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=[/tex] a 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.
[tex]f \approx[/tex] $\square$
Answer (2)
Substitute the given values d = 138 and v = 40 into the formula d ( v ) = 64.4 f 2.15 v 2 .
Solve for f : 138 = 64.4 f 2.15 ( 40 ) 2 .
Calculate f = 64.4 ( 138 ) 2.15 ( 40 ) 2 = 8887.2 3440 ≈ 0.38705 .
Round the value of f to the nearest hundredth: f ≈ 0.39 .
Explanation
Understanding the Problem We are given the formula for the stopping distance of a truck: d ( v ) = 64.4 f 2.15 v 2 where: d = stopping distance in feet v = initial velocity in miles per hour f = a constant related to friction We are given that when the truck's initial velocity on dry pavement is 40 mph, its stopping distance is 138 ft. We need to find the value of f , rounded to the nearest hundredth.
Substituting the Values We substitute the given values into the formula: 138 = 64.4 f 2.15 ( 40 ) 2 Now we solve for f .
Isolating f First, we can multiply both sides by f to get: 138 f = 64.4 2.15 ( 40 ) 2 Then, we divide both sides by 138 to isolate f :
f = 64.4 ( 138 ) 2.15 ( 40 ) 2
Calculating f Now we calculate the value of f :
f = 64.4 × 138 2.15 × 1600 = 8887.2 3440 ≈ 0.38705 Rounding to the nearest hundredth, we get f ≈ 0.39 .
Final Answer Therefore, the value of f rounded to the nearest hundredth is 0.39.
Examples
Understanding the stopping distance of vehicles is crucial for road safety. The formula and calculations we used can be applied in real-world scenarios such as designing traffic signals, setting speed limits, and determining safe following distances. For instance, traffic engineers can use this formula to estimate the minimum distance required between traffic lights to prevent accidents. Similarly, drivers can use this understanding to maintain a safe distance from the vehicle in front, especially in varying road conditions.
By substituting the values for stopping distance and initial velocity into the formula for stopping distance, we isolate and solve for f , finding it to be approximately 0.39 when rounded to the nearest hundredth.
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