https://file.aiquickdraw.com/mathbot/file/1729437803083mcd09crh.jpg

The velocity-time graph in the image represents the motion of a particle. Let's break down the problem and solve the given questions step by step:

### 1. **What is the greatest acceleration of the particle?**

- **Acceleration** is the rate of change of velocity, which is the slope of the velocity-time graph.
- From the graph, observe that the greatest change in velocity occurs between **6s and 10s**. The velocity changes from **4 m/s** to **10 m/s** over this time period.
  
  \[
  \text{Acceleration} = \frac{\Delta v}{\Delta t} = \frac{10 \, \text{m/s} - 4 \, \text{m/s}}{10 \, \text{s} - 6 \, \text{s}} = \frac{6 \, \text{m/s}}{4 \, \text{s}} = 1.5 \, \text{m/s}^2
  \]
  **Greatest acceleration = 1.5 m/s².**

### 2. **What is the total distance traveled by the particle?**

- The distance traveled is the **area under the velocity-time graph**.
- The graph consists of three distinct areas: two triangles and one rectangle.

  - **Area from 0 to 2 seconds** (triangle):
    \[
    \text{Area} = \frac{1}{2} \times 2 \, \text{s} \times 4 \, \text{m/s} = 4 \, \text{m}
    \]
  - **Area from 2 to 6 seconds** (rectangle):
    \[
    \text{Area} = 4 \, \text{m/s} \times 4 \, \text{s} = 16 \, \text{m}
    \]
  - **Area from 6 to 10 seconds** (triangle):
    \[
    \text{Area} = \frac{1}{2} \times 4 \, \text{s} \times (10 \, \text{m/s} - 4 \, \text{m/s}) = \frac{1}{2} \times 4 \times 6 = 12 \, \text{m}
    \]

  Now, sum the areas:
  \[
  \text{Total distance} = 4 \, \text{m} + 16 \, \text{m} + 12 \, \text{m} = 32 \, \text{m}
  \]
  **Total distance traveled = 32 m**.

### 3. **What is the average speed of the particle?**

- The **average speed** is the total distance traveled divided by the total time taken.
  
  \[
  \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{32 \, \text{m}}{10 \, \text{s}} = 3.2 \, \text{m/s}
  \]
  **Average speed = 3.2 m/s**.

### Final Answers:
1. **Greatest acceleration** = 1.5 m/s²
2. **Total distance traveled** = 32 m
3. **Average speed** = 3.2 m/s

Would you like more details or explanations? 

Here are 5 related questions for further exploration:
1. How do you find instantaneous velocity from a velocity-time graph?
2. What happens to the distance if the velocity-time graph includes negative velocities?
3. How can you calculate displacement from a velocity-time graph?
4. How would the graph change if there was constant deceleration?
5. What would happen to the average speed if the particle traveled in the negative direction?

**Tip:** Remember that the area under a velocity-time graph always represents displacement or distance traveled, depending on whether the velocity is positive or negative!

What is the greatest acceleration, total distance traveled, and average speed of the particle based on the velocity-time graph?

Solve the velocity-time graph problem by determining the greatest acceleration, total distance traveled, and average speed. This exercise uses kinematic equations and concepts, including the area under the graph to find the distance, and applies to Grades 9-11 physics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation