AHS Physics worksheet titled "Displacement and Velocity" featuring six problems related to physics concepts, including calculations and explanations.
Displacement and Velocity Worksheet for AHS Physics with problems and answers.
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Step-by-step solution for: Displacement and Velocity Worksheet - AHS Physics Name: Date ...
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Show Answer Key & Explanations
Step-by-step solution for: Displacement and Velocity Worksheet - AHS Physics Name: Date ...
Let’s solve each problem step by step, showing all work and explaining the physics concepts involved.
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Problem 1: Calculate the total displacement of a mouse walking along a ruler...
> Begins at x = 5 cm
> - Walks to x = 12 cm → displacement = +7 cm
> - Then walks a displacement of -8 cm → new position = 12 + (-8) = 4 cm
> - Lastly, walks to x = 7 cm → displacement from 4 cm to 7 cm = +3 cm
✔ Total displacement is defined as final position minus initial position:
> Final position = 7 cm
> Initial position = 5 cm
> Displacement D = 7 cm – 5 cm = 2 cm
✔️ Answer: D = 2 cm
*(Note: The red answer on the worksheet is correct.)*
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Problem 2: Find the average velocity (in m/s) of a bicycler that starts 150 meters north of town and is 1200 meters north of town after 30.0 minutes.
Average velocity = total displacement / total time
- Displacement = final position – initial position = 1200 m – 150 m = 1050 m north
- Time = 30.0 minutes = 30 × 60 = 1800 seconds
> Average velocity = 1050 m / 1800 s = 0.583 m/s north
⚠️ But the worksheet says “35 m/min North” — let’s check that:
> 1050 m / 30 min = 35 m/min North → This is also correct, but it’s in m/min, not m/s as requested.
The question specifically asks for m/s.
✔ So the correct answer in m/s is:
> 0.583 m/s North (or approximately 0.58 m/s North)
*(The worksheet answer is technically correct in units of m/min, but violates the unit request of m/s.)*
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Problem 3: Explain what is wrong with the following statement: A man walked at an average velocity of 5.2 m/s.
Velocity is a vector quantity — it must include both magnitude AND direction.
The statement gives only magnitude (5.2 m/s) and no direction (e.g., north, east, etc.).
Also, “walked” implies motion over time, but without specifying direction, it’s incomplete as a vector.
✔ Correct explanation:
> Velocity is a vector and must include direction. The statement only gives speed (magnitude), not velocity. Additionally, while “walked” implies motion, velocity requires directional information to be complete.
*(Worksheet answer: “It says velocity but it doesn’t have a direction and the action is not present.” — partially correct; “action is not present” is unclear — better to say “direction is missing.”)*
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Problem 4: A school bus takes 0.53 hours to reach the school from your house. If the average speed of the bus is 19 km/h, what is the displacement of the bus during the trip?
This is tricky — speed is scalar, displacement is vector.
We are given average speed and time, so we can find distance traveled:
> Distance = speed × time = 19 km/h × 0.53 h = 10.07 km
But displacement is the straight-line distance from start to end point.
❗ Unless told the bus traveled in a straight line, we cannot assume displacement equals distance.
However, the worksheet answer assumes straight-line motion: “if it traveled in a straight line”.
✔ So, if we assume straight-line motion (as implied by the worksheet’s answer), then:
> Displacement = 10.07 km ≈ 10.1 km (from house to school)
But strictly speaking, without that assumption, we cannot determine displacement from speed alone.
*(Worksheet answer: “The displacement of the bus during the trip is 9.5km from the house if it traveled in a straight line” — this is incorrect mathematically. 19 × 0.53 = 10.07, not 9.5.)*
✘ Correction:
> Displacement = 19 km/h × 0.53 h = 10.07 km (≈ 10.1 km) if traveling in a straight line
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Problem 5: A girl participating in cross-country spends the afternoon practicing, and ends the practice completely tired from her hard work, despite the fact that her average velocity during the practice was 0.0 m/s. Explain how this situation is possible.
Average velocity = total displacement / total time
If she ended at the same place she started (e.g., ran loops or back-and-forth routes), then displacement = 0, so average velocity = 0.
But she still covered a lot of distance, so she did a lot of work (energy expenditure) — hence, tired.
✔ Explanation:
> Average velocity depends only on displacement (change in position). If she returned to her starting point, displacement is zero, so average velocity is zero — even though she ran many kilometers and exerted herself physically.
*(Worksheet answer is correct: “she started and ended practice at the same spot.”)*
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Problem 6: A hiker is at the bottom of a canyon facing the canyon wall closest to her. She is 280.5 meters from the wall and the sound of her voice travels at 340 m/s at that location. How long after she shouts will she hear her echo?
Echo means sound travels to the wall AND back to her.
So total distance sound travels = 2 × 280.5 m = 561 meters
Speed of sound = 340 m/s
Time = distance / speed = 561 m / 340 m/s = ?
> 561 ÷ 340 = 1.65 seconds
✔ Answer: 1.65 seconds
*(Note: Worksheet has no answer written here — this is the correct solution.)*
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## ✔ FINAL ANSWERS SUMMARY:
1. D = 2 cm ✔️ (correct)
2. 0.583 m/s North *(worksheet answer is in wrong units — should be m/s, not m/min)*
3. Velocity requires direction — statement omits it. *(worksheet answer slightly unclear — improve wording)*
4. Displacement = 10.07 km (if straight line) — *worksheet answer 9.5 km is mathematically wrong*
5. She returned to start → displacement = 0 → avg velocity = 0, but distance ≠ 0 → she got tired. ✔️
6. t = 1.65 seconds — *not answered on worksheet*
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Let me know if you’d like these formatted into a clean worksheet answer key!
