Handwritten solutions to a distance and displacement worksheet, showing calculations for various travel paths.
A close-up of a "Distance and Displacement Worksheet" with handwritten calculations and answers for four physics problems involving distance and displacement, including a compass rose and grid paper instructions.
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Step-by-step solution for: SOLUTION: Distancr and displacement practice - Studypool
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Show Answer Key & Explanations
Step-by-step solution for: SOLUTION: Distancr and displacement practice - Studypool
You're working on a Distance and Displacement Worksheet, and you’ve already attempted some problems — but there are errors in your answers. Let’s go through each problem step-by-step to correct them and explain the difference between distance and displacement.
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## 📌 Key Concepts:
- Distance = Total length of the path traveled (scalar — only magnitude, no direction).
- Displacement = Straight-line distance from starting point to ending point (vector — has magnitude AND direction).
---
# ✔ Problem 1: Chanice on her scooter
> Chanice drives 7 km north, then 5 km east.
Just add the segments:
7 km + 5 km = 12 km
✔ Distance = 12 km
This is the straight-line distance from start to finish — forms a right triangle.
Use Pythagoras’ Theorem:
> c² = a² + b²
> c² = 7² + 5² = 49 + 25 = 74
> c = √74 ≈ 8.60 km
Direction: Since she went north then east, her final position is northeast of start. To find exact direction:
> tan(θ) = opposite/adjacent = 5/7 → θ = tan⁻¹(5/7) ≈ 35.5° East of North
> So displacement is 8.60 km at 35.5° East of North (or “NE” for short)
✘ Your answer said “8.60 km to SW” — that’s wrong! She ended up NE of start, not SW.
✔ Displacement = 8.60 km [35.5° East of North]
---
# ✔ Problem 2: Anthony walking to pizza place
> Walks 1 km east, then 1 km south, then 1 km east again.
Add all segments:
1 + 1 + 1 = 3 km
✔ Distance = 3 km
Let’s track his net movement:
- East: 1 km + 1 km = 2 km east
- South: 1 km south
So he ends up 2 km east and 1 km south of start → forms a right triangle.
Displacement magnitude:
> c² = 2² + 1² = 4 + 1 = 5
> c = √5 ≈ 2.24 km
Direction: tan(θ) = 1/2 → θ = tan⁻¹(0.5) ≈ 26.6° South of East
→ So displacement is 2.24 km [26.6° South of East]
✘ You wrote “0 km to SE” — that’s incorrect. He didn’t return to start!
✔ Displacement = 2.24 km [26.6° South of East]
---
# ✔ Problem 3: Justin’s fishing trip
> Boats 12 km south, then 4 km west, then 1 km north.
Add all segments:
12 + 4 + 1 = 17 km
✔ Distance = 17 km
Net movement:
- North-South: 12 km south – 1 km north = 11 km south
- East-West: 4 km west
So final position: 11 km south and 4 km west → right triangle.
Magnitude:
> c² = 11² + 4² = 121 + 16 = 137
> c = √137 ≈ 11.70 km
Direction: tan(θ) = 4/11 → θ = tan⁻¹(4/11) ≈ 20.0° West of South
→ So displacement is 11.70 km [20.0° West of South]
✘ You wrote “12.6 km to SW” — close, but wrong calculation. Also, √161 is not correct here.
✔ Displacement = 11.70 km [20.0° West of South]
---
# ✔ Problem 4: Tara on camel safari
> Travels 5 km north, then 3 km east, then 1 km north again.
Add all segments:
5 + 3 + 1 = 9 km
✔ Distance = 9 km
Net movement:
- North: 5 km + 1 km = 6 km north
- East: 3 km east
Right triangle again.
Magnitude:
> c² = 6² + 3² = 36 + 9 = 45
> c = √45 = 3√5 ≈ 6.71 km
Direction: tan(θ) = 3/6 = 0.5 → θ = tan⁻¹(0.5) ≈ 26.6° East of North
→ So displacement is 6.71 km [26.6° East of North]
✔ Displacement = 6.71 km [26.6° East of North]
---
## 🧾 Summary of Correct Answers:
| Problem | Distance | Displacement |
|---------|----------|--------------|
| 1 | 12 km | 8.60 km [35.5° E of N] |
| 2 | 3 km | 2.24 km [26.6° S of E] |
| 3 | 17 km | 11.70 km [20.0° W of S] |
| 4 | 9 km | 6.71 km [26.6° E of N] |
---
## 💡 Tips for Plotting on Grid Paper:
- Use graph paper with 1 cm = 1 km.
- Start at origin (0,0).
- For each segment, draw arrow in correct direction.
