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The problem you're working on involves a newspaper carrier who rides a bicycle and changes her speed over time. Let's break it down step by step to solve it:

### Problem Summary:
1. **Initial Speed**: The carrier starts biking at **10 km/h**.
2. **Slows Down**: She slows down to **5 km/h** over **7 seconds**.
3. **Maintains Speed**: She travels at **5 km/h** for **3 seconds**.
4. **Stops**: She slows to a stop over **6 seconds**.

### Let's address each part of the question:

#### a) **Graph of Speed over Time**:
The graph of speed over time would be a piecewise function with different slopes. Here's a general outline:

- **First 7 seconds**: The speed decreases linearly from **10 km/h** to **5 km/h**.
- **Next 3 seconds**: Speed remains constant at **5 km/h** (flat line).
- **Next 6 seconds**: The speed decreases linearly from **5 km/h** to **0 km/h**.

This would create a graph with:
1. A **downward sloping line** from 10 km/h to 5 km/h for the first 7 seconds.
2. A **horizontal line** at 5 km/h for the next 3 seconds.
3. Another **downward sloping line** from 5 km/h to 0 km/h for the final 6 seconds.

#### b) **Average Rate of Change in Speed Over the First 7 Seconds**:
The average rate of change in speed is calculated as the change in speed divided by the time interval.

- Initial speed = 10 km/h, Final speed = 5 km/h
- Time interval = 7 seconds

\[
\text{Average Rate of Change} = \frac{5 \text{ km/h} - 10 \text{ km/h}}{7 \text{ s}} = \frac{-5 \text{ km/h}}{7 \text{ s}} \approx -0.714 \text{ km/h per second}
\]

#### c) **Average Rate of Change in Speed from 7 to 12 Seconds**:
Between the 7th and 12th seconds, the speed decreases from **5 km/h** to **0 km/h** over a 5-second interval.

\[
\text{Average Rate of Change} = \frac{0 \text{ km/h} - 5 \text{ km/h}}{5 \text{ s}} = \frac{-5 \text{ km/h}}{5 \text{ s}} = -1 \text{ km/h per second}
\]

#### d) **Instantaneous Rate of Change at 12 seconds**:
At 12 seconds, the newspaper carrier is in the middle of decelerating from **5 km/h** to **0 km/h** over 6 seconds. The instantaneous rate of change in speed would be the slope of the linear part of the graph at that time. This is simply the rate of change in speed for the 6-second interval:

\[
\text{Instantaneous Rate of Change at 12 s} = -1 \text{ km/h per second}
\]

### Do you want a more detailed explanation on any part, or the graph drawn? Let me know!

Here are 5 related questions for further practice:
1. How would the graph change if the deceleration was not linear but exponential?
2. What would the total distance covered by the carrier during this time be?
3. How would you compute the total time if she decelerated to 0 in one phase instead of two?
4. How can you determine the carrier's acceleration in each phase?
5. If the bike speedometer had an error of ±0.5 km/h, how would that affect the calculations?

**Tip**: Always remember that the slope of a speed-time graph gives you the acceleration or deceleration rate!

A newspaper carrier delivers papers on her bicycle. She bikes to the first neighborhood at a rate of 10 km/h. She slows down at a constant rate over a period of 7 s, to a speed of 5 km/h, so that she can deliver her papers. After traveling at this rate for 3 s, she sees one of her customers and decides to stop. She slows at a constant rate until she stops. It takes her 6 s to stop. The problem includes parts: a) Graph of the carrier's speed over time. b) Average rate of change in speed over the first 7 s. c) Average rate of change in speed from second 7 to 12 seconds. d) Instantaneous rate of change in speed at 12 s.

Analyze a newspaper carrier's speed-time graph as she decelerates from 10 km/h to 5 km/h, stops, and the rate of change in speed over various intervals. Suitable for high school students, this problem teaches concepts like average and instantaneous rate of change, speed-time graphs, and linear deceleration.

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