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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)

  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}`  

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}`

Show Worked Solution

`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.)}`

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.)}`