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if a rock is dropped from a height of 175 ft, its position \\(t\\) seco…

Question

if a rock is dropped from a height of 175 ft, its position \\(t\\) seconds after it is dropped until it hits the ground is given by the function \\(s(t) = -16t^2 + 175\\).

round values below to 3 decimal places.

how long does it take the rock to hit the ground?

seconds

find the average velocity of the rock from when it is released until when it hits the ground.

feet per second

what time after the rock is thrown will its instantaneous velocity be equal to its average velocity? (apply the mean value theorem)

seconds after it is thrown

Explanation:

Response

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"concepts_used": [
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"Average Rate of Change",
"Mean Value Theorem"
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<reasoning>

Find the time when the rock hits the ground

\[

$$\begin{aligned} s(t) &= -16t^2 + 175 = 0 \\ 16t^2 &= 175 \\ t^2 &= \frac{175}{16} = 10.9375 \\ t &= \sqrt{10.9375} \approx 3.307189 \approx 3.307 \text{ seconds} \end{aligned}$$

\]

Calculate the average velocity

\[

$$\begin{aligned} v_{\text{avg}} &= \frac{s(t_{\text{ground}}) - s(0)}{t_{\text{ground}} - 0} \\ v_{\text{avg}} &= \frac{0 - 175}{3.307189 - 0} \approx -52.91502 \approx -52.915 \text{ feet per second} \end{aligned}$$

\]

Apply the Mean Value Theorem to find the time of equal instantaneous velocity

\[

$$\begin{aligned} v(t) &= s'(t) = -32t \\ -32t &= v_{\text{avg}} \\ -32t &= -52.91502 \\ t &= \frac{-52.91502}{-32} \approx 1.65359 \approx 1.654 \text{ seconds} \end{aligned}$$

\]
</reasoning>

<answer>

Question 1

How long does it take the rock to hit the ground?
<blank>3.307</blank> seconds

Question 2

Find the average velocity of the rock from when it is released until when it hits the ground.
<blank>-52.915</blank> feet per second

Question 3

What time after the rock is thrown will its instantaneous velocity be equal to its average velocity? (Apply the Mean Value Theorem)
<blank>1.654</blank> seconds after it is thrown
</answer>

<post_analysis>
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"Mathematics",
"Calculus",
"Mean Value Theorem"
]
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Answer:

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"concepts_used": [
"Position Velocity Acceleration Relations",
"Average Rate of Change",
"Mean Value Theorem"
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"new_concepts": [],
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<reasoning>

Find the time when the rock hits the ground

\[

$$\begin{aligned} s(t) &= -16t^2 + 175 = 0 \\ 16t^2 &= 175 \\ t^2 &= \frac{175}{16} = 10.9375 \\ t &= \sqrt{10.9375} \approx 3.307189 \approx 3.307 \text{ seconds} \end{aligned}$$

\]

Calculate the average velocity

\[

$$\begin{aligned} v_{\text{avg}} &= \frac{s(t_{\text{ground}}) - s(0)}{t_{\text{ground}} - 0} \\ v_{\text{avg}} &= \frac{0 - 175}{3.307189 - 0} \approx -52.91502 \approx -52.915 \text{ feet per second} \end{aligned}$$

\]

Apply the Mean Value Theorem to find the time of equal instantaneous velocity

\[

$$\begin{aligned} v(t) &= s'(t) = -32t \\ -32t &= v_{\text{avg}} \\ -32t &= -52.91502 \\ t &= \frac{-52.91502}{-32} \approx 1.65359 \approx 1.654 \text{ seconds} \end{aligned}$$

\]
</reasoning>

<answer>

Question 1

How long does it take the rock to hit the ground?
<blank>3.307</blank> seconds

Question 2

Find the average velocity of the rock from when it is released until when it hits the ground.
<blank>-52.915</blank> feet per second

Question 3

What time after the rock is thrown will its instantaneous velocity be equal to its average velocity? (Apply the Mean Value Theorem)
<blank>1.654</blank> seconds after it is thrown
</answer>

<post_analysis>
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