Question

1. a cyclist travels 45 km in 1 hour 15 minutes. 3 marks

(a) calculate average speed in km/h.
(b) convert your answer to m/s (3 s.f.).

1. a train travels at 90 km/h. how long does it take to travel 225 km? give your answer in hours and minutes. 2 marks

Explanation:

<pre_analysis>
{
"quality": "clear",
"question_count": 5,
"skills_matched": [
"step_cot",
"explain_then_answer",
"tutor",
"mastery_adaptive_brevity"
],
"needs_drawing": false,
"concepts_used": [
"Significant Figures",
"Metric Conversion",
"Derived Units"
],
"new_concepts": [
"Speed Calculations",
"Unit Conversion Rates"
],
"current_concepts": [
"Significant Figures",
"Metric Conversion",
"Derived Units",
"Speed Calculations",
"Unit Conversion Rates"
]
}
</pre_analysis>

<reasoning>

Analyze student's claim on significant figures

Using the Significant Figures knowledge point
\[

$$\begin{aligned} &\text{Number: } 4.50 \times 10^2\\ &\text{Significant digits in coefficient } 4.50: \{4, 5, 0\}\\ &\text{Count: } 3 \text{ significant figures} \end{aligned}$$

\]

Convert metric units for measurements

Using the Metric Conversion knowledge point
\[

$$\begin{aligned} &\text{(a) } 4.75 \text{ km} \times 1000 \text{ m/km} = 4750 \text{ m}\\ &\text{(b) } 350 \text{ mL} \div 1000 \text{ mL/L} = 0.35 \text{ L}\\ &\text{(c) } 2.3 \text{ kg} \times 1000 \text{ g/kg} = 2300 \text{ g}\\ &\text{(d) } 85000 \text{ cm}^2 \div 10000 \text{ cm}^2/\text{m}^2 = 8.5 \text{ m}^2 \end{aligned}$$

\]

Calculate average speed in km/h

Using the Derived Units knowledge point
\[

$$\begin{aligned} &d = 45 \text{ km}\\ &t = 1 \text{ h } 15 \text{ min} = 1.25 \text{ h}\\ &v = \frac{45 \text{ km}}{1.25 \text{ h}} = 36 \text{ km/h} \end{aligned}$$

\]

Convert speed to m/s with 3 s.f.

Using the Derived Units and Significant Figures knowledge points
\[

$$\begin{aligned} &v = 36 \text{ km/h} = \frac{36 \times 1000 \text{ m}}{3600 \text{ s}} = 10 \text{ m/s}\\ &\text{To 3 significant figures: } 10.0 \text{ m/s} \end{aligned}$$

\]

Calculate travel time in hours and minutes

We find the time taken for the train journey.
\[

$$\begin{aligned} &d = 225 \text{ km},\quad v = 90 \text{ km/h}\\ &t = \frac{225 \text{ km}}{90 \text{ km/h}} = 2.5 \text{ h}\\ &2.5 \text{ h} = 2 \text{ hours and } 30 \text{ minutes} \end{aligned}$$

\]
</reasoning>

<answer>

Question 3

No, the student is incorrect. In scientific notation, all digits in the coefficient are significant. The coefficient \(4.50\) has three significant figures (the \(4\), the \(5\), and the trailing zero after the decimal point). Therefore, \(4.50 \times 10^2\) has 3 significant figures.

Question 4

(a) \(4.75\text{ km} = 4750\text{ m}\)
(b) \(350\text{ mL} = 0.35\text{ L}\)
(c) \(2.3\text{ kg} = 2300\text{ g}\)
(d) \(85\,000\text{ cm}^2 = 8.5\text{ m}^2\)

Question 8

(a) \(36\text{ km/h}\)
(b) \(10.0\text{ m/s}\)

Question 9

2 hours and 30 minutes
</answer>

<post_analysis>
{
"subject": "Mathematics",
"question_type": "Multi-part",
"knowledge_point": [
"Mathematics",
"Algebra",
"Derived Units"
]
}
</post_analysis>

Answer:

<pre_analysis>
{
"quality": "clear",
"question_count": 5,
"skills_matched": [
"step_cot",
"explain_then_answer",
"tutor",
"mastery_adaptive_brevity"
],
"needs_drawing": false,
"concepts_used": [
"Significant Figures",
"Metric Conversion",
"Derived Units"
],
"new_concepts": [
"Speed Calculations",
"Unit Conversion Rates"
],
"current_concepts": [
"Significant Figures",
"Metric Conversion",
"Derived Units",
"Speed Calculations",
"Unit Conversion Rates"
]
}
</pre_analysis>

<reasoning>

Analyze student's claim on significant figures

Using the Significant Figures knowledge point
\[

$$\begin{aligned} &\text{Number: } 4.50 \times 10^2\\ &\text{Significant digits in coefficient } 4.50: \{4, 5, 0\}\\ &\text{Count: } 3 \text{ significant figures} \end{aligned}$$

\]

Convert metric units for measurements

Using the Metric Conversion knowledge point
\[

$$\begin{aligned} &\text{(a) } 4.75 \text{ km} \times 1000 \text{ m/km} = 4750 \text{ m}\\ &\text{(b) } 350 \text{ mL} \div 1000 \text{ mL/L} = 0.35 \text{ L}\\ &\text{(c) } 2.3 \text{ kg} \times 1000 \text{ g/kg} = 2300 \text{ g}\\ &\text{(d) } 85000 \text{ cm}^2 \div 10000 \text{ cm}^2/\text{m}^2 = 8.5 \text{ m}^2 \end{aligned}$$

\]

Calculate average speed in km/h

Using the Derived Units knowledge point
\[

$$\begin{aligned} &d = 45 \text{ km}\\ &t = 1 \text{ h } 15 \text{ min} = 1.25 \text{ h}\\ &v = \frac{45 \text{ km}}{1.25 \text{ h}} = 36 \text{ km/h} \end{aligned}$$

\]

Convert speed to m/s with 3 s.f.

Using the Derived Units and Significant Figures knowledge points
\[

$$\begin{aligned} &v = 36 \text{ km/h} = \frac{36 \times 1000 \text{ m}}{3600 \text{ s}} = 10 \text{ m/s}\\ &\text{To 3 significant figures: } 10.0 \text{ m/s} \end{aligned}$$

\]

Calculate travel time in hours and minutes

We find the time taken for the train journey.
\[

$$\begin{aligned} &d = 225 \text{ km},\quad v = 90 \text{ km/h}\\ &t = \frac{225 \text{ km}}{90 \text{ km/h}} = 2.5 \text{ h}\\ &2.5 \text{ h} = 2 \text{ hours and } 30 \text{ minutes} \end{aligned}$$

\]
</reasoning>

<answer>

Question 3

No, the student is incorrect. In scientific notation, all digits in the coefficient are significant. The coefficient \(4.50\) has three significant figures (the \(4\), the \(5\), and the trailing zero after the decimal point). Therefore, \(4.50 \times 10^2\) has 3 significant figures.

Question 4

(a) \(4.75\text{ km} = 4750\text{ m}\)
(b) \(350\text{ mL} = 0.35\text{ L}\)
(c) \(2.3\text{ kg} = 2300\text{ g}\)
(d) \(85\,000\text{ cm}^2 = 8.5\text{ m}^2\)

Question 8

(a) \(36\text{ km/h}\)
(b) \(10.0\text{ m/s}\)

Question 9

2 hours and 30 minutes
</answer>

<post_analysis>
{
"subject": "Mathematics",
"question_type": "Multi-part",
"knowledge_point": [
"Mathematics",
"Algebra",
"Derived Units"
]
}
</post_analysis>