SOLUTION: Distancr and displacement practice - Studypool - Free Printable

Let's solve each problem step by step, explaining the concepts of distance and displacement.

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🔹 Key Concepts:

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

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

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## ✔ Problem 1: Chanice’s Scooter Ride

> Chanice drives her scooter 7 km north, then 5 km east.

Step 1: Distance

Total path covered:
$$
7\,\text{km} + 5\,\text{km} = \boxed{12\,\text{km}}
$$

Step 2: Displacement

She moves 7 km north and 5 km east → forms a right triangle.

Use Pythagoras:
$$
\text{Displacement} = \sqrt{(7)^2 + (5)^2} = \sqrt{49 + 25} = \sqrt{74} \approx \boxed{8.60\,\text{km}}
$$

Direction: From origin to northeast → Northeast, but more precisely, we can say:

- The angle θ from north toward east:
$$
\tan\theta = \frac{5}{7} \Rightarrow \theta \approx 35.5^\circ \text{ east of north}
$$

So displacement is 8.60 km at 35.5° east of north.

✔ Answer:
- Distance: 12 km
- Displacement: 8.60 km, 35.5° east of north

> Note: The handwritten answer says "8.60 km to SW" — this is incorrect. It should be NE, not SW.

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## ✔ Problem 2: Anthony’s Walk

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

Step 1: Distance

Total path:
$$
1 + 1 + 1 = \boxed{3\,\text{km}}
$$

Step 2: Displacement

Break into components:
- East: 1 km + 1 km = 2 km east
- South: 1 km

So net movement: 2 km east, 1 km south

Now find straight-line displacement:
$$
\text{Displacement} = \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5} \approx \boxed{2.24\,\text{km}}
$$

Direction: South of east
$$
\tan\theta = \frac{1}{2} \Rightarrow \theta \approx 26.6^\circ \text{ south of east}
$$

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

> Handwritten answer says "0 km to SE" — this is wrong. He did move, so displacement ≠ 0.

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## ✔ Problem 3: Justin’s Fishing Trip

> Boat goes: 12 km south, 4 km west, 1 km north.

Step 1: Distance

Add all segments:
$$
12 + 4 + 1 = \boxed{17\,\text{km}}
$$

Step 2: Displacement

Break into components:
- North-South: 12 km south – 1 km north = 11 km south
- East-West: 4 km west = 4 km west

Now use Pythagoras:
$$
\text{Displacement} = \sqrt{(11)^2 + (4)^2} = \sqrt{121 + 16} = \sqrt{137} \approx \boxed{11.7\,\text{km}}
$$

Direction: South-west
$$
\tan\theta = \frac{4}{11} \Rightarrow \theta \approx 19.8^\circ \text{ west of south}
$$

✔ Answer:
- Distance: 17 km
- Displacement: 11.7 km, 19.8° west of south

> Handwritten answer says "12.6 km to SW" — close but incorrect. Should be ~11.7 km.

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## ✔ Problem 4: Tara’s Camel Safari

> She travels: 5 km north, 3 km east, 1 km north.

Step 1: Distance

Total path:
$$
5 + 3 + 1 = \boxed{9\,\text{km}}
$$

Step 2: Displacement

Net movement:
- North: 5 km + 1 km = 6 km north
- East: 3 km

So displacement:
$$
\text{Displacement} = \sqrt{(6)^2 + (3)^2} = \sqrt{36 + 9} = \sqrt{45} \approx \boxed{6.71\,\text{km}}
$$

Direction: North-east
$$
\tan\theta = \frac{3}{6} = 0.5 \Rightarrow \theta \approx 26.6^\circ \text{ east of north}
$$

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

> Handwritten answer is incomplete ("S all..."), so likely wrong.

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## ✔ Summary Table:

| Problem | Distance | Displacement |
|--------|----------|--------------|
| 1 | 12 km | 8.60 km, 35.5° E of N |
| 2 | 3 km | 2.24 km, 26.6° S of E |
| 3 | 17 km | 11.7 km, 19.8° W of S |
| 4 | 9 km | 6.71 km, 26.6° E of N |

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📌 Final Notes:

- Always add all segments for distance.
- For displacement, find net movement in x and y directions, then use vector math (Pythagoras and trigonometry).
- Direction matters! Use compass bearings or angles relative to cardinal directions.

Let me know if you'd like diagrams plotted on grid paper!