Question

18 mark for review
the scatterplot shows the distance y, in feet, it took for a certain vehicle to stop after
applying the brakes when traveling at speed x, in miles per hour.
which of the following best models the stopping distance, in feet, of the vehicle after
applying the brakes as a function of speed, in miles per hour?
a ( y = 0.068x^2 )
b ( y = 0.068x^2 + 30 )
c ( y = 0.68x^2 )
d ( y = 0.68x^2 + 30 )

Explanation:

Step1: Analyze the scatterplot trend

The scatterplot shows a quadratic trend (curved upward), so a quadratic model \( y = ax^2 + b \) is appropriate. We can test \( x = 20 \) (from the plot, around \( x = 20 \), \( y \) is small, maybe around 25 - 30? Wait, let's calculate for each option.

Step2: Test \( x = 20 \) in each option

• Option A: \( y = 0.068(20)^2 = 0.068 \times 400 = 27.2 \)
• Option B: \( y = 0.068(20)^2 + 30 = 27.2 + 30 = 57.2 \)
• Option C: \( y = 0.68(20)^2 = 0.68 \times 400 = 272 \) (too big, since at \( x = 20 \), the point is low, not 272)
• Option D: \( y = 0.68(20)^2 + 30 = 272 + 30 = 302 \) (also too big)

Now check \( x = 60 \):

• Option A: \( y = 0.068(60)^2 = 0.068 \times 3600 = 244.8 \), which is close to the point at \( x = 60 \) (around 225 - 250, so 244.8 is reasonable)
• Option B: \( 0.068(60)^2 + 30 = 244.8 + 30 = 274.8 \), which is higher than the plot's point at \( x = 60 \) (the point is around 225 - 250, 244.8 is better)
• Options C and D are way too big (e.g., \( x = 60 \) would be 2448 and 2478, which is way above the y - axis max of 300? Wait no, the y - axis is up to 300, but the points at \( x = 60 \) are around 225 - 250. Wait, my mistake earlier: at \( x = 60 \), the point is around 225 - 250. Let's recalculate Option A at \( x = 60 \): \( 0.068\times60^2 = 0.068\times3600 = 244.8 \), which is close. Option C: \( 0.68\times60^2 = 0.68\times3600 = 2448 \), which is way above 300, so C and D are out. Now between A and B: at \( x = 20 \), the point is low, around 25 - 30, Option A gives ~27, which matches. Option B gives ~57, which is higher than the plot's point at \( x = 20 \) (the point at \( x = 20 \) is near y = 30? Wait the plot's y - axis: at x = 20, the point is around y = 30? Wait no, the first point is at x = 20, y is around 30? Wait the y - axis has 0, 75, 150, 225, 300. So at x = 20, the point is between 0 and 75, closer to 30. So Option A's 27.2 is close, Option B's 57.2 is higher. At x = 60, Option A gives 244.8, which is close to the point (around 225 - 250, so 244.8 is good). So Option A is better.

Answer:

A. \( y = 0.068x^2 \)