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This worksheet features a distance-time graph illustrating a car's journey, accompanied by questions to test understanding of velocity and motion.

Hand-drawn distance-time graph showing car motion segments labeled O through G with physics questions below.

Hand-drawn distance-time graph showing car motion segments labeled O through G with physics questions below.

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Show Answer Key & Explanations Step-by-step solution for: This worksheet is about position-time and velocity-time graphs ...
Let's analyze the position vs. time graph provided in the image and answer each question step by step.

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Understanding the Graph



- The vertical axis represents distance (in kilometers).
- The horizontal axis represents time (in hours).
- The graph shows how the position of a car changes over time.
- The slope of the graph at any point gives the velocity:
- Positive slope: moving forward (+x direction)
- Zero slope: stopped
- Negative slope: moving backward

We'll go through each question:

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1) What is the car doing from “O” to “A” and from “F” to “G”?



- From O to A:
The graph shows a straight line going upward from (0,0) to (1,24). This means the car is moving forward at a constant velocity (since the slope is constant).

- From F to G:
The graph goes from (11,4) to (15,20), again with a positive slope, so the car is moving forward at a constant velocity.

Answer:
The car is accelerating or moving at constant velocity? Wait — actually, since it's a straight line, it's moving at constant velocity.

So:
👉 From O to A: The car is moving forward at constant speed.
👉 From F to G: The car is moving forward at constant speed.

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2) What is happening between “A” to “B”, “C” to “D”, and “E” to “F”?



Look at the segments:

- A to B: From (1,24) to (4,24) → horizontal line → no change in position
→ Car is stopped

- C to D: From (6,16) to (8,16) → also horizontal → stopped

- E to F: From (9,4) to (11,4) → horizontal → stopped

Answer:
The car is stopped during all these intervals.

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3) What is the car’s average velocity between “O” and “A”?



- Distance traveled = 24 km – 0 km = 24 km
- Time taken = 1 hour – 0 hours = 1 hour
- Average velocity = displacement / time = 24 km / 1 h = 24 km/h

Answer: 24 km/h

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4) How long is the car stopped?



Find all flat (horizontal) segments where velocity = 0:

1. A to B: 4 – 1 = 3 hours
2. C to D: 8 – 6 = 2 hours
3. E to F: 11 – 9 = 2 hours

Total stopped time = 3 + 2 + 2 = 7 hours

Answer: 7 hours

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5) What is the car’s average velocity between “B” and “C”?



- Position at B: 24 km
- Position at C: 16 km
- So displacement = 16 – 24 = –8 km
- Time: 6 – 4 = 2 hours
- Average velocity = displacement / time = –8 km / 2 h = –4 km/h

Negative sign means moving backward.

Answer: –4 km/h (or 4 km/h backward)

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6) What is the car’s average velocity between “D” and “E”?



- D: (8,16), E: (9,4)
- Displacement = 4 – 16 = –12 km
- Time = 9 – 8 = 1 hour
- Average velocity = –12 km / 1 h = –12 km/h

Answer: –12 km/h

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7) How far did the car travel from “O” to “B”?



This asks for distance traveled, not displacement.

- From O to A: 24 km (forward)
- From A to B: no movement → 0 km

So total distance = 24 km

Note: Even though the car stops at A to B, it doesn't move.

Answer: 24 km

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8) What is the car doing from “B” to “C” and “D” to “E”?



- B to C: From (4,24) to (6,16) → downward slope → moving backward at constant speed (straight line)
- D to E: From (8,16) to (9,4) → also downward → moving backward at constant speed

Answer:
The car is moving backward at constant velocity in both cases.

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9) When is the car exceeding a 10 km/hr speed limit?



Speed = absolute value of velocity. We need to find when |velocity| > 10 km/h.

Check each segment:

#### a) O to A:
Velocity = 24 km/h → > 10 km/hexceeding

→ Time: 0 to 1 hour

#### b) A to B:
Stopped → 0 km/h → not exceeding

#### c) B to C:
Velocity = –4 km/h → speed = 4 km/h → not exceeding

#### d) C to D:
Stopped → 0 km/h → not exceeding

#### e) D to E:
Velocity = –12 km/h → speed = 12 km/h → > 10 km/hexceeding

→ Time: 8 to 9 hours

#### f) E to F:
Stopped → 0 km/h → not exceeding

#### g) F to G:
From (11,4) to (15,20):
Displacement = 20 – 4 = 16 km
Time = 15 – 11 = 4 h
Velocity = 16/4 = 4 km/h → speed = 4 km/h → not exceeding

Answer:
The car exceeds 10 km/h during:
- 0 to 1 hour (O to A)
- 8 to 9 hours (D to E)

So:
👉 Between 0–1 hours and 8–9 hours

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Final Answers:



1) Moving forward at constant speed from O to A and from F to G
2) Car is stopped during A to B, C to D, and E to F
3) 24 km/h
4) 7 hours
5) –4 km/h
6) –12 km/h
7) 24 km
8) Moving backward at constant speed
9) From 0 to 1 hour and from 8 to 9 hours

Let me know if you'd like this summarized in a table or explained further!
Parent Tip: Review the logic above to help your child master the concept of velocity time graph worksheet.
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