Analyzing a velocity-time graph to determine motion characteristics such as direction, acceleration, and distance traveled.

Problem Analysis

The task involves analyzing a velocity-time graph to answer questions about motion, acceleration, and distance traveled. Let's break down the solution step by step.

#### Graph Analysis
The velocity-time graph shows the velocity of an object as a function of time. Key points to observe:
1. Velocity: Positive values indicate motion in the positive direction, and negative values indicate motion in the negative direction.
2. Acceleration: The slope of the velocity-time graph represents acceleration.
- A positive slope indicates positive acceleration.
- A negative slope indicates negative acceleration.
- A zero slope indicates constant velocity (no acceleration).

#### Answering Each Question

---

a. Uniform motion in the positive direction?

- Uniform motion means constant velocity (zero acceleration).
- From the graph, the segment where the velocity is constant and positive is between 30 minutes and 40 minutes.
- Answer: [30, 40]

---

b. Uniform motion in the negative direction?

- Uniform motion in the negative direction means constant negative velocity.
- From the graph, there is no segment where the velocity is constant and negative.
- Answer: None

---

c. Rest?

- Rest means zero velocity.
- From the graph, the velocity is zero at 40 minutes.
- Answer: [40]

---

d. Positive acceleration with positive velocity?

- Positive acceleration means a positive slope, and positive velocity means the velocity is above the zero line.
- From the graph, this occurs between 0 minutes and 10 minutes.
- Answer: [0, 10]

---

e. Negative velocity with positive acceleration?

- Negative velocity means the velocity is below the zero line.
- Positive acceleration means a positive slope.
- From the graph, this occurs between 40 minutes and 50 minutes.
- Answer: [40, 50]

---

f. Negative acceleration with positive velocity?

- Negative acceleration means a negative slope.
- Positive velocity means the velocity is above the zero line.
- From the graph, this occurs between 10 minutes and 30 minutes.
- Answer: [10, 30]

---

g. Negative velocity with negative acceleration?

- Negative velocity means the velocity is below the zero line.
- Negative acceleration means a negative slope.
- From the graph, there is no segment where both conditions are met.
- Answer: None

---

h. Zero velocity and negative acceleration?

- Zero velocity means the velocity is at the zero line.
- Negative acceleration means a negative slope.
- From the graph, this occurs at 40 minutes (the point where the velocity crosses zero and the slope is negative).
- Answer: [40]

---

i. Positive acceleration and zero velocity?

- Positive acceleration means a positive slope.
- Zero velocity means the velocity is at the zero line.
- From the graph, this occurs at 0 minutes (the point where the velocity starts from zero and the slope is positive).
- Answer: [0]

---

j. When does the person change direction?

- A change in direction occurs when the velocity crosses the zero line.
- From the graph, this happens at 40 minutes.
- Answer: [40]

---

k. What is the distance traveled from [0, 10]?

- Distance is the area under the velocity-time graph, considering only the magnitude (absolute value) of the velocity.
- From 0 to 10 minutes, the velocity is positive and increasing linearly.
- The area under the curve is a triangle:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 60 = 300 \, \text{m}
\]
- Answer: 300 m

---

l. What is the distance traveled from [0, 15]?

- From 0 to 10 minutes, the distance is 300 m (as calculated above).
- From 10 to 15 minutes, the velocity is constant at 60 m/min.
- The distance traveled in this interval is:
\[
\text{Distance} = \text{velocity} \times \text{time} = 60 \times 5 = 300 \, \text{m}
\]
- Total distance from 0 to 15 minutes:
\[
300 + 300 = 600 \, \text{m}
\]
- Answer: 600 m

---

m. What is the distance traveled from [30, 55]?

- From 30 to 40 minutes, the velocity is decreasing linearly from 20 m/min to 0 m/min.
- The area under the curve is a triangle:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 20 = 100 \, \text{m}
\]
- From 40 to 50 minutes, the velocity is negative and increasing linearly from 0 m/min to -40 m/min.
- The area under the curve is a triangle (distance is the absolute value):
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 40 = 200 \, \text{m}
\]
- Total distance from 30 to 55 minutes:
\[
100 + 200 = 300 \, \text{m}
\]
- Answer: 300 m

---

Final Answers

\[
\boxed{
\begin{array}{ll}
\text{a.} & [30, 40] \\
\text{b.} & \text{None} \\
\text{c.} & [40] \\
\text{d.} & [0, 10] \\
\text{e.} & [40, 50] \\
\text{f.} & [10, 30] \\
\text{g.} & \text{None} \\
\text{h.} & [40] \\
\text{i.} & [0] \\
\text{j.} & [40] \\
\text{k.} & 300 \, \text{m} \\
\text{l.} & 600 \, \text{m} \\
\text{m.} & 300 \, \text{m} \\
\end{array}
}
\]