91Ó°ÊÓ

Sports car: An automotive engineer is testing how quickly a newly designed sports car accelerates from rest. He has collected data giving the velocity, in miles per hour, as a function of the time, in seconds, since the car was at rest. Here is a table giving a portion of his data: \begin{tabular}{|c|c|} \hline Time & Velocity \\ \hline \(2.0\) & \(27.9\) \\ \hline \(2.5\) & \(33.8\) \\ \hline \(3.0\) & \(39.7\) \\ \hline \(3.5\) & \(45.6\) \\ \hline \end{tabular} a. By calculating differences, show that the data in this table can be modeled by a linear function. b. What is the slope for the linear function modeling velocity as a function of time? Explain in practical terms the meaning of the slope. c. Use the data in the table to find the formula for velocity as a linear function of time that is valid over the time period covered in the table. d. What would your formula from part \(\mathrm{c}\) give for the velocity of the car at time 0 ? What does this say about the validity of the linear formula over the initial segment of the experiment? Explain your answer in practical terms. e. Assume that the linear formula you found in part \(\mathrm{c}\) is valid from 2 seconds through 5 seconds. The marketing department wants to know from the engineer how to complete the following statement: "This car goes from 0 to \(60 \mathrm{mph}\) in seconds." How should they fill in the blank?

Step by step solution

01

Calculate Differences

We need to check whether the change in velocity is consistent. Calculate the differences between the velocities for each time interval. From 2.0 to 2.5 seconds: \[ 33.8 - 27.9 = 5.9 \]From 2.5 to 3.0 seconds: \[ 39.7 - 33.8 = 5.9 \]From 3.0 to 3.5 seconds: \[ 45.6 - 39.7 = 5.9 \]The differences are constant, indicating the data can be modeled by a linear function.

02

Calculate the Slope

The slope of the linear function is the rate of change of velocity with respect to time. Using any two points, let's calculate the slope using the formula: \[ \text{slope} = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1} \]Using the first two points, \[ \text{slope} = \frac{33.8 - 27.9}{2.5 - 2.0} = \frac{5.9}{0.5} = 11.8 \]The slope is 11.8 mph per second, meaning the car's speed increases by 11.8 mph every second.

03

Create the Linear Model

Using the slope and a point, find the linear formula for velocity as a function of time. Using the point \((2.0, 27.9)\), \[ v(t) = mt + b \]\[ 27.9 = 11.8 \times 2.0 + b \]Solve for \(b\):\[ 27.9 = 23.6 + b \]\[ b = 27.9 - 23.6 = 4.3 \]Thus, the linear model is \[ v(t) = 11.8t + 4.3 \]

04

Evaluate the Model at Zero

Use the linear model to find the velocity at time \(t = 0\):\[ v(0) = 11.8 \times 0 + 4.3 = 4.3 \text{ mph} \]The model suggests a velocity of 4.3 mph at \(t = 0\), which is incorrect since the car was at rest. This implies the linear model isn't valid when \(t < 2\) seconds.

05

Calculate Time for 0 to 60 mph

Assume the linear formula applies. Set \(v(t) = 60\) to solve for \(t\):\[ 60 = 11.8t + 4.3 \]\[ 60 - 4.3 = 11.8t \]\[ 55.7 = 11.8t \]\[ t = \frac{55.7}{11.8} \approx 4.72 \text{ seconds} \]The car accelerates to 60 mph in approximately 4.72 seconds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Slope Interpretation

When analyzing data, especially in physics, interpreting the slope of a linear function is essential. The slope tells us several things about the relationship between two quantities. In this exercise, the change in velocity over time is studied. Here, the slope represents how quickly velocity changes as time progresses.
The slope is calculated using the formula: \( \text{slope} = \frac{\Delta v}{\Delta t} \), where \( \Delta v \) is the change in velocity and \( \Delta t \) is the change in time. In practical terms, the slope is the rate of acceleration. For our sports car, the slope was found to be 11.8 mph per second, which means that every second the velocity of the car increases by 11.8 mph.

Understanding the slope helps us to grasp whether the car's acceleration is steady over the given time frame. This steady acceleration is a key characteristic when modeling using a linear function.

Velocity-Time Relation

The relationship between velocity and time can be expressed through linear equations when acceleration is constant. In this exercise, a linear model is used to describe how the car's velocity changes with time.
Linear functions take the form \( v(t) = mt + b \), where \( v(t) \) is the velocity at time t, \( m \) is the slope, and \( b \) is the y-intercept. The slope, already calculated, shows the rate of velocity increase. The y-intercept \( b \) is the theoretical velocity at \( t = 0 \).

• When \( t = 0 \), in a perfect linear world, it tells us the starting speed of the vehicle.
• However, the model indicated a velocity at rest time (t = 0) of 4.3 mph, which points out that a linear model isn't always fitting for every time range.

The linear relationship simplifies how engineers and scientists predict future states of an object’s motion from initial data. However, it is important to know when a model’s limitations occur, namely when stretching beyond its effective range.

Automotive Testing

Automotive testing involves capturing data such as speed and time under controlled conditions, which are then analyzed to enhance performance. In this scenario, testing the sports car aids in understanding its acceleration characteristics.

Data points are gathered during such tests to ensure the car behaves as expected, providing the manufacturer with invaluable insights into design effectiveness. It is crucial to:

• Understand how consistent acceleration can be modeled.
• Identify whether a simple model applies over various intervals.
• Predict how adjustments may affect car behavior.

Analyzing data to understand performance helps in safety tuning and features enhancement, ultimately improving the driving experience.
Although we used a linear model for predictions, the initial testing revealed limitations at \( t = 0 \), emphasizing not only the usefulness but also the discernment required in model application.

Linear Modeling

Linear modeling utilizes straight-line equations to approximate real-world scenarios. It is especially useful in automotive testing, where engineers simulate and predict outcomes based on controlled variables.
In this case, the equation \( v(t) = 11.8t + 4.3 \) models the car’s velocity. Linear models facilitate understanding because:

• They simplify complex systems into single equations.
• Offer clear visual interpretations through graphs.
• Allow for easier adjustments and recalibrations in the design process.

However, while linear models are powerful, they have constraints, such as the misinterpretation below the minimum data threshold (at time zero). Thus, using a model requires awareness of its applicability and boundaries.
Such understanding is pivotal not only for predicting outcomes but also for correcting discrepancies, ensuring reliability and accuracy during automotive development.