Problem 32 - Calculus

Question

According to Car and Driver, an Alfa Romeo going at 70 mph requires 177 feet to stop. Assuming that the stopping distance is proportional to the square of velocity, find the stopping distances required by an Alfa Romeo going at 35 mph and at 140 mph (its top speed).

Step-by-Step Solution

Answer

At 35 mph, the stopping distance is 44.22 feet; at 140 mph, it is 707.56 feet.

Step 1: Understand the Proportional Relationship

The problem states that the stopping distance is proportional to the square of the velocity. Mathematically, this can be represented as \( d = k \cdot v^2 \), where \( d \) is the stopping distance, \( v \) is the velocity, and \( k \) is the proportionality constant.

Step 2: Find the Proportionality Constant

Given that at 70 mph, the stopping distance is 177 feet, we can use this information to find \( k \). Substitute the values into the equation: \( 177 = k \cdot 70^2 \). Solve for \( k \):\[ k = \frac{177}{70^2} = \frac{177}{4900} = 0.0361 \text{ ft/mph}^2 \]

Step 3: Calculate Stopping Distance at 35 mph

Substitute 35 mph and the value of \( k \) into the equation to find the stopping distance:\[ d_{35} = 0.0361 \cdot 35^2 = 0.0361 \cdot 1225 = 44.2225 \text{ feet} \]

Step 4: Calculate Stopping Distance at 140 mph

Substitute 140 mph and the value of \( k \) into the equation to find the stopping distance:\[ d_{140} = 0.0361 \cdot 140^2 = 0.0361 \cdot 19600 = 707.56 \text{ feet} \]

Key Concepts

Understanding Proportional Relationships

In the context of this exercise, we are exploring how some quantities relate to each other in a proportional manner. Proportional relationships occur when two quantities increase or decrease at the same rate. In simple terms, if one quantity doubles, the other quantity doubles as well.
For this problem, the stopping distance (\( d \)) of a car is proportional to the square of its velocity (\( v^2 \)). This means as the velocity of the car increases, the stopping distance does not just increase linearly but rather by the square of the velocity.
This relationship can be represented using a formula:

• \( d = k \cdot v^2 \)

Where \( k \) is the proportionality constant. The constant \( k \) helps us connect the velocity squared to the stopping distance in a meaningful way, helping us translate velocities into stopping distances. It's crucial to understand that this constant remains the same for similar situations, meaning once we calculate \( k \), we can use it for different velocities.

The Process of Stopping Distance Calculation

Stopping distance calculation is an essential aspect of many real-world applications, particularly in automotive safety. Calculating the stopping distance involves understanding the effect of velocity on the distance required to bring a vehicle to a stop.
The process begins by identifying the proportional relationship, where the stopping distance is multiplied by the square of the velocity, with a constant of proportionality \( k \). Once we have this constant from known data, like a stopping distance at a specific velocity, calculating stopping distances for other speeds is straightforward.

• First, find \( k \) using a given stopping distance and velocity, solve: \( k = \frac{d}{v^2} \).
• Next, apply \( k \) to calculate the stopping distance for any other velocity: \( d = k \cdot v^2 \).

It's important to highlight that using the square of the velocity in this calculation means that an increase in speed dramatically increases stopping distance. This principle underscores the significant impact speed has on safety, particularly in driving scenarios.

Exploring the Velocity and Distance Relationship

The connection between velocity and distance is central to understanding how a vehicle behaves when coming to a stop. With stopping distance proportional to the square of velocity, any increment in speed drastically amplifies the stopping distance needed.
This does not just illustrate the utility in predicting stopping distances but also emphasizes the exponential nature of speed's impact.

• A reduction in velocity from 70 mph to 35 mph doesn't just halve the stopping distance; it reduces it by a factor of four - thanks to the square relationship. As calculated: \( d_{35} = 44.22 \) feet, which is significantly smaller than \( d_{70} = 177 \) feet.
• Conversely, increasing the velocity from 70 mph to 140 mph quadruples the stopping distance to \( d_{140} = 707.56 \) feet.

Understanding this relationship helps in designing safer driving practices and vehicles, as it clarifies how much distance is truly needed when varying speeds are involved. This further emphasizes why speed limits and braking distances are crucial aspects of road safety.