Worksheet titled "Distance and Displacement Worksheet" featuring nine problems that require students to calculate distance and displacement based on different travel paths and directions.

Let's solve each problem from the Distance and Displacement Worksheet step by step.

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🔹 Key Concepts:

- Distance: Total path traveled (scalar, always positive).
- Displacement: Straight-line distance from start to end point (vector, includes direction).

We'll use a coordinate system:
- East = +x direction
- North = +y direction
- West = -x
- South = -y

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1. Joey drives his Skidoo 7 km north, then 5 km east.

- Distance:
$ 7\,\text{km} + 5\,\text{km} = 12\,\text{km} $

- Displacement:
Use Pythagoras' theorem:
$ \sqrt{(7)^2 + (5)^2} = \sqrt{49 + 25} = \sqrt{74} \approx 8.6\,\text{km} $
Direction: Northeast (specifically, arctan(5/7) ≈ 35.5° east of north)

✔ Answer:
- Distance: 12 km
- Displacement: ≈8.6 km northeast

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2. Anthony walks 1 km east, 1 km south, then 1 km east again.

- Distance:
$ 1 + 1 + 1 = 3\,\text{km} $

- Displacement:
Net movement:
- East: $1 + 1 = 2\,\text{km}$
- South: $1\,\text{km}$
So displacement vector: $ (2\,\text{km}, -1\,\text{km}) $
Magnitude: $ \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \approx 2.24\,\text{km} $
Direction: Southeast (arctan(1/2) ≈ 26.6° south of east)

✔ Answer:
- Distance: 3 km
- Displacement: ≈2.24 km southeast

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3. Justin goes 12 km south, 4 km west, 1 km north.

- Distance:
$ 12 + 4 + 1 = 17\,\text{km} $

- Displacement:
Net movement:
- South: $12 - 1 = 11\,\text{km}$ → so $-11\,\text{km}$ in y-direction
- West: $-4\,\text{km}$ in x-direction
Vector: $(-4, -11)$
Magnitude: $ \sqrt{(-4)^2 + (-11)^2} = \sqrt{16 + 121} = \sqrt{137} \approx 11.7\,\text{km} $
Direction: Southwest (arctan(11/4) ≈ 70.0° south of west)

✔ Answer:
- Distance: 17 km
- Displacement: ≈11.7 km southwest

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4. Preston travels 5 km north, 3 km east, 1 km north.

- Distance:
$ 5 + 3 + 1 = 9\,\text{km} $

- Displacement:
Net:
- North: $5 + 1 = 6\,\text{km}$
- East: $3\,\text{km}$
Vector: $(3, 6)$
Magnitude: $ \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} \approx 6.71\,\text{km} $
Direction: Northeast (arctan(6/3) = 63.4° north of east)

✔ Answer:
- Distance: 9 km
- Displacement: ≈6.71 km northeast

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5. Neil travels 8 m east, 4 m north.

- Distance:
$ 8 + 4 = 12\,\text{m} $

- Displacement:
Vector: $(8, 4)$
Magnitude: $ \sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} \approx 8.94\,\text{m} $
Direction: arctan(4/8) = 26.6° north of east

✔ Answer:
- Distance: 12 m
- Displacement: ≈8.94 m northeast

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6. Alan flies: 400 km south → 700 km west → 1200 km south

- Distance:
$ 400 + 700 + 1200 = 2300\,\text{km} $

- Displacement:
Net movement:
- South: $400 + 1200 = 1600\,\text{km}$ → $-1600$ in y
- West: $-700$ in x
Vector: $(-700, -1600)$
Magnitude: $ \sqrt{(-700)^2 + (-1600)^2} = \sqrt{490000 + 2560000} = \sqrt{3050000} \approx 1746.4\,\text{km} $
Direction: Southwest (arctan(1600/700) ≈ 66.8° south of west)

✔ Answer:
- Distance: 2300 km
- Displacement: ≈1746 km southwest

> *(Use 1 cm = 100 km → draw 23 cm for distance, ~17.5 cm for displacement)*

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7. Brandon: 12 km north, 6 km east, 10 km north

- Distance:
$ 12 + 6 + 10 = 28\,\text{km} $

- Displacement:
Net:
- North: $12 + 10 = 22\,\text{km}$
- East: $6\,\text{km}$
Vector: $(6, 22)$
Magnitude: $ \sqrt{6^2 + 22^2} = \sqrt{36 + 484} = \sqrt{520} \approx 22.8\,\text{km} $
Direction: arctan(22/6) ≈ 74.2° north of east

✔ Answer:
- Distance: 28 km
- Displacement: ≈22.8 km northeast

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8. Alex rides: 700 m north, 300 m east, 400 m north, 600 m west, 1200 m south, 300 m east, 100 m north

Convert all to meters:

- North: $700 + 400 + 100 = 1200\,\text{m}$
- South: $1200\,\text{m}$ → net north-south: $1200 - 1200 = 0$
- East: $300 + 300 = 600\,\text{m}$
- West: $600\,\text{m}$ → net east-west: $600 - 600 = 0$

So net displacement = 0 m

- Distance:
Sum all:
$700 + 300 + 400 + 600 + 1200 + 300 + 100 = 3600\,\text{m}$

✔ Answer:
- Distance: 3600 m
- Displacement: 0 m (back to start)

> *(Use 1 cm = 100 m → draw 36 cm for distance, 0 cm for displacement)*

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9. Stephen travels 3 km south, then 4 km east. How far to return?

This is asking for the magnitude of displacement from start to end — that’s how far he must go back.

- Net movement:
- South: $-3\,\text{km}$
- East: $+4\,\text{km}$
Displacement magnitude: $ \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\,\text{km} $

So he needs to travel 5 km in the opposite direction (i.e., northwest, specifically 53.1° north of west).

✔ Answer:
- He needs to go 5 km to get back.

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✔ Final Answers Summary:

| Problem | Distance | Displacement |
|--------|----------|--------------|
| 1 | 12 km | ≈8.6 km NE |
| 2 | 3 km | ≈2.24 km SE |
| 3 | 17 km | ≈11.7 km SW |
| 4 | 9 km | ≈6.71 km NE |
| 5 | 12 m | ≈8.94 m NE |
| 6 | 2300 km | ≈1746 km SW |
| 7 | 28 km | ≈22.8 km NE |
| 8 | 3600 m | 0 m |
| 9 | N/A | 5 km (return) |

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📌 Instructions for Plotting on Grid Paper:

For each question:
1. Start at origin (0,0).
2. Draw vectors for each leg of the journey.
3. Label directions and lengths.
4. For displacement, draw a straight line from start to end.
5. Use scale if needed (e.g., 1 cm = 1 km or 100 km as specified).

Let me know if you'd like diagrams or printable versions!