Speed-time graph depicting four distinct phases of motion: A (acceleration), B (constant speed), C (acceleration), and D (deceleration).

The image shows a speed-time graph with four distinct regions labeled A, B, C, and D. The x-axis represents time in seconds (s), and the y-axis represents speed in meters per second (m/s). We are likely being asked to analyze motion based on this graph—common tasks include calculating distance traveled, acceleration, or displacement.

Let’s go step by step to interpret and solve what is typically expected from such a graph.

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🔍 Step 1: Understand the Graph

- Speed-Time Graph: The area under the curve gives the distance traveled.
- The slope of the graph gives acceleration:
- Positive slope → acceleration
- Zero slope → constant speed
- Negative slope → deceleration

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📊 Break Down Each Segment

#### Segment A (0 to 10 s)
- Speed increases from 0 m/s to 15 m/s
- This is uniform acceleration
- Shape: Right triangle

Acceleration = change in speed / time
= (15 - 0) / (10 - 0) = 1.5 m/s²

Distance (Area A) = Area of triangle
= ½ × base × height = ½ × 10 × 15 = 75 meters

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#### Segment B (10 to 20 s)
- Speed remains constant at 15 m/s
- This is constant velocity

Acceleration = 0 m/s² (no change in speed)

Distance (Area B) = Rectangle
= base × height = (20 - 10) × 15 = 10 × 15 = 150 meters

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#### Segment C (20 to 30 s)
- Speed increases from 15 m/s to 25 m/s
- Uniform acceleration

Acceleration = (25 - 15) / (30 - 20) = 10 / 10 = 1.0 m/s²

Distance (Area C) = Trapezoid
= ½ × (sum of parallel sides) × height
= ½ × (15 + 25) × 10 = ½ × 40 × 10 = 200 meters

Alternatively: average speed × time = (15 + 25)/2 × 10 = 20 × 10 = 200 m

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#### Segment D (30 to 50 s)
- Speed decreases from 25 m/s to 0 m/s
- Uniform deceleration

Acceleration = (0 - 25) / (50 - 30) = -25 / 20 = -1.25 m/s²

Distance (Area D) = Triangle
= ½ × base × height = ½ × 20 × 25 = 250 meters

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✔ Total Distance Traveled

Add all areas:

- A: 75 m
- B: 150 m
- C: 200 m
- D: 250 m

Total distance = 75 + 150 + 200 + 250 = 675 meters

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🧠 Summary of Key Results

| Segment | Time Interval | Speed Change | Acceleration | Distance |
|--------|----------------|---------------|----------------|----------|
| A | 0–10 s | 0 → 15 m/s | 1.5 m/s² | 75 m |
| B | 10–20 s | Constant 15 m/s | 0 m/s² | 150 m |
| C | 20–30 s | 15 → 25 m/s | 1.0 m/s² | 200 m |
| D | 30–50 s | 25 → 0 m/s | -1.25 m/s² | 250 m |

✔ Total Distance Traveled = 675 meters

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💡 Final Answer

> The object travels a total distance of 675 meters over 50 seconds.

If the task was to find total distance, that's the answer. If it was to find acceleration in each phase, or displacement (which equals distance here since speed is positive), we’ve covered all.

Let me know if you'd like to calculate average speed, total displacement, or anything else!