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The Fastest (and Most Expensive) Car! The table shows test data for the Bugatti Veyron, the fastest car made. The car is moving in a straight line (the \(x\)-axis). \(\begin{array}{llll}{\text { Time }(s)} & {0} & {2.1} & {20.0} & {53} \\\ {\text { Speed (mijh) }} & {0} & {60} & {200} & {253}\end{array}\) (a) Make a \(v_{x}-t\) graph of this car's velocity \((\) in \(\mathrm{mi} / \mathrm{h})\) as a function of time. Is its acceleration constant? (b) Calculate the car's average acceleration ( in \(\mathrm{m} / \mathrm{s}^{2} )\) between ( i 0 and \(2.1 \mathrm{s} ;\) (ii) 2.1 \(\mathrm{s}\) and 20.0 \(\mathrm{s}\); (iii) 20.0 s and 53 s. Are these results consistent with your graph in part (a)? (Before you decide to buy this car, it might be helpful to know that only 300 will be built, it runs out of gas in 12 minutes at top speed, and it costs \(\$ 1.25\) million!)

Step by step solution

01

Plot the v-t Graph

To plot the graph of velocity (\(v_x\) in \(\text{mi/h}\)) vs time (\(t\) in seconds), use the given data points: (0, 0), (2.1, 60), (20, 200), and (53, 253). Place time on the x-axis and velocity on the y-axis, and connect the points with straight lines. Since acceleration is the slope of the \(v_x-t\) graph, if the graph were a straight line, the acceleration would be constant. Check for straight line segments to infer constant acceleration within any period.

02

Check Graph for Constant Acceleration

Upon examining the graph plotted in Step 1, observe the slopes of the straight-line segments. Different slopes in different segments imply that acceleration is not constant over the entire period but may be constant in smaller time intervals.

03

Calculate Average Acceleration from 0 to 2.1 s

Use the formula for average acceleration \(a_{avg} = \frac{\Delta v}{\Delta t}\), where \(\Delta v\) is the change in velocity and \(\Delta t\) is the change in time. From 0 to 2.1 s, \(\Delta v = 60 \text{ mi/h} - 0 \text{ mi/h} = 60 \text{ mi/h}\), and \(\Delta t = 2.1 \text{ s}\). Convert 60 mi/h to m/s: \(60 \times 0.44704 \approx 26.82 \text{ m/s}\). \(a_{avg} = \frac{26.82 \text{ m/s}}{2.1 \text{ s}} \approx 12.77 \text{ m/s}^2\).

04

Calculate Average Acceleration from 2.1 to 20 s

Apply the same formula: \(\Delta v = 200 \text{ mi/h} - 60 \text{ mi/h} = 140 \text{ mi/h}\), and \(\Delta t = 20 - 2.1 = 17.9 \text{ s}\). Convert 140 mi/h to m/s: \(140 \times 0.44704 \approx 62.5856 \text{ m/s}\). \(a_{avg} = \frac{62.5856 \text{ m/s}}{17.9 \text{ s}} \approx 3.50 \text{ m/s}^2\).

05

Calculate Average Acceleration from 20 to 53 s

Again use \(a_{avg} = \frac{\Delta v}{\Delta t}\): \(\Delta v = 253 \text{ mi/h} - 200 \text{ mi/h} = 53 \text{ mi/h}\), and \(\Delta t = 53 - 20 = 33 \text{ s}\). Convert 53 mi/h to m/s: \(53 \times 0.44704 \approx 23.6941 \text{ m/s}\). \(a_{avg} = \frac{23.6941 \text{ m/s}}{33 \text{ s}} \approx 0.72 \text{ m/s}^2\).

06

Compare Results with Graph

The average accelerations found in the various intervals (12.77, 3.50, 0.72 \(\text{ m/s}^2\)) indicate that the car's acceleration is not constant over the entire period, which matches the observation from the graph that shows varying slopes between points.

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Key Concepts

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

Velocity-Time Graph

A velocity-time graph is a crucial tool in understanding how an object's velocity changes over time. For this exercise, you have to plot the velocity (\(v_x\) in \(\text{mi/h}\)) of the Bugatti Veyron as a function of time (\(t\) in seconds). The time forms the \(x\)-axis, and the velocity forms the \(y\)-axis.

Using the data points \((0, 0), (2.1, 60), (20, 200),\) and \((53, 253)\), plot them on the graph. Connect these points with straight lines to see the trend in velocity changes.

• From 0 to 2.1 s, the car speeds up to 60 \(\text{mi/h}\)
• From 2.1 s to 20 s, the velocity increases sharply to 200 \(\text{mi/h}\)
• And from 20 s to 53 s, it gradually reaches the top speed of 253 \(\text{mi/h}\)

Each segment's slope on the graph indicates the car's acceleration rate within that period.

Average Acceleration

Average acceleration (\(a_{avg}\)) measures how the velocity changes with respect to time. To calculate, use the formula \(a_{avg} = \frac{\Delta v}{\Delta t}\) where \(\Delta v\) is the change in velocity, and \(\Delta t\) is the change in time.

For the Bugatti Veyron, these calculations were:

• From 0 to 2.1 s, the change in velocity was 60 \(\text{mi/h}\)-to-\(26.82\text{ m/s}\); providing an average acceleration of \(12.77 \, \text{m/s}^2\)
• From 2.1 s to 20 s, the velocity changed by 140 \(\text{mi/h}\)-to-\(62.59\text{ m/s}\); giving \(3.50 \, \text{m/s}^2\)
• From 20 s to 53 s, a smaller change of 53 \(\text{mi/h}\)-to-\(23.69\text{ m/s}\); yielding \(0.72 \, \text{m/s}^2\)

These values express how acceleration varied significantly through different time intervals.

Constant Acceleration

In physics, constant acceleration means the speed changes at a consistent rate over time. This results in a straight-line graph when plotting velocity against time. However, for the Bugatti Veyron during this test, acceleration is not constant across the entire period.

The varying slopes of the plotted \(v_x-t\) graph from the previous section indicate that acceleration changes in each segment. Between data points, the graph showed steep, moderate, and gentle slopes, confirming non-constant acceleration. Even though it may be constant within smaller intervals, the entire range considered shows inconsistent rates.

When the graph doesn't form a single straight line for the entire duration, the acceleration changes over time. This shows the dynamic nature of the car's performance during its test.

Motion in One Dimension

Motion in one dimension refers to movement along a single straight path, like the test track for the Bugatti Veyron. It's the simplest kind of motion, where only speed and direction along one axis are concerned.

This scenario reduces the complexities of real-world 3-dimensional motion. The car's journey, defined by only its speed between the start and end, depicts the heart of one-dimensional motion.

In this exercise, time stamps and speed changes were specifically charted along the \(x\)-axis. The straightforward, linear dimension provides a clear view of kinematic concepts, including velocity changes and acceleration along a singular path's entirety.

Understanding one-dimensional motion is crucial as it forms the basis for analyzing more complex movements in physics, such as those involving turns or curves.