Lesson 2.1 - Distance & Displacement - Classful - Free Printable

Let’s solve this step by step. The key concepts are:

- Distance = total length of the path traveled (scalar, always positive).
- Displacement = straight-line distance from start to end point + direction (vector).

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## ✔ PART 1: For each picture below, find the distance and displacement.

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1. Rectangular Path

Path: Start at bottom-left star → right 15 km → up 6 km → left 15 km → down 6 km → back to start.

- Distance = sum of all segments = 15 + 6 + 15 + 6 = 42 km
- Displacement = start and end at same point → 0 km

✔ Answer:
> Distance = 42 km
> Displacement = 0 km

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2. Triangle Path

Start at bottom-left star → right 15 cm → up 9 cm → diagonal back 12 cm to start? Wait — that doesn’t make sense geometrically.

Actually, looking at the triangle: it’s a right triangle with legs 9 cm and 15 cm? But hypotenuse is labeled 12 cm — which is impossible because 9² + 15² = 81 + 225 = 306 → √306 ≈ 17.5 cm ≠ 12 cm.

➡️ This must be a typo or mislabeling. But since the problem shows a path going along two sides (9 cm and 15 cm) and then directly back along the hypotenuse (12 cm), we’ll go with what’s drawn.

So path: start → right 15 cm → up 9 cm → diagonal back 12 cm to start.

Wait — if you go right 15, up 9, then back 12 to start, that implies the 12 cm side connects the top-right corner to the start — but that’s not a straight line unless it’s a different shape.

Actually, re-examining: the arrow goes from start → up 9 cm → diagonal 12 cm → right 15 cm? No — the arrows show:

- Start → right 15 cm
- Then up 9 cm
- Then diagonal 12 cm back to start?

That would only work if the 12 cm is the hypotenuse of a 9-15 triangle — which isn’t possible. So likely, the 12 cm is the hypotenuse, and the 9 cm and 15 cm are the legs — but 9²+15²≠144.

This is inconsistent. But perhaps the diagram intends:

Start → along base 15 cm → up 9 cm → back along hypotenuse 12 cm? Still impossible.

Alternatively, maybe the path is: start → up 9 cm → diagonal 12 cm → right 15 cm? Doesn’t close.

Given the confusion, let’s assume the intended path is: start → right 15 cm → up 9 cm → back along hypotenuse 12 cm to start — even though it’s geometrically incorrect. In that case:

- Distance = 15 + 9 + 12 = 36 cm
- Displacement = back to start → 0 cm

But if the triangle is meant to be a right triangle with legs 9 cm and 12 cm, hypotenuse 15 cm — then the labeling is swapped.

Looking again: the vertical leg is 9 cm, horizontal is 15 cm, diagonal is 12 cm — this is impossible.

➡️ Most likely, the 12 cm is mislabeled. It should be ~17.5 cm. But since we have to go with given numbers, and the path ends where it started (implied by the “X” at start and end), we’ll assume displacement is zero.

✔ Answer:
> Distance = 36 cm
> Displacement = 0 cm

*(Note: If this were a real physics problem, we’d flag the inconsistency. But for educational purposes, assuming closed loop.)*

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3. Parallelogram Path

Start at bottom-right star → left 20 m → up-left 11 m → right 10 m → down-right 11 m → back to start? Wait — the arrows show:

Start → left 20 m → up 11 m → right 10 m → down 11 m → back to start? Not quite.

Actually, looking at the diagram:

- Start at bottom-right star
- Go left 20 m to bottom-left
- Go up-left 11 m to top-left
- Go right 10 m to top-right
- Go down-right 11 m to start

This forms a parallelogram? Let’s check:

- Bottom: 20 m
- Top: 10 m? That doesn’t match.
Actually, the top side is 10 m, bottom is 20 m — so it’s a trapezoid? But the two slanted sides are both 11 m.

Path: start → left 20 m → up 11 m → right 10 m → down 11 m → back to start.

Wait — after going right 10 m, you’re not directly above start. Then down 11 m — does that bring you back?

If the figure is symmetric, and the two slanted sides are equal, and top is 10 m, bottom is 20 m, then the horizontal offset is 5 m on each side.

So when you go down 11 m from top-right, you land 5 m left of start — not back to start. So displacement ≠ 0.

This is getting messy. Let’s calculate displacement vectorially.

Assume coordinate system:

- Start at (0, 0)
- Go left 20 m → (-20, 0)
- Go up 11 m → (-20, 11) — wait, no, the 11 m is diagonal? The label says “11 m” along the slanted side.

Actually, the diagram shows:

- Bottom side: 20 m (horizontal)
- Left slant: 11 m (upward to left? Or just upward?)
- Top side: 10 m (horizontal)
- Right slant: 11 m (downward to right)

So it’s a trapezoid with parallel sides 20 m and 10 m, and non-parallel sides 11 m each.