- Final displacement = straight line from start to end point.
- Measure length with ruler or calculate using Pythagoras.
---
Let me know if you want help drawing these paths on grid paper or need more practice! 😊
---
## 📌 Key Concepts:
- Distance = Total length of the path traveled (scalar — only magnitude, no direction).
- Displacement = Straight-line distance from starting point to ending point (vector — has magnitude AND direction).
---
# ✔ Problem 1: Chanice on her scooter
> Chanice drives 7 km north, then 5 km east.
➤ Distance:
Just add the segments:
7 km + 5 km = 12 km
✔ Distance = 12 km
➤ Displacement:
This is the straight-line distance from start to finish — forms a right triangle.
Use Pythagoras’ Theorem:
> c² = a² + b²
> c² = 7² + 5² = 49 + 25 = 74
> c = √74 ≈ 8.60 km
Direction: Since she went north then east, her final position is northeast of start. To find exact direction:
> tan(θ) = opposite/adjacent = 5/7 → θ = tan⁻¹(5/7) ≈ 35.5° East of North
> So displacement is 8.60 km at 35.5° East of North (or “NE” for short)
✘ Your answer said “8.60 km to SW” — that’s wrong! She ended up NE of start, not SW.
✔ Displacement = 8.60 km [35.5° East of North]
---
# ✔ Problem 2: Anthony walking to pizza place
> Walks 1 km east, then 1 km south, then 1 km east again.
➤ Distance:
Add all segments:
1 + 1 + 1 = 3 km
✔ Distance = 3 km
➤ Displacement:
Let’s track his net movement:
- East: 1 km + 1 km = 2 km east
- South: 1 km south
So he ends up 2 km east and 1 km south of start → forms a right triangle.
Displacement magnitude:
> c² = 2² + 1² = 4 + 1 = 5
> c = √5 ≈ 2.24 km
Direction: tan(θ) = 1/2 → θ = tan⁻¹(0.5) ≈ 26.6° South of East
→ So displacement is 2.24 km [26.6° South of East]
✘ You wrote “0 km to SE” — that’s incorrect. He didn’t return to start!
✔ Displacement = 2.24 km [26.6° South of East]
---
# ✔ Problem 3: Justin’s fishing trip
> Boats 12 km south, then 4 km west, then 1 km north.
➤ Distance:
Add all segments:
12 + 4 + 1 = 17 km
✔ Distance = 17 km
➤ Displacement:
Net movement:
- North-South: 12 km south – 1 km north = 11 km south
- East-West: 4 km west
So final position: 11 km south and 4 km west → right triangle.
Magnitude:
> c² = 11² + 4² = 121 + 16 = 137
> c = √137 ≈ 11.70 km
Direction: tan(θ) = 4/11 → θ = tan⁻¹(4/11) ≈ 20.0° West of South
→ So displacement is 11.70 km [20.0° West of South]
✘ You wrote “12.6 km to SW” — close, but wrong calculation. Also, √161 is not correct here.
✔ Displacement = 11.70 km [20.0° West of South]
---
# ✔ Problem 4: Tara on camel safari
> Travels 5 km north, then 3 km east, then 1 km north again.
➤ Distance:
Add all segments:
5 + 3 + 1 = 9 km
✔ Distance = 9 km
➤ Displacement:
Net movement:
- North: 5 km + 1 km = 6 km north
- East: 3 km east
Right triangle again.
Magnitude:
> c² = 6² + 3² = 36 + 9 = 45
> c = √45 = 3√5 ≈ 6.71 km
Direction: tan(θ) = 3/6 = 0.5 → θ = tan⁻¹(0.5) ≈ 26.6° East of North
→ So displacement is 6.71 km [26.6° East of North]
✔ Displacement = 6.71 km [26.6° East of North]
---
## 🧾 Summary of Correct Answers:
| Problem | Distance | Displacement |
|---------|----------|--------------|
| 1 | 12 km | 8.60 km [35.5° E of N] |
| 2 | 3 km | 2.24 km [26.6° S of E] |
| 3 | 17 km | 11.70 km [20.0° W of S] |
| 4 | 9 km | 6.71 km [26.6° E of N] |
---
## 💡 Tips for Plotting on Grid Paper:
- Use graph paper with 1 cm = 1 km.
- Start at origin (0,0).
- For each segment, draw arrow in correct direction.
- Final displacement = straight line from start to end point.
- Measure length with ruler or calculate using Pythagoras.
---
Let me know if you want help drawing these paths on grid paper or need more practice! 😊
Parent Tip: Review the logic above to help your child master the concept of distance and displacement worksheet with answers.
