Math Problem Statement

An emergency flare is shot into the air. Its height, in meters, above the ground at various times in its flight is given in the following table. Determine the average rate of change in the height of the flare during each interval. Use your results to explain what is happening to the height of the flare during each interval.

Solution

Let's work through the question provided in the image.

Problem Overview:

We have a table that shows the height of a flare over time. The question asks us to calculate the average rate of change in height of the flare during two intervals and then interpret what happens to the flare's height.

Table Summary:

  • Time (s): 0.0,0.5,1.0,1.5,2.0,2.5,3.0,3.5,4.00.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5, 4.0
  • Height (m): 2.00,15.75,27.00,35.75,42.00,45.75,47.00,45.75,42.002.00, 15.75, 27.00, 35.75, 42.00, 45.75, 47.00, 45.75, 42.00

Part (a) - Average Rate of Change

The average rate of change is given by the formula: Average Rate of Change=ΔHeightΔTime=H2H1t2t1\text{Average Rate of Change} = \frac{\Delta \text{Height}}{\Delta \text{Time}} = \frac{H_2 - H_1}{t_2 - t_1}

  1. Interval 1: 1.0t2.01.0 \leq t \leq 2.0

    • H1=27.00mH_1 = 27.00 \, \text{m} at t1=1.0st_1 = 1.0 \, \text{s}
    • H2=42.00mH_2 = 42.00 \, \text{m} at t2=2.0st_2 = 2.0 \, \text{s}

    Average Rate of Change=42.0027.002.01.0=15.001.0=15.00m/s\text{Average Rate of Change} = \frac{42.00 - 27.00}{2.0 - 1.0} = \frac{15.00}{1.0} = 15.00 \, \text{m/s}

  2. Interval 2: 3.0t4.03.0 \leq t \leq 4.0

    • H1=47.00mH_1 = 47.00 \, \text{m} at t1=3.0st_1 = 3.0 \, \text{s}
    • H2=42.00mH_2 = 42.00 \, \text{m} at t2=4.0st_2 = 4.0 \, \text{s}

    Average Rate of Change=42.0047.004.03.0=5.001.0=5.00m/s\text{Average Rate of Change} = \frac{42.00 - 47.00}{4.0 - 3.0} = \frac{-5.00}{1.0} = -5.00 \, \text{m/s}

Part (b) - Interpret the Results

  • Interval 1 ( 1.0t2.01.0 \leq t \leq 2.0 ): The positive average rate of change indicates that the flare is rising, with an upward velocity of 15 m/s.
  • Interval 2 ( 3.0t4.03.0 \leq t \leq 4.0 ): The negative rate of change shows that the flare is descending, at a rate of -5 m/s, meaning it is falling back toward the ground.

Do you need more details or have any questions?

Here are five related questions to deepen your understanding:

  1. How does the height change between t=2.5st = 2.5 \, \text{s} and t=3.5st = 3.5 \, \text{s}?
  2. What is the physical interpretation of the rate of change becoming negative?
  3. Can you calculate the total time the flare spends rising?
  4. How would you calculate the maximum height the flare reaches?
  5. What other methods could you use to estimate the flare's velocity?

Tip: The average rate of change gives a general sense of movement, but to fully understand the motion, it’s helpful to calculate rates for smaller intervals!

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Math Problem Analysis

Mathematical Concepts

Average Rate of Change
Linear Functions
Kinematics

Formulas

(Average Rate of Change) = (ΔHeight)/(ΔTime) = (H2 - H1)/(t2 - t1)

Theorems

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Suitable Grade Level

Grades 9-11