Question

a model rocket was launched from the ground. the height of the rocket was recorded as a function of time.

\

$$\begin{array}{|c|c|} \\hline \\text{time (seconds)} & \\text{height (feet)} \\\\ \\hline 0 & 0 \\\\ \\hline 2 & 10 \\\\ \\hline 4 & 15 \\\\ \\hline 6 & 17 \\\\ \\hline 8 & 15 \\\\ \\hline \\end{array}$$

find the average rate of change in the height between 0 and 6 seconds. round the answer to the nearest tenth.

• 17 feet per second
• 17.6 feet per second
• 2.8 feet per second
• 6 feet per second

Explanation:

Identify the values from the table

$$
LATEXBLOCK0
$$

Apply the average rate of change formula

$$ \text{Average Rate of Change} = \frac{h(6) - h(0)}{6 - 0} = \frac{17 - 0}{6 - 0} = \frac{17}{6} $$

Calculate and round the final value

$$ \frac{17}{6} \approx 2.8333 \approx 2.8\text{ feet per second} $$

Answer:

• (A) 17 feet per second
• (B) 17.6 feet per second
• (C) 2.8 feet per second (Correct answer)
• (D) 6 feet per second