https://file.aiquickdraw.com/mathbot/file/1726313208553fy5i4dri.jpg

Let's go through the questions based on the speed-time graph provided.

### Observations from the graph:
- The graph has 3 segments: **OA**, **AB**, and **BC**.
- **OA**: A straight line showing increasing speed from 0 to 6 m/s over 4 seconds.
- **AB**: A horizontal line showing constant speed (6 m/s) from 4 to 10 seconds.
- **BC**: A line showing decreasing speed from 6 m/s to 0 m/s from 10 to 16 seconds.

### Answers to the Questions:
**a) What type of motion is represented by OA?**
- The motion represented by **OA** is **uniform acceleration**, as the speed is increasing at a constant rate over time.

**b) What type of motion is represented by AB?**
- The motion represented by **AB** is **uniform motion**, because the speed is constant (6 m/s) throughout this period (no acceleration or deceleration).

**c) What type of motion is represented by BC?**
- The motion represented by **BC** is **uniform retardation** (or deceleration), as the speed is decreasing at a constant rate over time.

**d) Find out the acceleration of the body.**
- Acceleration is calculated from the slope of the **OA** segment.
  \[
  a = \frac{\Delta v}{\Delta t} = \frac{6 \, \text{m/s} - 0 \, \text{m/s}}{4 \, \text{seconds}} = \frac{6}{4} = 1.5 \, \text{m/s}^2
  \]
  The acceleration of the body is **1.5 m/s²**.

**e) Calculate the retardation of the body.**
- Retardation (negative acceleration) is calculated from the slope of the **BC** segment.
  \[
  r = \frac{\Delta v}{\Delta t} = \frac{0 \, \text{m/s} - 6 \, \text{m/s}}{16 \, \text{s} - 10 \, \text{s}} = \frac{-6}{6} = -1 \, \text{m/s}^2
  \]
  The retardation of the body is **1 m/s²**.

**f) Find out the distance traveled by the body from A to B.**
- The distance from **A** to **B** is given by the area under the graph from **A** to **B**.
  \[
  \text{Distance} = \text{Speed} \times \text{Time} = 6 \, \text{m/s} \times (10 \, \text{s} - 4 \, \text{s}) = 6 \times 6 = 36 \, \text{m}
  \]
  The distance traveled from **A** to **B** is **36 meters**.

### Do you have any further questions or need more details?

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Here are 5 relative questions for expanding your understanding:
1. How do we calculate the total distance traveled by the body?
2. What would the graph look like if the body had a non-uniform acceleration?
3. How does the slope of the graph indicate acceleration or deceleration?
4. What is the difference between uniform and non-uniform motion?
5. How can we find the average speed from the speed-time graph?

**Tip:** The area under a speed-time graph represents the distance traveled by the object. For non-linear graphs, you might need to break the area into simple shapes to calculate the total distance.

Study the speed-time graph of a body given here and answer the following questions: a) What type of motion is represented by OA? b) What type of motion is represented by AB? c) What type of motion is represented by BC? d) Find out the acceleration of the body. e) Calculate the retardation of the body. f) Find out the distance travelled by the body from A to B.

This problem analyzes the motion of a body based on a speed-time graph, covering concepts such as uniform acceleration, retardation, and distance calculation. It includes detailed solutions for each part of the problem, suitable for students in Grades 9-10.

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