Question

the table shows the distance that two trains are from center station at various times of the day.

how do the speeds of the trains compare?

train 1 is traveling 1 mile per hour faster than train 2.
train 1 is traveling 6 miles per hour faster than train 2.
train 2 is traveling 4 miles per hour faster than train 1.
train 2 is traveling 5 miles per hour faster than train 1.

Explanation:

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<reasoning>

Calculate elapsed times

Using the Rate of Change knowledge point
\[

$$\begin{aligned} &\Delta t_1 = 12:00 - 11:30 = 30\text{ minutes} = 0.5\text{ hours}\\ &\Delta t_2 = 12:45 - 12:00 = 45\text{ minutes} = 0.75\text{ hours} \end{aligned}$$

\]

Calculate Train 1 speed

Using the Rate of Change knowledge point
\[

$$\begin{aligned} &v_{1,a} = \frac{31 - 13}{0.5} = \frac{18}{0.5} = 36\text{ mph}\\ &v_{1,b} = \frac{58 - 31}{0.75} = \frac{27}{0.75} = 36\text{ mph} \end{aligned}$$

\]

Calculate Train 2 speed

Using the Rate of Change knowledge point
\[

$$\begin{aligned} &v_{2,a} = \frac{27 - 7}{0.5} = \frac{20}{0.5} = 40\text{ mph}\\ &v_{2,b} = \frac{57 - 27}{0.75} = \frac{30}{0.75} = 40\text{ mph} \end{aligned}$$

\]

Compare the speeds

We compare the constant speeds of both trains:

• Train 1 speed: \(36\text{ mph}\)
• Train 2 speed: \(40\text{ mph}\)

Difference:
\[
40 - 36 = 4\text{ mph}
\]
Thus, Train 2 is traveling 4 miles per hour faster than Train 1.
</reasoning>

<answer>
<mcq-option>Train 1 is traveling 1 mile per hour faster than train 2.</mcq-option>
<mcq-option>Train 1 is traveling 6 miles per hour faster than train 2.</mcq-option>
<mcq-correct>Train 2 is traveling 4 miles per hour faster than train 1.</mcq-correct>
<mcq-option>Train 2 is traveling 5 miles per hour faster than train 1.</mcq-option>
</answer>

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]
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</post_analysis>

Answer:

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<reasoning>

Calculate elapsed times

Using the Rate of Change knowledge point
\[

$$\begin{aligned} &\Delta t_1 = 12:00 - 11:30 = 30\text{ minutes} = 0.5\text{ hours}\\ &\Delta t_2 = 12:45 - 12:00 = 45\text{ minutes} = 0.75\text{ hours} \end{aligned}$$

\]

Calculate Train 1 speed

Using the Rate of Change knowledge point
\[

$$\begin{aligned} &v_{1,a} = \frac{31 - 13}{0.5} = \frac{18}{0.5} = 36\text{ mph}\\ &v_{1,b} = \frac{58 - 31}{0.75} = \frac{27}{0.75} = 36\text{ mph} \end{aligned}$$

\]

Calculate Train 2 speed

Using the Rate of Change knowledge point
\[

$$\begin{aligned} &v_{2,a} = \frac{27 - 7}{0.5} = \frac{20}{0.5} = 40\text{ mph}\\ &v_{2,b} = \frac{57 - 27}{0.75} = \frac{30}{0.75} = 40\text{ mph} \end{aligned}$$

\]

Compare the speeds

We compare the constant speeds of both trains:

• Train 1 speed: \(36\text{ mph}\)
• Train 2 speed: \(40\text{ mph}\)

Difference:
\[
40 - 36 = 4\text{ mph}
\]
Thus, Train 2 is traveling 4 miles per hour faster than Train 1.
</reasoning>

<answer>
<mcq-option>Train 1 is traveling 1 mile per hour faster than train 2.</mcq-option>
<mcq-option>Train 1 is traveling 6 miles per hour faster than train 2.</mcq-option>
<mcq-correct>Train 2 is traveling 4 miles per hour faster than train 1.</mcq-correct>
<mcq-option>Train 2 is traveling 5 miles per hour faster than train 1.</mcq-option>
</answer>

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