Math Problem Statement

Watch the video in Blackboard and record the data provided. Use the data to complete the tables and graph both Car A and Car B on the total distance/total time graph and velocity/interval graph.

Solution

To help you complete this exercise, let's start by calculating the missing values for both cars in the data tables.

For Car A:

  1. Total Distance: This is cumulative, adding each distance to the previous total.
  2. Average Velocity: This is calculated by dividing the distance of each interval by the time for that interval.

Here's how we calculate each row for Car A:

  • Interval 1:

    • Total Distance: 4.724.72 meters
    • Average Velocity: 4.72 m3 s1.57 m/s\frac{4.72 \text{ m}}{3 \text{ s}} \approx 1.57 \text{ m/s}
  • Interval 2:

    • Total Distance: 4.72+3.99=8.714.72 + 3.99 = 8.71 meters
    • Average Velocity: 3.99 m3 s1.33 m/s\frac{3.99 \text{ m}}{3 \text{ s}} \approx 1.33 \text{ m/s}
  • Interval 3:

    • Total Distance: 8.71+4.72=13.438.71 + 4.72 = 13.43 meters
    • Average Velocity: 4.72 m3 s1.57 m/s\frac{4.72 \text{ m}}{3 \text{ s}} \approx 1.57 \text{ m/s}
  • Interval 4:

    • Total Distance: 13.43+3.96=17.3913.43 + 3.96 = 17.39 meters
    • Average Velocity: 3.96 m3 s1.32 m/s\frac{3.96 \text{ m}}{3 \text{ s}} \approx 1.32 \text{ m/s}
  • Interval 5:

    • Total Distance: 17.39+3.96=21.3517.39 + 3.96 = 21.35 meters
    • Average Velocity: 3.96 m3 s1.32 m/s\frac{3.96 \text{ m}}{3 \text{ s}} \approx 1.32 \text{ m/s}
  • Interval 6:

    • Total Distance: 21.35+4.17=25.5221.35 + 4.17 = 25.52 meters
    • Average Velocity: 4.17 m3 s1.39 m/s\frac{4.17 \text{ m}}{3 \text{ s}} \approx 1.39 \text{ m/s}

For Car B:

Similarly, calculate each row.

  • Interval 1:

    • Total Distance: 7.017.01 meters
    • Average Velocity: 7.01 m3 s2.34 m/s\frac{7.01 \text{ m}}{3 \text{ s}} \approx 2.34 \text{ m/s}
  • Interval 2:

    • Total Distance: 7.01+8.45=15.467.01 + 8.45 = 15.46 meters
    • Average Velocity: 8.45 m3 s2.82 m/s\frac{8.45 \text{ m}}{3 \text{ s}} \approx 2.82 \text{ m/s}
  • Interval 3:

    • Total Distance: 15.46+9.54=25.0015.46 + 9.54 = 25.00 meters
    • Average Velocity: 9.54 m3 s3.18 m/s\frac{9.54 \text{ m}}{3 \text{ s}} \approx 3.18 \text{ m/s}

Now that we have calculated the missing values, I’ll provide them in a structured format for both tables. Would you like me to present the completed tables and assist with plotting the graphs, or do you have any specific questions about these calculations?

Here are some questions to further your understanding:

  1. Why is it important to calculate the average velocity for each interval?
  2. How does cumulative total distance help in understanding the motion of each car?
  3. What does comparing total distances at the end of each interval reveal about the performance of Car A vs. Car B?
  4. How would you interpret differences in average velocities between the two cars?
  5. How might this data change if the interval times were different?

Tip: Always ensure units are consistent when calculating velocity, as it simplifies interpretation and reduces errors.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Distance
Velocity
Average Velocity
Graphing

Formulas

Total Distance = Sum of all distances at each interval
Average Velocity = Distance / Time

Theorems

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Suitable Grade Level

Grades 9-12