Practice worksheet for calculating distance and displacement with diagrams and word problems.

Problem Analysis and Solution

The task involves calculating distance and displacement for various paths described in the images and word problems. Let's solve each part step by step.

---

#### Part 1: For each picture below, find the distance and the displacement of the given path.

##### Problem 1:
- Path Description: The path is a rectangle with sides of 6 km and 15 km.
- Distance: The total distance traveled is the sum of all the segments:
\[
\text{Distance} = 6 + 15 + 6 + 15 = 42 \text{ km}
\]
- Displacement: Displacement is the straight-line distance from the starting point to the ending point. Since the path forms a closed loop (starting and ending at the same point), the displacement is:
\[
\text{Displacement} = 0 \text{ km}
\]

##### Problem 2:
- Path Description: The path is a right triangle with legs of 9 cm and 12 cm, and the hypotenuse is 15 cm.
- Distance: The total distance traveled is the sum of all the segments:
\[
\text{Distance} = 9 + 12 + 15 = 36 \text{ cm}
\]
- Displacement: Displacement is the straight-line distance from the starting point to the ending point, which is the hypotenuse of the triangle:
\[
\text{Displacement} = 15 \text{ cm}
\]

##### Problem 3:
- Path Description: The path is a triangle with sides of 10 m, 11 m, and 20 m.
- Distance: The total distance traveled is the sum of all the segments:
\[
\text{Distance} = 10 + 11 + 20 = 41 \text{ m}
\]
- Displacement: Displacement is the straight-line distance from the starting point to the ending point. Since the path forms a closed loop (starting and ending at the same point), the displacement is:
\[
\text{Displacement} = 0 \text{ m}
\]

---

#### Part 2: For each path described, find the distance and displacement.

##### Problem 4:
- Path Description: Frank starts at point D, walks to point C, then to B, then to A.
- Distance: The total distance traveled is the sum of the lengths of each segment:
\[
\text{Distance} = 25 + 15 + 15 + 25 = 80 \text{ mm}
\]
- Displacement: Displacement is the straight-line distance from the starting point (D) to the ending point (A). Since D and A are directly opposite each other, the displacement is:
\[
\text{Displacement} = 50 \text{ mm}
\]

##### Problem 5:
- Path Description: Jose runs from point D to point A, then back to D, then to C.
- Distance: The total distance traveled is the sum of the lengths of each segment:
\[
\text{Distance} = 100 + 100 + 100 = 300 \text{ m}
\]
- Displacement: Displacement is the straight-line distance from the starting point (D) to the ending point (C). The displacement is:
\[
\text{Displacement} = 100 \text{ m}
\]

##### Problem 6:
- Path Description: Mark drives from C to B, then back to B, then to C.
- Distance: The total distance traveled is the sum of the lengths of each segment:
\[
\text{Distance} = 1.5 + 1.5 + 1.5 + 1.5 = 6 \text{ km}
\]
- Displacement: Displacement is the straight-line distance from the starting point (C) to the ending point (C). Since the path forms a closed loop, the displacement is:
\[
\text{Displacement} = 0 \text{ km}
\]

---

#### Part 3: Find the distance and displacement in each word problem by first drawing a picture.

##### Problem 7:
- Path Description: A whale swims East for 300 m, then back West for 100 m.
- Distance: The total distance traveled is the sum of the lengths of each segment:
\[
\text{Distance} = 300 + 100 = 400 \text{ m}
\]
- Displacement: Displacement is the straight-line distance from the starting point to the ending point. The whale ends up 200 m East of the starting point:
\[
\text{Displacement} = 200 \text{ m, East}
\]

##### Problem 8:
- Path Description: Coach Park walks 5 m North, then 10 m West, then 5 m South.
- Distance: The total distance traveled is the sum of the lengths of each segment:
\[
\text{Distance} = 5 + 10 + 5 = 20 \text{ m}
\]
- Displacement: Displacement is the straight-line distance from the starting point to the ending point. The coach ends up 10 m West of the starting point:
\[
\text{Displacement} = 10 \text{ m, West}
\]

##### Problem 9:
- Path Description: A cross-country runner runs 1 km NE, then 1 km SE, then 1 km SW, and finishes by running 1 km NW.
- Distance: The total distance traveled is the sum of the lengths of each segment:
\[
\text{Distance} = 1 + 1 + 1 + 1 = 4 \text{ km}
\]
- Displacement: Displacement is the straight-line distance from the starting point to the ending point. Since the path forms a closed loop (starting and ending at the same point), the displacement is:
\[
\text{Displacement} = 0 \text{ km}
\]

---

Final Answers:

#### Part 1:
1. Distance = 42 km, Displacement = 0 km
2. Distance = 36 cm, Displacement = 15 cm
3. Distance = 41 m, Displacement = 0 m

#### Part 2:
4. Distance = 80 mm, Displacement = 50 mm
5. Distance = 300 m, Displacement = 100 m
6. Distance = 6 km, Displacement = 0 km

#### Part 3:
7. Distance = 400 m, Displacement = 200 m, East
8. Distance = 20 m, Displacement = 10 m, West
9. Distance = 4 km, Displacement = 0 km

Boxed Final Answer:

\[
\boxed{
\begin{array}{ll}
\text{Part 1:} & \text{1. Distance = 42 km, Displacement = 0 km} \\
& \text{2. Distance = 36 cm, Displacement = 15 cm} \\
& \text{3. Distance = 41 m, Displacement = 0 m} \\
\text{Part 2:} & \text{4. Distance = 80 mm, Displacement = 50 mm} \\
& \text{5. Distance = 300 m, Displacement = 100 m} \\
& \text{6. Distance = 6 km, Displacement = 0 km} \\
\text{Part 3:} & \text{7. Distance = 400 m, Displacement = 200 m, East} \\
& \text{8. Distance = 20 m, Displacement = 10 m, West} \\
& \text{9. Distance = 4 km, Displacement = 0 km} \\
\end{array}
}
\]