The distance it takes a truck to stop can be modeled by the function
[tex]d(v)=\frac{2.15 v^2}{64.4 f}[/tex]
[tex]d=[/tex] stopping distance in feet
[tex]v=[/tex] initial velocity in miles per hour
[tex]f= a[/tex] constant related to friction
When the truck's initial velocity on dry pavement is 40 mph, its stopping distance is 138 ft.
Determine the value of [tex]f[/tex], rounded to the nearest hundredth.
Choose the quadratic model for the situation.
A. [tex]d(v)=\frac{2.15 v^2}{.039}[/tex]
B. [tex]d(v)=\frac{2.15 v^2}{64.79}[/tex]
C. [tex]d(v)=\frac{2.15 v^2}{25.116}[/tex]
Answer (2)
Substitute the given value of f into the denominator of the function.
Calculate the denominator: 64.4 × 0.39 = 25.116 .
The correct quadratic model is d ( v ) = 25.116 2.15 v 2 .
Therefore, the answer is d ( v ) = 25.116 2.15 v 2 .
Explanation
Understanding the Problem We are given the function d ( v ) = 64.4 f 2.15 v 2 , where d is the stopping distance in feet, v is the initial velocity in miles per hour, and f is a constant related to friction. We are given that f ≈ 0.39 . We need to choose the correct quadratic model for the situation by substituting f = 0.39 into the original equation.
Calculating the Denominator We need to calculate the value of 64.4 × 0.39 . The result of this multiplication will be the denominator of the correct quadratic model.
Finding the Correct Model The result of the multiplication is 64.4 × 0.39 = 25.116 . Therefore, the correct quadratic model is d ( v ) = 25.116 2.15 v 2 .
Examples
Understanding stopping distances is crucial in traffic safety and urban planning. For example, engineers use these calculations to determine safe distances between traffic lights or to design roads that minimize accidents. By understanding the relationship between speed, friction, and stopping distance, we can create safer driving environments and reduce the risk of collisions. This knowledge is also useful for drivers to make informed decisions about speed and following distance, especially in varying road conditions.
The value of the friction constant f is approximately 0.39. Substituting this value into the stopping distance formula gives the correct quadratic model as d ( v ) = 25.116 2.15 v 2 , which corresponds to option C. Hence, the answer is d ( v ) = 25.116 2.15 v 2 .
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