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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 braking distance = k × (speed)² where k is the constant of variation - HSC - SSCE Mathematics Standard - Question 22 - 2023 - Paper 1

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Question 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--braking-distance-=-k-×-(speed)²--where-k-is-the-constant-of-variation-HSC-SSCE Mathematics Standard-Question 22-2023-Paper 1.png

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 braking distance = k ×... show full transcript

Worked Solution & Example Answer: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 braking distance = k × (speed)² where k is the constant of variation - HSC - SSCE Mathematics Standard - Question 22 - 2023 - Paper 1

Step 1

Find the value of k.

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Answer

Given the equation for braking distance:

extbrakingdistance=k×(speed)2 ext{braking distance} = k \times (\text{speed})^2

We know that the braking distance when the speed is 50 km/h is 20 m:

20=k×(50)220 = k \times (50)^2

First, we calculate (50)²:

(50)2=2500(50)^2 = 2500

Now substituting this value back into the equation gives:

20=k×250020 = k \times 2500

To solve for k, we divide both sides by 2500:

k=202500k = \frac{20}{2500}

Calculating this results in:

k=0.008k = 0.008

Step 2

What is the braking distance when the speed of the car is 90 km/h?

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Answer

Using the value of k found in part (a), we can substitute it back into the braking distance formula:

extbrakingdistance=0.008×(speed)2 ext{braking distance} = 0.008 \times (\text{speed})^2

Substituting the speed of 90 km/h into the equation:

extbrakingdistance=0.008×(90)2 ext{braking distance} = 0.008 \times (90)^2

Calculating (90)²:

(90)2=8100(90)^2 = 8100

Thus, the equation becomes:

extbrakingdistance=0.008×8100 ext{braking distance} = 0.008 \times 8100

Calculating that gives:

extbrakingdistance=64.8 m ext{braking distance} = 64.8 \text{ m}

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