Physics worksheet focusing on displacement, velocity, and echo problems for high school students.

Problem 1: Calculate the total displacement of a mouse walking along a ruler.

#### Given:
- Initial position: \( x = 5 \, \text{cm} \)
- First movement: Walks to \( x = 12 \, \text{cm} \)
- Second movement: Walks a displacement of \( -8 \, \text{cm} \)
- Third movement: Walks to \( x = 7 \, \text{cm} \)

#### Solution:
1. First movement: The mouse walks from \( x = 5 \, \text{cm} \) to \( x = 12 \, \text{cm} \).
- Displacement: \( \Delta x_1 = 12 \, \text{cm} - 5 \, \text{cm} = 7 \, \text{cm} \)

2. Second movement: The mouse walks a displacement of \( -8 \, \text{cm} \).
- New position: \( x = 12 \, \text{cm} + (-8 \, \text{cm}) = 4 \, \text{cm} \)
- Displacement: \( \Delta x_2 = -8 \, \text{cm} \)

3. Third movement: The mouse walks to \( x = 7 \, \text{cm} \).
- Displacement: \( \Delta x_3 = 7 \, \text{cm} - 4 \, \text{cm} = 3 \, \text{cm} \)

4. Total displacement: The total displacement is the sum of all individual displacements.
\[
\Delta x_{\text{total}} = \Delta x_1 + \Delta x_2 + \Delta x_3 = 7 \, \text{cm} + (-8 \, \text{cm}) + 3 \, \text{cm} = 2 \, \text{cm}
\]

#### Final Answer:
\[
\boxed{2 \, \text{cm}}
\]

---

Problem 2: Find the average velocity of a bicyclist.

#### Given:
- Initial position: 150 meters north of town
- Final position: 1200 meters north of town
- Time taken: 30.0 minutes

#### Solution:
1. Displacement: The displacement is the change in position.
\[
\Delta x = \text{Final position} - \text{Initial position} = 1200 \, \text{m} - 150 \, \text{m} = 1050 \, \text{m}
\]

2. Time: Convert time from minutes to seconds.
\[
t = 30.0 \, \text{minutes} \times 60 \, \text{s/minute} = 1800 \, \text{s}
\]

3. Average velocity: The average velocity is given by:
\[
v_{\text{avg}} = \frac{\Delta x}{t} = \frac{1050 \, \text{m}}{1800 \, \text{s}} = 0.5833 \, \text{m/s}
\]

#### Final Answer:
\[
\boxed{0.583 \, \text{m/s}}
\]

---

Problem 3: Explain what is wrong with the statement.

#### Statement:
"A man walked at an average velocity of 5.2 m/s."

#### Explanation:
The statement is incomplete because velocity is a vector quantity, which means it has both magnitude and direction. The given statement only provides the magnitude (5.2 m/s) but does not specify the direction. Without the direction, the statement is incomplete and ambiguous.

#### Final Answer:
\[
\boxed{\text{The statement is incomplete because it does not specify the direction of the velocity.}}
\]

---

Problem 4: Calculate the displacement of the school bus.

#### Given:
- Time taken: 0.53 hours
- Average speed: 19 km/h

#### Solution:
1. Distance traveled: The distance traveled is given by:
\[
\text{Distance} = \text{Speed} \times \text{Time} = 19 \, \text{km/h} \times 0.53 \, \text{h} = 10.07 \, \text{km}
\]

2. Displacement: Since the problem does not mention any changes in direction, we assume the bus travels in a straight line. Therefore, the displacement is equal to the distance traveled.
\[
\text{Displacement} = 10.07 \, \text{km}
\]

#### Final Answer:
\[
\boxed{10.07 \, \text{km}}
\]

---

Problem 5: Explain how a girl can have an average velocity of 0 m/s during practice.

#### Given:
- A girl participates in cross-country practice and ends up at the same location as she started.
- Her average velocity is 0 m/s.

#### Explanation:
Average velocity is defined as the total displacement divided by the total time:
\[
v_{\text{avg}} = \frac{\Delta x}{t}
\]
If the girl starts and ends at the same location, her total displacement (\( \Delta x \)) is zero. Therefore, regardless of the time taken (\( t \)), the average velocity will be:
\[
v_{\text{avg}} = \frac{0}{t} = 0 \, \text{m/s}
\]

This situation is possible because displacement depends only on the initial and final positions, not on the path taken. Even if she runs around extensively during practice, as long as she returns to her starting point, her net displacement is zero.

#### Final Answer:
\[
\boxed{\text{The girl's average velocity is 0 m/s because her total displacement is zero (she ends up at the same location as she started).}}
\]

---

Problem 6: Calculate the time it takes for the hiker to hear the echo.

#### Given:
- Distance from the hiker to the canyon wall: 280.5 meters
- Speed of sound: 340 m/s

#### Solution:
1. Echo mechanism: When the hiker shouts, the sound travels to the canyon wall and reflects back to the hiker. The total distance traveled by the sound is twice the distance to the wall:
\[
\text{Total distance} = 2 \times 280.5 \, \text{m} = 561 \, \text{m}
\]

2. Time to hear the echo: The time taken for the sound to travel this distance is given by:
\[
t = \frac{\text{Total distance}}{\text{Speed of sound}} = \frac{561 \, \text{m}}{340 \, \text{m/s}} = 1.65 \, \text{s}
\]

#### Final Answer:
\[
\boxed{1.65 \, \text{s}}
\]

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Final Summary of Answers:

1. \(\boxed{2 \, \text{cm}}\)
2. \(\boxed{0.583 \, \text{m/s}}\)
3. \(\boxed{\text{The statement is incomplete because it does not specify the direction of the velocity.}}\)
4. \(\boxed{10.07 \, \text{km}}\)
5. \(\boxed{\text{The girl's average velocity is 0 m/s because her total displacement is zero (she ends up at the same location as she started).}}\)
6. \(\boxed{1.65 \, \text{s}}\)