(a) Sketching the Position Function Graph:

You are asked to sketch a graph of your position as a function of time based on the following events:

1. You leave your dorm at 8:30 am, and the time starts at $t = 0$.
2. You walk at a constant rate of 100 feet per minute for 10 minutes, meaning you walk $100 \times 10 = 1000$ feet.
3. After 10 minutes (at $t = 10$), you stop and chat for 5 minutes, during which your position stays constant.
4. At $t = 15$, you continue walking at the same pace until you arrive at class at 8:58 am. So, the total time you walk after chatting is $28 - 15 = 13$ minutes. During these 13 minutes, you walk $100 \times 13 = 1300$ feet.
5. Thus, your total displacement from the dorm is $1000 + 1300 = 2300$ feet.

The graph would look like this:

• X-axis (time $t$): Starts from $t = 0$ (8:30 am) and ends at $t = 28$ minutes (8:58 am).
• Y-axis (position): Starts from 0 feet (your dorm) and reaches 2300 feet (your class).
• The graph would have three sections:
  1. From $t = 0$ to $t = 10$, a straight line with slope 100 (indicating constant velocity).
  2. From $t = 10$ to $t = 15$, a horizontal line (position remains constant at 1000 feet).
  3. From $t = 15$ to $t = 28$, another straight line with slope 100 (constant velocity again).

(b) Analyzing the Graph:

Now, let's answer the following questions based on the graph:

i. What was your displacement over the time interval [0, 28]?

Displacement is the net change in position from the initial to the final point. You start at 0 feet (your dorm) and end at 2300 feet (your class). Therefore, your displacement is: $\text{Displacement} = 2300 - 0 = 2300 \text{ feet}.$

ii. What was your total distance traveled over the time interval [0, 28]?

Total distance traveled is the sum of the distances you walked, without considering direction. You walked:

• 1000 feet in the first 10 minutes,
• then 1300 feet after the 5-minute stop.

So, the total distance traveled is: $\text{Total distance} = 1000 + 1300 = 2300 \text{ feet}.$

iii. What was your average velocity over the time interval [0, 28]?

Average velocity is the total displacement divided by the total time: $\text{Average velocity} = \frac{\text{Displacement}}{\text{Time}} = \frac{2300 \text{ feet}}{28 \text{ minutes}} \approx 82.14 \text{ feet per minute}.$

iv. What was your average speed over the time interval [0, 28]?

Average speed is the total distance traveled divided by the total time: $\text{Average speed} = \frac{\text{Total distance}}{\text{Time}} = \frac{2300 \text{ feet}}{28 \text{ minutes}} \approx 82.14 \text{ feet per minute}.$ Note that in this case, the average speed and average velocity are the same because there is no change in direction.

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Would you like further clarification or details on any part of this explanation?

Here are five related questions to explore further:

1. What would change if you had walked slower after chatting?
2. How would the graph change if you took more breaks during the walk?
3. How would you find the instantaneous velocity at a specific point?
4. What if you sped up after chatting? How would that affect average speed and velocity?
5. How does average speed differ from instantaneous speed?

Tip: Average velocity depends on displacement, while average speed depends on total distance, so they can differ when direction changes occur!