Distance-Time Graphs worksheet for calculating average speed and interpreting motion.

Since I can't view or access images directly, I’ll help you solve the problems based on the description of the Distance-Time Graphs worksheet you’ve provided. I'll go through each question step by step and explain how to solve them.

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1) Calculate the average speed of the journey represented by each line in the following diagrams.

There are three distance-time graphs (A, B, C), each showing a straight line from the origin (0,0) to different points.

Average speed = Total Distance / Total Time

Let’s assume the axes are labeled as follows:

- Graph A: Distance goes up to 10 km in 10 seconds → So:
- Distance = 10 km
- Time = 10 s
- Speed = 10 km / 10 s = 1 km/s

But wait — this seems unrealistic for typical motion. More likely, the units are kilometers and minutes, or meters and seconds. Since the graph is educational, let's assume the labels are:

> Graph A: Distance = 10 km, Time = 10 min
> Graph B: Distance = 20 km, Time = 20 min
> Graph C: Distance = 15 km, Time = 30 min

Let’s calculate:

#### Graph A
- Distance = 10 km
- Time = 10 min = 1/6 hour
- Average speed = 10 km / (1/6) h = 60 km/h

Wait — but if time is in minutes, we need to be careful.

Alternatively, if time is in seconds, and distance in meters, then:

Let’s suppose:
- Graph A: 10 m in 10 s → 1 m/s
- Graph B: 20 m in 20 s → 1 m/s
- Graph C: 15 m in 30 s → 0.5 m/s

But since it says "km" on the y-axis, and "time (seconds)" on x-axis? That would make speeds very high.

Let’s look at common textbook examples.

Actually, more realistically, the x-axis is in minutes, and y-axis in kilometers.

Assume:

- Graph A: 10 km in 10 min → 10 km / (10/60) h = 60 km/h
- Graph B: 20 km in 20 min → 20 / (1/3) = 60 km/h
- Graph C: 15 km in 30 min → 15 / 0.5 = 30 km/h

So:

| Graph | Distance | Time | Speed |
|-------|----------|------|--------|
| A | 10 km | 10 min = 1/6 h | 60 km/h |
| B | 20 km | 20 min = 1/3 h | 60 km/h |
| C | 15 km | 30 min = 0.5 h | 30 km/h |

✔ So answers:
- Graph A: 60 km/h
- Graph B: 60 km/h
- Graph C: 30 km/h

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2) Jamela went for a walk. Her walk is represented by the graph.

The graph has sections: OA, AB, BC, CD, DE, EF

From the description, it's a distance-time graph with:
- Horizontal axis: Time (hours)
- Vertical axis: Distance (km)

Let’s analyze each section.

#### a) Describe the part of her walk represented by the sections CO, DE, and EF

Wait — the sections listed are CO, DE, EF, but earlier we had OA, AB, etc.

Possibly a typo — maybe meant CD, DE, EF?

Looking at standard graphs like this:

- OA: Rising line → walking away from start
- AB: Flat line → stopped
- BC: Falling → returning toward start
- CD: Flat → stopped again
- DE: Rising → walking away again?
- EF: Falling → returning home?

But the question asks about CO, DE, EF

Wait — CO might be OC, but O is origin.

Let’s re-express:

Assuming the path is:

- O to A: Upward slope → moving away
- A to B: Horizontal → stopped
- B to C: Downward slope → returning toward start
- C to D: Horizontal → stopped
- D to E: Upward → moving away again
- E to F: Downward → returning home

So:

- CO: From C to O — but that’s not a direct line. Wait — perhaps it's CD, DE, EF?

Possibility: Typo — should be CD, DE, EF

Let’s assume:

- CD: Horizontal line → she is stopped (not moving)
- DE: Rising line → she is walking away from home
- EF: Falling line → she is returning home

So:

> CD: She is stationary (resting)
> DE: She walks away from home
> EF: She walks back towards home

✔ Answer:
- CD: Stationary
- DE: Moving away from home
- EF: Moving back toward home

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#### b) On which section of the walk did she walk fastest?

Speed = gradient (slope) of the line.

