Problem 29 - University Calculus: Early Transcendentals

Question

Based on data from the U.S. Bureau of Public Roads, a model for the total stopping distance of a moving car in terms of its speed is $$ s=1.1 v+0.054 v^{2} $$ where \(s\) is measured in \(\mathrm{ft}\) and \(v\) in mph. The linear term \(1.1 v\) models the distance the car travels during the time the driver perceives a need to stop until the brakes are applied, and the quadratic term \(0.054 v^{2}\) models the additional braking distance once they are applied. Find \(d s / d v\) at \(v=35\) and \(v=70 \mathrm{mph},\) and interpret the meaning of the derivative.

Step-by-Step Solution

Answer

At 35 mph, \(ds/dv\) is 4.88 ft/mph, and at 70 mph, it is 8.66 ft/mph. The derivative shows how stopping distance increases faster as speed increases.

Step 1: Understanding the Formula

The given formula for the stopping distance is \( s = 1.1v + 0.054v^2 \). This expresses the total stopping distance \( s \) as a function of speed \( v \). The first term, \( 1.1v \), represents the distance covered as the driver reacts, while the second term, \( 0.054v^2 \), accounts for the braking distance.

Step 2: Differentiate the Function with Respect to Speed

To find the rate of change of stopping distance with respect to speed, we differentiate the function \( s = 1.1v + 0.054v^2 \) with respect to \( v \). The derivative \( \frac{ds}{dv} \) is computed as follows:\[ \frac{d}{dv}(1.1v + 0.054v^2) = 1.1 + 2 \times 0.054v = 1.1 + 0.108v \].

Step 3: Evaluate the Derivative at v=35 mph

Substitute \( v = 35 \) into the derivative \( \frac{ds}{dv} = 1.1 + 0.108v \):\[ \frac{ds}{dv} \Big|_{v=35} = 1.1 + 0.108 \times 35 = 1.1 + 3.78 = 4.88 \].This means that at 35 mph, the stopping distance increases by approximately 4.88 feet for every additional mph.

Step 4: Evaluate the Derivative at v=70 mph

Substitute \( v = 70 \) into the derivative \( \frac{ds}{dv} = 1.1 + 0.108v \):\[ \frac{ds}{dv} \Big|_{v=70} = 1.1 + 0.108 \times 70 = 1.1 + 7.56 = 8.66 \].Thus, at 70 mph, the stopping distance increases by approximately 8.66 feet for every additional mph.

Step 5: Interpret the Derivative

The derivative \( \frac{ds}{dv} \) represents the rate of change of stopping distance with respect to speed. At 35 mph, the stopping distance changes at a rate of 4.88 feet per mph, and at 70 mph, it changes at 8.66 feet per mph. This indicates that as speed increases, the rate at which the stopping distance increases also rises, highlighting the impact of speed on stopping distance.

Key Concepts

Stopping Distance

Stopping distance refers to the total length a vehicle travels from the moment a driver perceives the need to stop until the vehicle comes to a complete halt. It's composed of two main components:

• Reaction Distance: This is the distance traveled during the driver's reaction time, from realizing the need to stop to actually applying the brakes. In our formula, this is modeled by the term \(1.1v\), where \(v\) is the speed in mph.
• Braking Distance: This accounts for the distance covered from the time the brakes are applied until the car stops completely. This is captured by the term \(0.054v^2\) in our equation, highlighting that braking distance increases with the square of the speed.

Understanding stopping distance is crucial for safe driving, as it varies significantly with speed.
The faster a car is moving, the longer the stopping distance, due to increased braking distance.

Speed

Speed, denoted as \(v\) in the formula, has a direct impact on both components of stopping distance: reaction and braking distances.
Here is why speed matters so much:

• Linear Relation to Reaction Distance: The reaction distance \(1.1v\) shows a direct linear relationship. This means if you double the speed, the reaction distance is also doubled.
• Quadratic Relation to Braking Distance: The braking distance's quadratic term \(0.054v^2\) suggests that accelerating has a more exponential effect on how far the vehicle will travel while braking. This area of the formula emphasizes the importance of reducing speed, especially in situations where stopping quickly is necessary.

It's essential for drivers to recognize how increasing speed can significantly impact the stopping distance, reinforcing the need for mindful driving speeds, especially in adverse conditions.

Differentiation

Differentiation is the mathematical process of finding a derivative, which measures how a function changes as its input changes. In this context, we differentiate the stopping distance equation \( s = 1.1v + 0.054v^2 \) to understand how the stopping distance \(s\) changes concerning speed \(v\).
To differentiate the function, we apply the rules of differentiation:

• The derivative of \(1.1v\) with respect to \(v\) is simply 1.1.
• For the term \(0.054v^2\), the derivative is calculated using the power rule, resulting in \(2 imes 0.054v = 0.108v\).

Thus, the derivative of the stopping distance with respect to speed, \( \frac{ds}{dv} \), becomes \(1.1 + 0.108v\).
This expression gives us a dynamic tool to calculate changes in stopping distance at any given speed, providing a way to assess how incremental speed increases affect stopping capability.

Interpretation of Derivative

The derivative \( \frac{ds}{dv} \) from our function reveals how stopping distance increases per additional unit of speed. Understanding the interpretation of this derivative is key in grasping the real-world implication:

• At 35 mph: The value \( \frac{ds}{dv} = 4.88 \) means the stopping distance increases by 4.88 feet for each additional mph.
• At 70 mph: A greater value of \( \frac{ds}{dv} = 8.66 \) indicates that for every extra mph increase, the stopping distance extends by 8.66 feet.

The interpretation shows the relationship isn't linear; instead, as speeds rise, the stopping distance grows faster, a crucial factor in traffic safety. Drivers must be aware that doubling the speed doesn't just double stopping distance, but more drastically impacts it, exacerbating safety risks in high-speed contexts.