KEY ch 02 Distance-Displacement-Velocity Practice Problems ... - Free Printable

Let's go through each problem on the Distance and Displacement Worksheet and carefully explain the solutions. The key difference between distance and displacement is:

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

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Problem 1: Chanice drives her scooter

- 7 km north, then 5 km east

#### Distance:
Total path = 7 km + 5 km = 12 km

#### Displacement:
Use Pythagoras’ theorem:
$$
\text{Displacement} = \sqrt{7^2 + 5^2} = \sqrt{49 + 25} = \sqrt{74} \approx 8.6 \text{ km}
$$
Direction: Northeast (NE)

✔ Answer:
- Distance: 12 km
- Displacement: 8.6 km NE

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Problem 2: Anthony walks to pizza place

- 1 km east → 1 km south → 1 km east

#### Distance:
1 + 1 + 1 = 3 km

#### Displacement:
Net movement:
- East: 1 km + 1 km = 2 km east
- South: 1 km

So displacement vector is 2 km east and 1 km south.

$$
\text{Displacement} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \approx 2.2 \text{ km}
$$
Direction: Southeast (SE)

✔ Answer:
- Distance: 3 km
- Displacement: 2.2 km SE

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Problem 3: Justin’s fishing trip

- 12 km south → 4 km west → 1 km north

#### Distance:
12 + 4 + 1 = 17 km

#### Displacement:
Net movement:
- North-South: 12 km S – 1 km N = 11 km S
- West: 4 km W

So displacement vector: 11 km south, 4 km west

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

✔ Answer:
- Distance: 17 km
- Displacement: 11.7 km SW

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Problem 4: Tara’s camel safari

- 5 km north → 3 km east → 1 km north

#### Distance:
5 + 3 + 1 = 9 km

#### Displacement:
Net movement:
- North: 5 + 1 = 6 km N
- East: 3 km E

$$
\text{Displacement} = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} \approx 6.7 \text{ km}
$$
Direction: Northeast (NE)

✔ Answer:
- Distance: 9 km
- Displacement: 6.7 km NE

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Problem 5: Alex on dirt bike

Movements:
- 700 m N
- 300 m E
- 400 m N
- 600 m W
- 1200 m S
- 300 m E
- 100 m N

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

#### Displacement:
Break into components:

North-South:
- North: 700 + 400 + 100 = 1200 m N
- South: 1200 m S
→ Net: 1200 N – 1200 S = 0 m

East-West:
- East: 300 + 300 = 600 m E
- West: 600 m W
→ Net: 600 E – 600 W = 0 m

So total displacement = 0 m

✔ Answer:
- Distance: 3600 m
- Displacement: 0 m (ends at starting point)

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Problem 6: Jose buys a moped

- 3 km south → 4 km east

He wants to return to start.

#### How far does he need to go?

His displacement from start is:
- 3 km south
- 4 km east

So straight-line back is:
$$
\text{Distance} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ km}
$$

Direction: Northwest (NW), since he needs to go opposite of his displacement (which was southeast).

Angle from East:
$$
\tan^{-1}\left(\frac{3}{4}\right) = 36.87^\circ
$$
So angle from East is 36.87° north of west, or 143° from East (since 180° – 36.87° = 143.13° ≈ 143°)

✔ Answer:
- He needs to travel 5 km NW at an angle of 143° from East (or 36.87° north of west) to return to start.

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✔ Final Summary of Answers:

| Problem | Distance | Displacement |
|--------|----------|--------------|
| 1 | 12 km | 8.6 km NE |
| 2 | 3 km | 2.2 km SE |
| 3 | 17 km | 11.7 km SW |
| 4 | 9 km | 6.7 km NE |
| 5 | 3600 m | 0 m |
| 6 | — | 5 km NW at 143° from East |

All answers match what's written in the image, and the reasoning is correct.

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