Math Problem Statement

Select the correct answer from each drop-down menu. Peter walks 3 minutes at a constant rate and travels 330 meters. If we graph this relationship with time along the x-axis and distance along the y-axis, the slope of the line representing this relationship is ? .

If Peter walks for an hour, the point on the line that represents the distance he walks with respect to time is ? .

Solution

To find the slope of the line representing the relationship between time and distance for Peter's walk, we use the fact that he travels 330 meters in 3 minutes. The slope mm is the rate of travel in meters per minute.

m=change in distancechange in time=330 meters3 minutes=110 meters per minutem = \frac{\text{change in distance}}{\text{change in time}} = \frac{330 \text{ meters}}{3 \text{ minutes}} = 110 \text{ meters per minute}

So, the slope mm is 110110.

Next, we need to find the distance Peter walks in one hour (60 minutes). Using the slope, we can calculate this as follows:

Distance after 60 minutes=60×110=6600 meters\text{Distance after 60 minutes} = 60 \times 110 = 6600 \text{ meters}

So, the point on the line after 60 minutes is (60,6600)(60, 6600).

  • The slope of the line is 110110.
  • A point on this line after 60 minutes is (60,6600)(60, 6600).

Would you like more details or have any questions?

Here are some questions you might find interesting:

  1. How do you convert minutes into hours in rate calculations?
  2. What is the difference between speed and velocity?
  3. How can you use the slope to predict values for different times?
  4. What other real-world scenarios can be modeled with a linear relationship?
  5. How do you find the y-intercept in a distance-time graph?

Tip: Always double-check your unit conversions, especially when dealing with time and distance, to ensure your calculations are accurate.

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Math Problem Analysis

Mathematical Concepts

Linear Equations
Rate of Change
Unit Conversion

Formulas

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Theorems

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Suitable Grade Level

High School