Problem 137 - A Graphical Approach to Precalculus with Limits

Question

If air resistance is ignored, the braking distance \(D\) (in feet) for an automobile to change its velocity from \(V_{1}\) to \(V_{2}\) (feet per second) can be modeled by the equation $$D=\frac{1.05\left(V_{1}^{2}-V_{2}^{2}\right)}{64.4\left(K_{1}+K_{2}+\sin \theta\right)}$$ \(K_{1}\) is a constant determined by the efficiency of the brakes and tires, \(K_{2}\) is a constant determined by the rolling resistance of the automobile, and \(\theta\) is the grade of the highway. (Source: Mannering, \(\mathrm{F}\), and W. Kilareski, Principles of Highway Engineering and Traffic Analysis, Second Edition, John Wiley and Sons.) (a) Approximate the number of feet required to slow a car from 55 to 30 mph while traveling uphill on a grade of \(\theta=3.5^{\circ} .\) Let \(K_{1}=0.4\) and \(K_{2}=0.02 .\) (Hint: Change miles per hour to feet per second.) (b) Repeat part (a) with \(\theta=-2^{\circ}\) (c) How is the braking distance affected by the grade \(\theta ?\) Does this agree with your driving experience?

Step-by-Step Solution

Answer

(a) 148.98 ft; (b) 184.92 ft; (c) Braking is longer downhill, agreeing with experience.

Step 1: Convert Velocities to Feet per Second

Given velocities are in miles per hour (mph) and need to be converted to feet per second (ft/s). Use the conversion: 1 mph = 1.467 ft/s. For initial velocity ( V_{1}): 55 mph = 55 * 1.467 ft/s = 80.685 ft/s. For final velocity ( V_{2}): 30 mph = 30 * 1.467 ft/s = 44.01 ft/s.

Step 2: Calculate Distance for Uphill Slope ( θ=3.5°)

Use the braking distance formula for an uphill grade: \[D = \frac{1.05((80.685)^2 - (44.01)^2)}{64.4(0.4 + 0.02 + \sin(3.5^\circ))}\]Calculate \(\sin(3.5^\circ) \approx 0.061 \).Then substitute values into the formula: \[D = \frac{1.05((6510.892225) - (1936.4481))}{64.4(0.42 + 0.061)}= \frac{1.05(4574.444125)}{64.4(0.481)}\]\(D \approx 148.98\) ft.

Step 3: Calculate Distance for Downhill Slope ( θ=-2°)

Use the braking distance formula for a downhill grade: \[D = \frac{1.05((80.685)^2 - (44.01)^2)}{64.4(0.4 + 0.02 + \sin(-2^\circ))}\]Calculate \(\sin(-2^\circ) \approx -0.035\).Then substitute values into the formula:\[D = \frac{1.05(4574.444125)}{64.4(0.42 - 0.035)}= \frac{1.05(4574.444125)}{64.4(0.385)}\]\(D \approx 184.92\) ft.

Step 4: Analyze Effects and Reflect

The braking distance is longer downhill (\(184.92\) ft) compared to uphill (\(148.98\) ft). This makes sense because going downhill increases gravitational pull, making it harder to stop. This aligns with practical driving experience, where braking is smoother uphill due to gravity assisting with deceleration.

Key Concepts

Kinematic Equations

Kinematic equations are essential tools in the realm of physics, especially when analyzing motion-related problems, such as determining braking distances. These equations describe how objects move under the influence of various forces. For instance, when a car decelerates, we can use kinematic equations to understand the relationships between acceleration, velocity, distance, and time. In the braking distance scenario, the equation given allows us to calculate the necessary stoppage distance based on initial and final speeds. By using parameters specific to the vehicle's resistance and the road incline, we can predict how far a car will travel before coming to a stop. This formula highlights how different forces in motion can be mapped mathematically to assist with real-world driving situations. It serves as a practical application of kinematic principles.

Physics in Precalculus

Considering the simultaneous use of physics and precalculus elevates a student's understanding of mathematical relationships governing real-world phenomena. In the braking distance equation, we analyze various physical factors, such as velocity and incline, within a mathematical framework. This blending of concepts aids in solving complex real-life problems. Specifically, the problem involves trigonometric functions like sine, which help quantify to what degree inclines or declines influence a car's stopping distance. By mastering these techniques, students are equipped not only to solve textbook problems but also to apply fundamental physics principles in practical scenarios they encounter daily.

Velocity Conversion

Converting velocity units is a vital step in many physics problems to ensure consistent usage of units throughout calculations. In this exercise, we convert miles per hour (mph) to feet per second (ft/s) using the conversion factor where 1 mph is equivalent to 1.467 ft/s. This transformation is crucial since the formula for braking distance is derived based on feet per second. Without conversion, results would be incorrect or meaningless, demonstrating the importance of unit consistency in problem-solving. By practicing velocity conversions, students gain skills for accuracy in computations and communication in projects that involve different measurement systems.

Effects of Incline on Motion

Inclines dramatically influence an object's motion by altering its effective forces. When a car moves uphill, the gravitational force acts in opposition to the motion, assisting in deceleration, making stopping easier and faster. Conversely, moving downhill requires overcoming additional gravitational pull, lengthening the braking distance. This becomes evident in the exercise with braking distances calculated as shorter when ascending and longer when descending. Reflecting on this, such physics applications mirror everyday driving scenarios, underpinning the necessity for understanding how inclines affect vehicle dynamics. By grasping this concept, drivers are better prepared for safe navigation through varying road conditions.