Worksheet on displacement, velocity, and acceleration with diagrams and problems.

Problem Analysis:

The worksheet contains three main sections, each with specific tasks related to displacement, velocity, and acceleration. Let's solve each part step by step.

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#### Section 1: Paths A, B, and C
Question: Along which path would you have to move with the greatest speed to arrive at the destination in the same amount of time?

- Paths A, B, and C are different routes between two locations.
- The key here is that the time taken for all paths is the same, but the distances of the paths differ.
- Speed is defined as:
$$
\text{Speed} = \frac{\text{Distance}}{\text{Time}}
$$
Since the time is constant, the path with the longest distance will require the greatest speed.

Solution:
- Path C appears to be the longest among the three paths.
- Therefore, you would need to move with the greatest speed along Path C to cover its longer distance in the same amount of time as the other paths.

Answer: Path C

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#### Section 2: Running and Walking Trip
Given:
- Distance from your house to your friend's house: 3 miles.
- Time to run to your friend's house: 30 minutes.
- You immediately walk back home, taking 30 minutes on your return trip.

Questions:
1. What was the average speed (in mi/hr) for the entire trip?
2. What was the average velocity (in mi/hr) for the entire trip?

##### Part 2.1: Average Speed
- Definition of Average Speed: Total distance traveled divided by total time taken.
- Total Distance: You travel 3 miles to your friend's house and 3 miles back home, so:
$$
\text{Total Distance} = 3 \, \text{miles} + 3 \, \text{miles} = 6 \, \text{miles}
$$
- Total Time: You spend 30 minutes running and 30 minutes walking, so:
$$
\text{Total Time} = 30 \, \text{minutes} + 30 \, \text{minutes} = 60 \, \text{minutes} = 1 \, \text{hour}
$$
- Average Speed:
$$
\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{6 \, \text{miles}}{1 \, \text{hour}} = 6 \, \text{mi/hr}
$$

##### Part 2.2: Average Velocity
- Definition of Average Velocity: Displacement divided by total time taken.
- Displacement: Displacement is the net change in position. Since you start at your house, go to your friend's house, and return to your house, your net displacement is:
$$
\text{Displacement} = 0 \, \text{miles}
$$
- Average Velocity:
$$
\text{Average Velocity} = \frac{\text{Displacement}}{\text{Total Time}} = \frac{0 \, \text{miles}}{1 \, \text{hour}} = 0 \, \text{mi/hr}
$$

Answers:
1. Average Speed: $ \boxed{6 \, \text{mi/hr}} $
2. Average Velocity: $ \boxed{0 \, \text{mi/hr}} $

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#### Section 3: Cross-Country Skier
Given:
- The skier moves from location O to location D via intermediate points.
- Each leg of the journey takes 1 minute.
- Distances between points:
- O to A: 4 km
- A to B: 7 km
- B to C: 5 km
- C to D: 1 km
- Total time for the trip: 4 minutes.

Questions:
1. Calculate the average speed (in km/min) and the average velocity (in km/min) of the skier during the entire trip.
2. Calculate the average velocity (in km/min) of the second minute.

##### Part 3.1: Average Speed and Average Velocity for the Entire Trip

###### Step 1: Total Distance
- The total distance traveled is the sum of the distances between each pair of points:
$$
\text{Total Distance} = 4 \, \text{km} + 7 \, \text{km} + 5 \, \text{km} + 1 \, \text{km} = 17 \, \text{km}
$$

###### Step 2: Total Time
- The total time taken is 4 minutes.

###### Step 3: Average Speed
- Using the formula for average speed:
$$
\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{17 \, \text{km}}{4 \, \text{minutes}} = 4.25 \, \text{km/min}
$$

###### Step 4: Displacement
- Displacement is the straight-line distance from the starting point (O) to the ending point (D). From the diagram, the displacement is:
$$
\text{Displacement} = 17 \, \text{km} \quad (\text{since the path is a straight line in this case})
$$

###### Step 5: Average Velocity
- Using the formula for average velocity:
$$
\text{Average Velocity} = \frac{\text{Displacement}}{\text{Total Time}} = \frac{17 \, \text{km}}{4 \, \text{minutes}} = 4.25 \, \text{km/min}
$$

Answers:
1. Average Speed: $ \boxed{4.25 \, \text{km/min}} $
2. Average Velocity: $ \boxed{4.25 \, \text{km/min}} $

##### Part 3.2: Average Velocity During the Second Minute
- During the second minute, the skier moves from point A to point B.
- The displacement during this interval is the distance from A to B, which is:
$$
\text{Displacement} = 7 \, \text{km}
$$
- The time taken is 1 minute.
- Using the formula for average velocity:
$$
\text{Average Velocity} = \frac{\text{Displacement}}{\text{Time}} = \frac{7 \, \text{km}}{1 \, \text{minute}} = 7 \, \text{km/min}
$$

Answer:
$$
\boxed{7 \, \text{km/min}}
$$

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Final Answers:

1. Path C
2.
- Average Speed: $ \boxed{6 \, \text{mi/hr}} $
- Average Velocity: $ \boxed{0 \, \text{mi/hr}} $
3.
- Average Speed: $ \boxed{4.25 \, \text{km/min}} $
- Average Velocity: $ \boxed{4.25 \, \text{km/min}} $
- Average Velocity During the Second Minute: $ \boxed{7 \, \text{km/min}} $