---
Problem 1: Calculate the total displacement of a mouse walking along a ruler...
> Begins at x = 5 cm
> - Walks to x = 12 cm → displacement = +7 cm
> - Then walks a displacement of -8 cm → new position = 12 + (-8) = 4 cm
> - Lastly, walks to x = 7 cm → displacement from 4 cm to 7 cm = +3 cm
✔ Total displacement is defined as final position minus initial position:
> Final position = 7 cm
> Initial position = 5 cm
> Displacement D = 7 cm – 5 cm = 2 cm
✔️ Answer: D = 2 cm
*(Note: The red answer on the worksheet is correct.)*
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Problem 2: Find the average velocity (in m/s) of a bicycler that starts 150 meters north of town and is 1200 meters north of town after 30.0 minutes.
Average velocity = total displacement / total time
- Displacement = final position – initial position = 1200 m – 150 m = 1050 m north
- Time = 30.0 minutes = 30 × 60 = 1800 seconds
> Average velocity = 1050 m / 1800 s = 0.583 m/s north
⚠️ But the worksheet says “35 m/min North” — let’s check that:
> 1050 m / 30 min = 35 m/min North → This is also correct, but it’s in m/min, not m/s as requested.
The question specifically asks for m/s.
✔ So the correct answer in m/s is:
> 0.583 m/s North (or approximately 0.58 m/s North)
*(The worksheet answer is technically correct in units of m/min, but violates the unit request of m/s.)*
---
Problem 3: Explain what is wrong with the following statement: A man walked at an average velocity of 5.2 m/s.
Velocity is a vector quantity — it must include both magnitude AND direction.
The statement gives only magnitude (5.2 m/s) and no direction (e.g., north, east, etc.).
Also, “walked” implies motion over time, but without specifying direction, it’s incomplete as a vector.
✔ Correct explanation:
> Velocity is a vector and must include direction. The statement only gives speed (magnitude), not velocity. Additionally, while “walked” implies motion, velocity requires directional information to be complete.
*(Worksheet answer: “It says velocity but it doesn’t have a direction and the action is not present.” — partially correct; “action is not present” is unclear — better to say “direction is missing.”)*
---
Problem 4: A school bus takes 0.53 hours to reach the school from your house. If the average speed of the bus is 19 km/h, what is the displacement of the bus during the trip?
This is tricky — speed is scalar, displacement is vector.
We are given average speed and time, so we can find distance traveled:
> Distance = speed × time = 19 km/h × 0.53 h = 10.07 km
But displacement is the straight-line distance from start to end point.
❗ Unless told the bus traveled in a straight line, we cannot assume displacement equals distance.
However, the worksheet answer assumes straight-line motion: “if it traveled in a straight line”.
✔ So, if we assume straight-line motion (as implied by the worksheet’s answer), then:
> Displacement = 10.07 km ≈ 10.1 km (from house to school)
But strictly speaking, without that assumption, we cannot determine displacement from speed alone.
*(Worksheet answer: “The displacement of the bus during the trip is 9.5km from the house if it traveled in a straight line” — this is incorrect mathematically. 19 × 0.53 = 10.07, not 9.5.)*
✘ Correction:
> Displacement = 19 km/h × 0.53 h = 10.07 km (≈ 10.1 km) if traveling in a straight line
---
Problem 5: A girl participating in cross-country spends the afternoon practicing, and ends the practice completely tired from her hard work, despite the fact that her average velocity during the practice was 0.0 m/s. Explain how this situation is possible.
Average velocity = total displacement / total time
If she ended at the same place she started (e.g., ran loops or back-and-forth routes), then displacement = 0, so average velocity = 0.
But she still covered a lot of distance, so she did a lot of work (energy expenditure) — hence, tired.
✔ Explanation:
> Average velocity depends only on displacement (change in position). If she returned to her starting point, displacement is zero, so average velocity is zero — even though she ran many kilometers and exerted herself physically.
*(Worksheet answer is correct: “she started and ended practice at the same spot.”)*
---
Problem 6: A hiker is at the bottom of a canyon facing the canyon wall closest to her. She is 280.5 meters from the wall and the sound of her voice travels at 340 m/s at that location. How long after she shouts will she hear her echo?
Echo means sound travels to the wall AND back to her.
So total distance sound travels = 2 × 280.5 m = 561 meters
Speed of sound = 340 m/s
Time = distance / speed = 561 m / 340 m/s = ?
> 561 ÷ 340 = 1.65 seconds
✔ Answer: 1.65 seconds
*(Note: Worksheet has no answer written here — this is the correct solution.)*
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## ✔ FINAL ANSWERS SUMMARY:
1. D = 2 cm ✔️ (correct)
2. 0.583 m/s North *(worksheet answer is in wrong units — should be m/s, not m/min)*
3. Velocity requires direction — statement omits it. *(worksheet answer slightly unclear — improve wording)*
4. Displacement = 10.07 km (if straight line) — *worksheet answer 9.5 km is mathematically wrong*
5. She returned to start → displacement = 0 → avg velocity = 0, but distance ≠ 0 → she got tired. ✔️
6. t = 1.65 seconds — *not answered on worksheet*
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Let me know if you’d like these formatted into a clean worksheet answer key!
Parent Tip: Review the logic above to help your child master the concept of displacement velocity and acceleration worksheet.
