Speed-Time Graphs Worksheets, Questions and Revision | MME - Free Printable

The image shows a speed-time graph with four distinct regions labeled A, B, C, and D. The vertical axis represents speed in meters per second (m/s), and the horizontal axis represents time in seconds (s).

We are likely being asked to find something related to motion — such as distance traveled, acceleration, or total displacement — based on this graph.

Since no specific question is stated, I will solve for:

1. Total distance traveled (which is the area under the speed-time graph).
2. Acceleration in each segment.
3. Describe what happens in each region (A, B, C, D).

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🔹 Step 1: Understand the Graph

- Speed-Time Graph: Area under the curve = Distance traveled
- Slope of the line = Acceleration (change in speed over time)

Let’s analyze each segment.

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🔹 Region A: From t = 0 to t = 15 s

- Speed increases from 0 m/s to 15 m/s.
- This is a straight line → constant acceleration.
- Shape: Right triangle

#### Distance (Area A):
$$
\text{Area}_A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 15 \times 15 = 112.5 \text{ m}
$$

#### Acceleration:
$$
a = \frac{\Delta v}{\Delta t} = \frac{15 - 0}{15 - 0} = 1 \text{ m/s}^2
$$

---

🔹 Region B: From t = 15 to t = 25 s

- Speed remains constant at 15 m/s.
- Horizontal line → zero acceleration.

#### Distance (Area B):
$$
\text{Area}_B = \text{base} \times \text{height} = (25 - 15) \times 15 = 10 \times 15 = 150 \text{ m}
$$

#### Acceleration:
$$
a = 0 \text{ m/s}^2
$$

---

🔹 Region C: From t = 25 to t = 30 s

- Speed increases from 15 m/s to 25 m/s.
- Straight line → constant acceleration.

#### Distance (Area C):
Shape: Triangle on top of rectangle? Actually, it's a trapezoid or triangle?

Wait — we can treat it as a trapezoid or compute as area under the line.

But since it starts at 15 and ends at 25 over 5 seconds:

$$
\text{Area}_C = \frac{1}{2} \times (\text{initial} + \text{final}) \times \text{time} = \frac{1}{2} \times (15 + 25) \times 5 = \frac{1}{2} \times 40 \times 5 = 100 \text{ m}
$$

Alternatively, think of it as a triangle added to a rectangle:

- Rectangle: $15 \times 5 = 75$
- Triangle: $\frac{1}{2} \times 5 \times (25 - 15) = \frac{1}{2} \times 5 \times 10 = 25$
- Total: $75 + 25 = 100$ m ✔

#### Acceleration:
$$
a = \frac{25 - 15}{30 - 25} = \frac{10}{5} = 2 \text{ m/s}^2
$$

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🔹 Region D: From t = 30 to t = 50 s

- Speed decreases from 25 m/s to 0 m/s.
- Straight line → constant deceleration.

#### Distance (Area D):
Triangle shape:
$$
\text{Area}_D = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (50 - 30) \times 25 = \frac{1}{2} \times 20 \times 25 = 250 \text{ m}
$$

#### Acceleration (deceleration):
$$
a = \frac{0 - 25}{50 - 30} = \frac{-25}{20} = -1.25 \text{ m/s}^2
$$

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🔹 Summary Table

| Region | Time Interval | Speed Change | Acceleration | Distance |
|-------|----------------|--------------|---------------|----------|
| A | 0–15 s | 0 → 15 m/s | +1 m/s² | 112.5 m |
| B | 15–25 s | 15 → 15 m/s | 0 m/s² | 150 m |
| C | 25–30 s | 15 → 25 m/s | +2 m/s² | 100 m |
| D | 30–50 s | 25 → 0 m/s | -1.25 m/s² | 250 m |

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🔹 Final Answers

#### ✔ Total Distance Traveled:
$$
\text{Total Distance} = A + B + C + D = 112.5 + 150 + 100 + 250 = \boxed{612.5} \text{ meters}
$$

#### ✔ Total Time:
50 seconds

#### ✔ Average Speed:
$$
\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{612.5}{50} = \boxed{12.25} \text{ m/s}
$$

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🔹 Interpretation of Motion

- A: Object accelerates uniformly from rest to 15 m/s.
- B: Constant speed motion (no acceleration).
- C: Further acceleration to 25 m/s.
- D: Uniform deceleration until stopping.

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🟩 Conclusion

If the task was to find the total distance traveled, the answer is:

> 612.5 meters

If other quantities were requested (e.g., acceleration in each phase), they are also provided above.

Let me know if you had a specific question (e.g., "What is the acceleration during region C?" or "How far did it travel in the first 20 seconds?"), and I can tailor the answer accordingly!