Distance Time Graphs Worksheet | Fun and Engaging PDF Worksheets - Free Printable

Let's solve each part of the Distance-Time Graphs worksheet step by step.

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1) Work out the average speed of the journey represented by each line

Average speed = Distance ÷ Time

#### Graph 1:
- Distance = 200 km
- Time = 4 hours
→ Average speed = $ \frac{200}{4} = 50 $ km/h

✔ Answer: 50 km/h

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#### Graph 2:
- Distance = 400 m
- Time = 20 seconds
→ Average speed = $ \frac{400}{20} = 20 $ m/s

✔ Answer: 20 m/s

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#### Graph 3:
- Distance = 16 miles (from 0 to 16 on y-axis)
- Time = 2 hours
→ Average speed = $ \frac{16}{2} = 8 $ mph

✔ Answer: 8 mph

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✔ Summary for Question 1:

- Graph 1: 50 km/h
- Graph 2: 20 m/s
- Graph 3: 8 mph

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2) Jameka’s Walk

#### Graph Description:
- X-axis: Time (hours), from 0 to 8
- Y-axis: Distance from home (km), from 0 to 12

Points:
- A(0,0) → B(3,12) → C(4,12) → D(5,6) → E(6,6) → F(8,0)

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#### a. Describe sections CD, DE, and EF

- CD: From time 4 to 5 hours, distance drops from 12 km to 6 km.
- She is moving towards home, so she is walking back.
- This is a decreasing slope → returning home at constant speed.

- DE: From time 5 to 6 hours, distance stays at 6 km.
- No change in distance → stopped (resting).
- So, stationary at 6 km from home.

- EF: From time 6 to 8 hours, distance decreases from 6 km to 0 km.
- She walks back home at constant speed.

✔ Answer:
- CD: Walking back towards home
- DE: Stopped (at rest)
- EF: Walking back home

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#### b. On which section did she walk fastest?

Speed = gradient (slope) of the line.

We compare slopes:

- AB (0–3 hrs): From (0,0) to (3,12) → $ \frac{12}{3} = 4 $ km/h
- BC (3–4 hrs): Flat → 0 km/h (stopped)
- CD (4–5 hrs): From (4,12) to (5,6) → $ \frac{6}{1} = 6 $ km/h (speed = 6 km/h)
- DE (5–6 hrs): Flat → 0 km/h
- EF (6–8 hrs): From (6,6) to (8,0) → $ \frac{6}{2} = 3 $ km/h

Fastest speed = CD at 6 km/h

✔ Answer: Section CD

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#### c. Average speed for first 2 hours

From A to point between A and B:
- At t = 2 hours, distance = ? Let's read from graph:
- AB goes from (0,0) to (3,12), so slope = 4 km/h
- In 2 hours: distance = $ 4 \times 2 = 8 $ km

So:
- Distance = 8 km
- Time = 2 hours
- Average speed = $ \frac{8}{2} = 4 $ km/h

✔ Answer: 4 km/h

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✔ Summary for Question 2:

- a. CD: returning home; DE: stopped; EF: walking home
- b. Fastest: CD
- c. Average speed (first 2 hrs): 4 km/h

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3) Steam Train Journey

#### Graph:
- X-axis: Journey time (minutes), 0 to 90
- Y-axis: Distance from station (km), 0 to 20
- Points: A(0,0), B(30,10), C(40,10), D(50,20), E(90,0)

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

At 40 minutes (point C), distance = 10 km

So, it traveled 10 km in first 40 minutes.

✔ Answer: 10 km

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#### b. Speed between A and B (in km/h)

- A to B: from 0 to 30 minutes → 30 minutes = 0.5 hours
- Distance: from 0 to 10 km → 10 km

Speed = $ \frac{10}{0.5} = 20 $ km/h

✔ Answer: 20 km/h

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#### c. What does line DE represent?

- D(50,20) to E(90,0)
- Distance decreases from 20 km to 0 km
- Time: 50 to 90 minutes → 40 minutes
- It’s a straight line → constant speed
- So, the train is returning to the station at constant speed

✔ Answer: The train is returning to the station at a constant speed

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✔ Summary for Question 3:

- a. 10 km
- b. 20 km/h
- c. The train is returning to the station at a constant speed

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🟩 Final Answers:

#### 1) Average Speeds:
- Graph 1: 50 km/h
- Graph 2: 20 m/s
- Graph 3: 8 mph

#### 2) Jameka’s Walk:
- a.
- CD: Returning home
- DE: Stationary (stopped)
- EF: Walking back home
- b. CD (fastest)
- c. 4 km/h

#### 3) Steam Train:
- a. 10 km
- b. 20 km/h
- c. Returning to station at constant speed

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