Math Problem Statement

An orange falling from 28 feet has a height of s(t) = 28 − 16t2 feet when it has fallen for t seconds. Make a table to record the average rates that the orange is falling during the last second, half-second, quarter-second, and eighth-of-a-second of its fall. (Round your answers to three decimal places.)

Solution

The height of the orange as a function of time is given by:

s(t)=2816t2feets(t) = 28 - 16t^2 \, \text{feet}

To calculate the average rate of change of the height (which represents the average velocity) during the last second, half-second, quarter-second, and eighth-of-a-second of its fall, we need to compute the difference in height over the difference in time, i.e.,

Average Rate of Change (AROC)=s(t2)s(t1)t2t1\text{Average Rate of Change (AROC)} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}

The orange reaches the ground when s(t)=0s(t) = 0, so we first solve for tt by setting s(t)=0s(t) = 0:

2816t2=028 - 16t^2 = 0 t2=2816=1.75t^2 = \frac{28}{16} = 1.75 t=1.751.323secondst = \sqrt{1.75} \approx 1.323 \, \text{seconds}

Thus, the orange hits the ground at t1.323t \approx 1.323 seconds.

Now, we'll compute the average rate of change during different intervals before it hits the ground.

1. Last Second: [t=0.323 to t=1.323][t = 0.323 \text{ to } t = 1.323]

AROC=s(1.323)s(0.323)1.3230.323\text{AROC} = \frac{s(1.323) - s(0.323)}{1.323 - 0.323}

2. Last Half-Second: [t=0.823 to t=1.323][t = 0.823 \text{ to } t = 1.323]

AROC=s(1.323)s(0.823)1.3230.823\text{AROC} = \frac{s(1.323) - s(0.823)}{1.323 - 0.823}

3. Last Quarter-Second: [t=1.073 to t=1.323][t = 1.073 \text{ to } t = 1.323]

AROC=s(1.323)s(1.073)1.3231.073\text{AROC} = \frac{s(1.323) - s(1.073)}{1.323 - 1.073}

4. Last Eighth-of-a-Second: [t=1.198 to t=1.323][t = 1.198 \text{ to } t = 1.323]

AROC=s(1.323)s(1.198)1.3231.198\text{AROC} = \frac{s(1.323) - s(1.198)}{1.323 - 1.198}

Let's compute these values and round the answers to three decimal places.The average rates at which the orange is falling during the last second, half-second, quarter-second, and eighth-of-a-second of its fall are:

  1. Last second: 26.332feet per second-26.332 \, \text{feet per second}
  2. Last half-second: 34.332feet per second-34.332 \, \text{feet per second}
  3. Last quarter-second: 38.332feet per second-38.332 \, \text{feet per second}
  4. Last eighth-of-a-second: 40.332feet per second-40.332 \, \text{feet per second}

Would you like more details on these calculations or have any other questions?

Here are five follow-up questions:

  1. How would the rates change if the height function had different coefficients?
  2. What is the instantaneous rate of change at the time the orange hits the ground?
  3. How would the average rate of change vary with other fractions of a second?
  4. What is the maximum height the orange reaches before falling?
  5. How long would it take for the orange to hit the ground if dropped from a different height?

Tip: The average rate of change is an approximation of velocity, but as the time intervals get smaller, it approaches the instantaneous velocity!

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Math Problem Analysis

Mathematical Concepts

Calculus
Average Rate of Change
Quadratic Functions

Formulas

s(t) = 28 - 16t^2
Average Rate of Change (AROC) = (s(t2) - s(t1)) / (t2 - t1)

Theorems

Quadratic Equations
Rate of Change Theorem

Suitable Grade Level

Grades 11-12