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Let's solve this Distance and Displacement Practice Worksheet step by step.

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🔍 Understanding the Concepts:

- Distance: Total path length traveled (scalar quantity, no direction).
- Always positive.
- Sum of all movements.

- Displacement: Straight-line distance from start to end point, with direction (vector quantity).
- Can be zero or negative depending on direction.
- Use a coordinate system (e.g., east = +, west = –; north = +, south = –).

> We assume:
> - East and North are positive directions.
> - West and South are negative directions.

We'll use the formula:
> Displacement = Final Position – Initial Position

But since we’re moving in steps, we’ll calculate net movement in each direction.

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✔ Example Given (Row 1):

- Move 4 m east → +4
- Move 2 m west → –2
- Total Distance = 4 + 2 = 6 m
- Net Displacement = 4 – 2 = +2 m east

So displacement is 2 m east, which matches the table.

Now let’s fill in the rest.

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🧮 Solving Each Row:

#### Row 2:
- Move 4 m north → +4 (north)
- Move 2 m south → –2 (south)
- Distance = 4 + 2 = 6 m
- Displacement = 4 – 2 = +2 m north

✔ Answer:
- Distance: 6 m
- Displacement: 2 m north

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#### Row 3:
- Move 2 m east → +2
- Move 4 m west → –4
- Distance = 2 + 4 = 6 m
- Displacement = 2 – 4 = –2 m → 2 m west

✔ Answer:
- Distance: 6 m
- Displacement: 2 m west

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#### Row 4:
- Move 5 m east → +5
- Move 5 m west → –5
- Distance = 5 + 5 = 10 m
- Displacement = 5 – 5 = 0 m

✔ Answer:
- Distance: 10 m
- Displacement: 0 m (no direction, since it's zero)

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#### Row 5:
- Move 5 m south → –5
- Move 2 m north → +2
- Distance = 5 + 2 = 7 m
- Displacement = –5 + 2 = –3 m → 3 m south

✔ Answer:
- Distance: 7 m
- Displacement: 3 m south

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#### Row 6:
- Move 10 m west → –10
- Move 3 m east → +3
- Distance = 10 + 3 = 13 m
- Displacement = –10 + 3 = –7 m → 7 m west

✔ Answer:
- Distance: 13 m
- Displacement: 7 m west

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#### Row 7:
- Move 3 m east → +3
- Move 4 m north → +4
- These are perpendicular directions (east and north), so we can't just add them algebraically. But since they're in different directions, we treat them separately.

Wait — but displacement is a vector, so we need to find the straight-line distance from start to finish using Pythagoras.

Let’s break it down:

- Net movement:
- East: +3 m
- North: +4 m

This forms a right triangle.

- Displacement magnitude = √(3² + 4²) = √(9 + 16) = √25 = 5 m
- Direction: Use tangent: tan⁻¹(4/3) ≈ 53.1° north of east → we can say 5 m northeast (or more precisely, 5 m at 53° north of east)

But for simplicity in this worksheet, likely expect magnitude and direction.

However, if only one direction is allowed, then write as 5 m northeast.

But let's see what the worksheet expects.

Since it says “include direction,” and previous ones used compass directions, we’ll go with:

> Displacement: 5 m northeast (approximate)

But technically, it's 5 m at 53° north of east

But for this level, "5 m northeast" is acceptable.

- Distance = 3 + 4 = 7 m

✔ Answer:
- Distance: 7 m
- Displacement: 5 m northeast

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#### Row 8:
- Move 6 m east → +6
- Move 8 m south → –8
- Again, perpendicular directions.

- Distance = 6 + 8 = 14 m
- Net movement:
- East: +6
- South: –8
- Displacement magnitude = √(6² + 8²) = √(36 + 64) = √100 = 10 m
- Direction: tan⁻¹(8/6) = tan⁻¹(4/3) ≈ 53.1° south of east → 10 m southeast

✔ Answer:
- Distance: 14 m
- Displacement: 10 m southeast

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#### Row 9:
- Move 9 m north → +9
- Move 12 m west → –12
- Perpendicular again.

- Distance = 9 + 12 = 21 m
- Net movement:
- North: +9
- West: –12
- Displacement magnitude = √(9² + 12²) = √(81 + 144) = √225 = 15 m
- Direction: tan⁻¹(12/9) = tan⁻¹(4/3) ≈ 53.1° west of north → 15 m northwest? Wait!

Wait: 9 m north, 12 m west → so it's 12 m west, 9 m north → direction is west of north

So: 15 m, 53° west of north → or 15 m northwest (approximate)

✔ Answer:
- Distance: 21 m
- Displacement: 15 m, 53° west of north or 15 m northwest

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✔ Final Table Filled In:

| # | First Move | Second Move | Distance (m) | Displacement |
|---|------------|-------------|--------------|--------------|
| 1 | Move 4 m east | Move 2 m west | 6 | 2 m east |
| 2 | Move 4 m north | Move 2 m south | 6 | 2 m north |
| 3 | Move 2 m east | Move 4 m west | 6 | 2 m west |
| 4 | Move 5 m east | Move 5 m west | 10 | 0 m |
| 5 | Move 5 m south | Move 2 m north | 7 | 3 m south |
| 6 | Move 10 m west | Move 3 m east | 13 | 7 m west |
| 7 | Move 3 m east | Move 4 m north | 7 | 5 m northeast |
| 8 | Move 6 m east | Move 8 m south | 14 | 10 m southeast |
| 9 | Move 9 m north | Move 12 m west | 21 | 15 m, 53° west of north |

> Note: For rows 7–9, since the motions are in perpendicular directions, displacement is found using the Pythagorean Theorem.

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📌 Key Takeaway:

- Distance is total path length.
- Displacement is net change in position (straight line from start to end), including direction.
- When motion is in two directions (e.g., east and north), use vectors and Pythagoras.

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✔ Bonus: Scenario Where Displacement is Zero but Distance is Not

> Example: Walk 5 m east, then 5 m west.
- Distance = 10 m
- Displacement = 0 m (back to start)

That’s exactly Row 4!

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Let me know if you'd like help drawing the graphs or diagrams!