Worksheet with nine problems on distance and displacement calculations.

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: $700\,\text{km}$ → $-700$ 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

> Note: Scale given is 1 cm = 100 km → You can draw this on grid paper accordingly.

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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

Let’s break it down:

| Movement | Δx (east) | Δy (north) |
|------------------|-----------|------------|
| 700 m north | 0 | +700 |
| 300 m east | +300 | 0 |
| 400 m north | 0 | +400 |
| 600 m west | -600 | 0 |
| 1200 m south | 0 | -1200 |
| 300 m east | +300 | 0 |
| 100 m north | 0 | +100 |

Now sum:
- Δx: $ 300 - 600 + 300 = 0 $
- Δy: $ 700 + 400 - 1200 + 100 = 0 $

So net displacement = 0 m

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

✔ Answer:
- Distance: 3600 m
- Displacement: 0 m (he ends where he started)

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

- This forms a right triangle: 3 km south, 4 km east.
- To get back to start, he must travel the straight-line distance from final point to origin.

Use Pythagoras:
$ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\,\text{km} $

Direction: Northwest (arctan(3/4) ≈ 36.9° west of north)

✔ Answer:
- He needs to go 5 km to return to start.

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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 | — | 5 km (return) |

> 📌 For plotting on grid paper, use scale as suggested (e.g., 1 cm = 100 km or 1 cm = 100 m), and draw vectors accordingly.

Let me know if you'd like a visual sketch or help plotting them!