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at 10:17 a.m., you pass a police car at 55mph that is stopped on the fr…

Question

at 10:17 a.m., you pass a police car at 55mph that is stopped on the freeway. you pass a second police car at 55mph at 10:53 a.m., which is located 39 miles from the first police car. if the speed limit is 70mph, can the police cite you for speeding?

find your average velocity from the 10:17 a.m. to 10:53 a.m.

mph
round value above to 2 decimal places.

by the mvt, since the average velocity over the interval was select an answer than the posted speed limit of 70mph the police select an answer site you for speeding.

Explanation:

Response

<pre_analysis>
{
"quality": "clear",
"question_count": 1,
"skills_matched": [
"step_cot",
"blank_filling_formatter",
"explain_then_answer",
"quick"
],
"needs_drawing": false,
"concepts_used": [
"Mean Value Theorem",
"Average Rate of Change"
],
"new_concepts": [],
"current_concepts": [
"Mean Value Theorem",
"Average Rate of Change"
]
}
</pre_analysis>

<reasoning>

Calculate the elapsed time in hours

\[

$$\begin{aligned} \Delta t &= 53\text{ min} - 17\text{ min} = 36\text{ min} \\ \Delta t &= \frac{36}{60}\text{ hours} = 0.6\text{ hours} \end{aligned}$$

\]

Calculate the average velocity

\[

$$\begin{aligned} v_{\text{avg}} &= \frac{\Delta s}{\Delta t} \\ v_{\text{avg}} &= \frac{39\text{ miles}}{0.6\text{ hours}} = 65\text{ mph} \end{aligned}$$

\]

Apply the Mean Value Theorem

\[

$$\begin{aligned} v_{\text{avg}} &= 65\text{ mph} < 70\text{ mph} \end{aligned}$$

\]
</reasoning>

<answer>
Find your average velocity from the 10:17 a.m. to 10:53 a.m.

<blank>65</blank> mph
Round value above to 2 decimal places.

By the MVT, since the average velocity over the interval was <blank>less</blank> than the posted speed limit of 70mph the police <blank>cannot</blank> site you for speeding.
</answer>

<post_analysis>
{
"subject": "Mathematics",
"question_type": "Fill-in-the-blank",
"knowledge_point": [
"Mathematics",
"Calculus",
"Mean Value Theorem"
]
}
</post_analysis>

Answer:

<pre_analysis>
{
"quality": "clear",
"question_count": 1,
"skills_matched": [
"step_cot",
"blank_filling_formatter",
"explain_then_answer",
"quick"
],
"needs_drawing": false,
"concepts_used": [
"Mean Value Theorem",
"Average Rate of Change"
],
"new_concepts": [],
"current_concepts": [
"Mean Value Theorem",
"Average Rate of Change"
]
}
</pre_analysis>

<reasoning>

Calculate the elapsed time in hours

\[

$$\begin{aligned} \Delta t &= 53\text{ min} - 17\text{ min} = 36\text{ min} \\ \Delta t &= \frac{36}{60}\text{ hours} = 0.6\text{ hours} \end{aligned}$$

\]

Calculate the average velocity

\[

$$\begin{aligned} v_{\text{avg}} &= \frac{\Delta s}{\Delta t} \\ v_{\text{avg}} &= \frac{39\text{ miles}}{0.6\text{ hours}} = 65\text{ mph} \end{aligned}$$

\]

Apply the Mean Value Theorem

\[

$$\begin{aligned} v_{\text{avg}} &= 65\text{ mph} < 70\text{ mph} \end{aligned}$$

\]
</reasoning>

<answer>
Find your average velocity from the 10:17 a.m. to 10:53 a.m.

<blank>65</blank> mph
Round value above to 2 decimal places.

By the MVT, since the average velocity over the interval was <blank>less</blank> than the posted speed limit of 70mph the police <blank>cannot</blank> site you for speeding.
</answer>

<post_analysis>
{
"subject": "Mathematics",
"question_type": "Fill-in-the-blank",
"knowledge_point": [
"Mathematics",
"Calculus",
"Mean Value Theorem"
]
}
</post_analysis>