Quadratics (Std 2-X) Tagged With: Band 4, smc-830-20-Quadratics, smc-830-60-Proportional

v1 Algebra, STD2 A4 2023 HSC 22

The stopping distance of a motor bike, in metres, is directly proportional to the square of its speed in km/h, and can be represented by the equation

`text{stopping distance}\ = k xx text{(speed)}^2`

where `k` is the constant of variation.

The stopping distance for a motor bike travelling at 40 km/h is 16 m.

  1. Find the value of `k`.  (2 marks)

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  2. What is the stopping distance when the speed of the motor bike is 80 km/h?  (1 mark)

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a.    `k=0.01`

b.    `64.0\ text{m}`

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a.  `text{stopping distance}\ = k xx text{(speed)}^2`

`16` `=k xx 40^2`  
`k` `=16/40^2=0.001`  

 
b.    `text{Find stopping distance}\ (d)\ text{when speed = 80 km/h:}`

`d=0.01 xx 80^2=64.0\ text{m}`

Filed Under: Non-Linear: Exponential/Quadratics (Std 2-X) Tagged With: Band 4, smc-830-20-Quadratics, smc-830-60-Proportional

Algebra, STD2 A4 2023 HSC 22

The braking distance of a car, in metres, is directly proportional to the square of its speed in km/h, and can be represented by the equation

`text{braking distance}\ = k xx text{(speed)}^2`

where `k` is the constant of variation.

The braking distance for a car travelling at 50 km/h is 20 m.

  1. Find the value of `k`.  (2 marks)

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  2. What is the braking distance when the speed of the car is 90 km/h?  (1 mark)

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a.    `k=0.008`

b.    `64.8\ text{m}`

Show Worked Solution

a.  `text{braking distance}\ = k xx text{(speed)}^2`

`20` `=k xx 50^2`  
`k` `=20/50^2`  
  `=0.008`  

 
b.    `text{Find braking distance}\ (d)\ text{when speed = 90 km/h:}`

`d` `=0.008 xx 90^2`  
  `=64.8\ text{m}`  

Filed Under: Non-Linear: Exponential/Quadratics (Std 2) Tagged With: Band 4, smc-830-20-Quadratics, smc-830-60-Proportional

Algebra, STD2 A4 SM-Bank 2

Moses finds that for a Froghead eel, its mass is directly proportional to the square of its length.

An eel of this species has a length of 72 cm and a mass of 8250 grams.

What is the expected length of a Froghead eel with a mass of 10.2 kg? Give your answer to one decimal place.  (3 marks)

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`80.1\ text{cm}`

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`text(Mass) prop text(length)^2`

`m = kl^2`
 

`text(Find)\ k:`

`8250` `= k xx 72^2`
`k` `= 8250/72^2`
  `= 1.591…`

 
`text(When)\ \ l\ \ text(when)\ \ m = 10\ 200:`

`10\ 200` `= 1.591… xx l^2`
`l^2` `= (10\ 200)/(1.591…)`
`:. l` `= 80.069…`
  `= 80.1\ text{cm  (to 1 d.p.)}`

Filed Under: Non-Linear: Exponential/Quadratics (Std 2) Tagged With: Band 4, smc-830-20-Quadratics, smc-830-60-Proportional

Algebra, STD2 A4 2009 HSC 28c

The height above the ground, in metres, of a person’s eyes varies directly with the square of the distance, in kilometres, that the person can see to the horizon.

A person whose eyes are 1.6 m above the ground can see 4.5 km out to sea.

How high above the ground, in metres, would a person’s eyes need to be to see an island that is 15 km out to sea? Give your answer correct to one decimal place.   (3 marks)

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 `17.8\ text(m)\ \ text{(to 1 d.p.)}`

Show Worked Solution
♦♦ Mean mark 22%
CRITICAL STEP: Reading the first line of the question carefully and establishing the relationship `h=k d^2` is the key part of solving this question.

`h prop d^2`

`h=kd^2`

`text(When)\ h = 1.6,\ d = 4.5`

`1.6` `= k xx 4.5^2`
`:. k` `= 1.6/4.5^2`
  `= 0.07901` `…`

 

`text(Find)\ h\ text(when)\ d = 15`

`h` `= 0.07901… xx 15^2`
  `= 17.777…`
  `= 17.8\ text(m)\ \ \ text{(to 1 d.p.)}`

Filed Under: Exponential/Quadratic (Projectile), Non-Linear: Exponential/Quadratics (Std 2), Variation and Rates of Change Tagged With: Band 5, num-title-ct-patha, num-title-qs-hsc, smc-4239-40-a prop other, smc-830-20-Quadratics, smc-830-60-Proportional

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