Distance and Displacement Worksheet with six problems and a compass rose.

Let's solve each problem step by step, explaining the difference between distance and displacement:

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🔹 Key Concepts:

- Distance: The total path traveled (scalar quantity, always positive).
- Displacement: The straight-line distance from start to end point, with direction (vector quantity).

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

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1. Chance drives her scooter 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 (since both components are positive)

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

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

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

- Displacement:
Total movement:
- East: $ 1 + 1 = 2 $ km
- South: $ 1 $ km
So, displacement vector is 2 km east, 1 km south.

Magnitude:
$ \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5} \approx 2.24 \text{ km} $

Direction: Southeast (specifically, arctan(1/2) ≈ 26.6° south of east)

✔ Answer:
- Distance: 3 km
- Displacement: ≈2.24 km, 26.6° south of east

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

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

- Displacement:
- North-South: $ -12 + 1 = -11 $ km (11 km south)
- East-West: $ -4 $ km (4 km west)

So displacement vector: 11 km south, 4 km west

Magnitude:
$ \sqrt{(11)^2 + (4)^2} = \sqrt{121 + 16} = \sqrt{137} \approx 11.7 \text{ km} $

Direction: Southwest (arctan(4/11) ≈ 19.9° west of south)

✔ Answer:
- Distance: 17 km
- Displacement: ≈11.7 km, 19.9° west of south

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

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

- Displacement:
- North: $ 5 + 1 = 6 $ km
- East: $ 3 $ km

Magnitude:
$ \sqrt{(6)^2 + (3)^2} = \sqrt{36 + 9} = \sqrt{45} \approx 6.71 \text{ km} $

Direction: Northeast (arctan(3/6) = arctan(0.5) ≈ 26.6° east of north)

✔ Answer:
- Distance: 9 km
- Displacement: ≈6.71 km, 26.6° east of north

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5. 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 compute net movement in x (east-west) and y (north-south):

#### North-South (y-axis):
- +700 (north)
- +400 (north)
- –1200 (south)
- +100 (north)

Total: $ 700 + 400 + 100 - 1200 = 1200 - 1200 = 0 $ m → No net north-south movement

#### East-West (x-axis):
- +300 (east)
- –600 (west)
- +300 (east)

Total: $ 300 - 600 + 300 = 0 $ m → No net east-west movement

So he ends at the starting point!

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

- Displacement: Since he returns to start → 0 m

✔ Answer:
- Distance: 3600 m (or 3.6 km)
- Displacement: 0 m

> Note: For plotting on grid paper, scale 1 cm = 100 m → so each segment is scaled accordingly.

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6. Jose travels 3 km south, then 4 km east. How far to return to start?

- He is now 3 km south and 4 km east of start.

To return, he must go directly back — this is the displacement magnitude.

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

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

But question only asks: How far does he need to go to get back?

✔ Answer:
- 5 km

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✔ Final Answers Summary:

| Problem | Distance | Displacement |
|--------|----------|--------------|
| 1 | 12 km | ≈8.6 km NE |
| 2 | 3 km | ≈2.24 km, 26.6° S of E |
| 3 | 17 km | ≈11.7 km, 19.9° W of S |
| 4 | 9 km | ≈6.71 km, 26.6° E of N |
| 5 | 3600 m | 0 m |
| 6 | Not asked | 5 km (to return) |

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📌 Note for Plotting on Grid Paper:
For each problem, plot the movements step-by-step using a compass rose (N, E, S, W). Use consistent scale (e.g., 1 cm = 1 km or 1 cm = 100 m depending on problem). Draw vectors for each leg, then draw the straight line from start to end for displacement.

Let me know if you'd like diagrams drawn out!