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Chapter 3: Problem 43

Solve the given problems. The stopping distance \(d\) (in \(\mathrm{ft}\) ) of a car going \(v \mathrm{mi} / \mathrm{h}\) is given by \(d=v+0.05 v^{2} .\) Because \(d=f(v),\) find \(f(30), f(2 v),\) and \(f(60),\) using both \(f(v)\) and \(f(2 v)\)

Short Answer

Expert verified
The stopping distances are: \( f(30) = 75 \) ft, \( f(2v) = 2v + 0.2v^2 \), \( f(60) = 240 \) ft.

Step by step solution

01

Understand the function

The stopping distance function is given by the equation \( f(v) = v + 0.05v^2 \). This equation tells us how the stopping distance, \( d \), changes with different speeds, \( v \), in miles per hour.
02

Calculate f(30)

To find \( f(30) \), we substitute 30 for \( v \) in the function. \[ f(30) = 30 + 0.05(30)^2 \]Calculate \( 0.05(30)^2 \):\( 30^2 = 900 \)\( 0.05 \times 900 = 45 \)So, \( f(30) = 30 + 45 = 75 \).The stopping distance at 30 miles per hour is 75 feet.
03

Formula Derivation for f(2v)

To find \( f(2v) \), substitute \( 2v \) for \( v \) in the function:\[ f(2v) = 2v + 0.05(2v)^2 \]Expanding \( (2v)^2 \): \( (2v)^2 = 4v^2 \)Thus, \( f(2v) = 2v + 0.05 \times 4v^2 = 2v + 0.2v^2 \)
04

Calculate f(60) using derived formula

Utilize the derived formula for \( f(2v) \):\[ f(60) = 2(30) + 0.2 \times (30)^2 \]Calculate \( 0.2 \times 900 \):\( 0.2 \times 900 = 180 \)So, \( f(60) = 60 + 180 = 240 \).The stopping distance at 60 miles per hour is 240 feet.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Function
A quadratic function is a type of polynomial represented by the general form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). In this instance, the stopping distance function \( f(v) = v + 0.05v^2 \) is a quadratic function. The term involving \( v^2 \) indicates that the stopping distance increases at an accelerated rate as speed increases. Quadratic functions graph as a parabola on a coordinate plane. Here, the focus is on understanding how the variables in the equation affect the real-world stopping distance of a car.
Function Evaluation
Function evaluation involves substituting specific values into a function to calculate its output. This is what we do when finding \( f(30) \) or \( f(60) \). For instance, with the stopping distance function \( f(v) = v + 0.05v^2 \), to evaluate \( f(30) \), we plug 30 in place of \( v \), and compute \[ f(30) = 30 + 0.05 \times 30^2 \] which equals 75. This tells us the stopping distance at 30 miles per hour is 75 feet. Function evaluation is a powerful tool that allows us to predict specific outcomes based on given variables.
Substitution Method
The substitution method is an essential technique in algebra where you replace variables with specific values or expressions. In this problem, we see substitution used twice: first, substituting single values like 30 or 60 for \( v \), and secondly, substituting expressions like \( 2v \). While solving for \( f(2v) \), the process involves taking \( f(v) = v + 0.05v^2 \) and substituting \( 2v \) for \( v \), resulting in \( f(2v) = 2v + 0.2v^2 \). This method simplifies calculations and reveals how changes in speed affect stopping distances, demonstrating patterns or relationships within mathematical problems.

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Most popular questions from this chapter

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