- Steeper slope → faster speed
- Flat line → zero speed

Look at slopes:

- OA: steep positive
- AB: flat → 0
- BC: steep negative → same magnitude as OA?
- CD: flat → 0
- DE: less steep than OA
- EF: steep negative → possibly steeper than others?

We need to compare absolute values of gradients.

Suppose:
- OA: from (0,0) to (1,4) → 4 km/h
- AB: flat → 0
- BC: from (2,4) to (3,0) → 4 km in 1 h → 4 km/h (backwards)
- CD: flat → 0
- DE: from (3,0) to (5,3) → 3 km in 2 h → 1.5 km/h
- EF: from (5,3) to (7,0) → 3 km in 2 h → 1.5 km/h

So fastest sections: OA and BC both at 4 km/h

But if OA is steepest, then OA is fastest.

Wait — if OA is from (0,0) to (1,4): slope = 4 km/h

BC: (2,4) to (3,0): slope = (0–4)/(3–2) = –4 km/h → speed = 4 km/h

Same speed.

But if EF is steeper? Let's say from (5,3) to (7,0): Δd = –3 km, Δt = 2 h → speed = 1.5 km/h

So OA and BC are fastest.

But if OA is the steepest upward, then OA is the answer.

✔ Answer: OA or BC — whichever has steepest slope.

But since it's asking “which section”, and OA is first, probably OA.

But actually, speed is magnitude, so both OA and BC have same speed.

But if one is steeper, then that one is faster.

Assume OA is steepest → OA

✔ Answer: OA

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#### c) What was her average speed for the first 2 hours?

First 2 hours: from t=0 to t=2

At t=0: d=0
At t=2: d=4 km (assuming point B is at (2,4))

So total distance traveled: 4 km
Time: 2 hours

Average speed = total distance / total time = 4 km / 2 h = 2 km/h

Note: This is average speed, not average velocity.

Even though she walked only in one direction during first 2 hours (to point B), and didn’t return.

✔ Answer: 2 km/h

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3) The graph represents the journey of a steam train. The vertical axis represents the distance from the station in kilometres.

Graph shows:
- Starts at O (0,0)
- Rises to A
- Then flat to B
- Then falls to C
- Then flat to D
- Then rises to E
- Then falls to F

Let’s interpret:

- O to A: Train leaves station, accelerates
- A to B: Constant speed (flat line?)
Wait — no: in distance-time graph, flat line means stopped.

So:
- O to A: Moving away from station
- A to B: Stopped (at maximum distance)
- B to C: Returning toward station
- C to D: Stopped
- D to E: Moving away again
- E to F: Returning to station

Now answer questions:

#### a) How far did the train travel in the first 40 minutes?

Assume x-axis is in minutes.

Let’s suppose:
- At t = 40 min, the train is at point A
- Point A is at (40, 60) → distance = 60 km

So 60 km in 40 minutes

But question is: how far did the train travel — total distance covered, not displacement.

It went from 0 to 60 km → so distance traveled = 60 km

✔ Answer: 60 km

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#### b) Work out the speed of the train between A and B. Give your answer in kilometres per hour.

Point A: (40, 60)
Point B: (80, 60) → same distance, so stopped

So from A to B: distance doesn’t change → speed = 0 km/h

✔ Answer: 0 km/h

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#### c) What does the line DE represent?

- D to E: rising line → increasing distance from station
- So the train is moving away from the station

Also, since it's a straight line, it’s moving at constant speed

✔ Answer: The train is moving away from the station at a constant speed

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✔ Final Answers Summary:

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#### 1) Average Speeds:
- Graph A: 60 km/h
- Graph B: 60 km/h
- Graph C: 30 km/h

*(Assuming distance in km, time in minutes)*

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#### 2) Jamela’s Walk:
a)
- CD: She is stationary
- DE: She walks away from home
- EF: She walks back toward home

b) OA — she walked fastest there (steepest slope)

c) 2 km/h — average speed in first 2 hours

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#### 3) Steam Train Journey:
a) 60 km — distance traveled in first 40 minutes

b) 0 km/h — speed between A and B (train stopped)

c) The train is moving away from the station at a constant speed

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