To find displacement: start and end are the same point? The “X” is at start and end — so yes, it’s a closed loop.

Therefore:

- Distance = 20 + 11 + 10 + 11 = 52 m
- Displacement = 0 m

✔ Answer:
> Distance = 52 m
> Displacement = 0 m

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## ✔ PART 2: For each path described, find distance and displacement.

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4. Frank starts at D → C → B → A

Rectangle: AB = CD = 25 mm, AD = BC = 15 mm

Path: D → C → B → A

- D to C: 25 mm (right)
- C to B: 15 mm (up)
- B to A: 25 mm (left)

Total distance = 25 + 15 + 25 = 65 mm

Displacement: from D to A → straight line. Since D to A is vertical side = 15 mm up.

But direction? From D to A is North (if A is top-left, D is top-right — wait, labels:

A ———— D
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B ———— C

So A top-left, D top-right, B bottom-left, C bottom-right.

Frank: D → C → B → A

Start: D (top-right)
End: A (top-left)

Displacement: from D to A = left 25 mm → 25 mm West

✔ Answer:
> Distance = 65 mm
> Displacement = 25 mm West

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5. Jose runs D → A → D → C

Path: D → A → D → C

- D to A: 25 mm left
- A to D: 25 mm right
- D to C: 15 mm down

Total distance = 25 + 25 + 15 = 65 mm

Displacement: start at D, end at C → D to C is down 15 mm → 15 mm South

✔ Answer:
> Distance = 65 mm
> Displacement = 15 mm South

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6. Mark drives C → B → B → C

Path: C → B → B → C

- C to B: 15 mm up (since C bottom-right, B bottom-left? Wait — in rectangle:

A ———— D
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B ———— C

So C to B is left 25 mm? Wait — earlier we had AB = 25 mm (top/bottom), AD = 15 mm (sides).

In problem 6, the rectangle has:

A ———— D
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B ———— C

With AB = 100 m, BC = 50 m? Wait — no, problem 6 says:

> Mark drives from C to B to A, then back to B, then to C.

And the diagram shows:

A ———— D
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B ———— C

With AB = 3.5 km, BC = 1.5 km? Wait — labels:

“AB = 3.5 km”, “BC = 1.5 km”, “CD = 3.5 km”, “DA = 1.5 km”

So it’s a rectangle: width 3.5 km, height 1.5 km.

Mark: C → B → A → B → C

Path:

- C to B: left 3.5 km
- B to A: up 1.5 km
- A to B: down 1.5 km
- B to C: right 3.5 km

Total distance = 3.5 + 1.5 + 1.5 + 3.5 = 10 km

Displacement: start at C, end at C → 0 km

✔ Answer:
> Distance = 10 km
> Displacement = 0 km

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## ✔ PART 3: Word Problems — Draw picture, then find distance and displacement.

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7. Whale swims East 300 m, then West 100 m

Picture:
Start → → → → (300 m East) → ← ← (100 m West)

Final position: 300 - 100 = 200 m East of start.

- Distance = 300 + 100 = 400 m
- Displacement = 200 m East

✔ Answer:
> Distance = 400 m
> Displacement = 200 m East

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8. Coach Park: 5 m North, 10 m West, 5 m South

Picture:
Start → ↑ 5 m → ← 10 m → ↓ 5 m

Net movement: North and South cancel → only 10 m West.

- Distance = 5 + 10 + 5 = 20 m
- Displacement = 10 m West

✔ Answer:
> Distance = 20 m
> Displacement = 10 m West

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9. Cross country runner: 1 km NE, 1 km SE, 1 km SW, 1 km NW

This is tricky. Let’s break into components.

Assume NE = 45° from East, etc.

Each leg is 1 km at 45° intervals.

- NE: +x and +y components: (cos45, sin45) = (√2/2, √2/2)
- SE: +x, -y: (√2/2, -√2/2)
- SW: -x, -y: (-√2/2, -√2/2)
- NW: -x, +y: (-√2/2, √2/2)

Sum x-components: √2/2 + √2/2 - √2/2 - √2/2 = 0
Sum y-components: √2/2 - √2/2 - √2/2 + √2/2 = 0

So total displacement = 0 km

Distance = 1 + 1 + 1 + 1 = 4 km

✔ Answer:
> Distance = 4 km
> Displacement = 0 km

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## 📝 FINAL ANSWERS:

Part 1:

1. Distance = 42 km, Displacement = 0 km
2. Distance = 36 cm, Displacement = 0 cm *(assuming closed loop despite geometry issue)*
3. Distance = 52 m, Displacement = 0 m

Part 2:

4. Distance = 65 mm, Displacement = 25 mm West
5. Distance = 65 mm, Displacement = 15 mm South
6. Distance = 10 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

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Let me know if you want hand-drawn diagrams or vector explanations!