https://file.aiquickdraw.com/mathbot/file/17294373689814ezn4zgc.jpg

Let's break down the information from the velocity-time graph and the questions provided:

1. **Graph Overview**: The graph shows velocity (in m/s) on the y-axis and time (in seconds) on the x-axis. The particle starts at a velocity of 8 m/s at \(t = 0\) and decelerates uniformly to \(v = -2\) m/s at \(t = 8\) seconds.

2. **Acceleration of the Particle**:
   - Acceleration is the slope of the velocity-time graph.
   - The change in velocity (\(\Delta v\)) is \( -2 - 8 = -10 \, \text{m/s}\).
   - The time (\(\Delta t\)) over which this change happens is \(8 \, \text{s}\).
   - Thus, the acceleration is:
     \[
     a = \frac{\Delta v}{\Delta t} = \frac{-10}{8} = -1.25 \, \text{m/s}^2.
     \]
   It seems that the value in the box "0.5" is incorrect. The correct answer should be \(-1.25 \, \text{m/s}^2\).

3. **Total Distance Travelled**:
   - The distance travelled is the area under the velocity-time graph.
   - The area is divided into two regions: a triangle above the x-axis and one below the x-axis.
   - Area of the first triangle (from 0 to 6 seconds, above the x-axis):
     \[
     \text{Area} = \frac{1}{2} \times 6 \, \text{s} \times 8 \, \text{m/s} = 24 \, \text{m}.
     \]
   - Area of the second triangle (from 6 to 8 seconds, below the x-axis):
     \[
     \text{Area} = \frac{1}{2} \times 2 \, \text{s} \times 2 \, \text{m/s} = 2 \, \text{m}.
     \]
   - Total distance travelled = \(24 + 2 = 26 \, \text{m}\).
   The entered value of "16 m" seems incorrect. The correct distance should be \(26 \, \text{m}\).

4. **Displacement after 8 seconds**:
   - The displacement is the net area under the graph (accounting for direction).
   - The area above the x-axis (from 0 to 6 seconds) is \(+24 \, \text{m}\), while the area below the x-axis (from 6 to 8 seconds) is \(-2 \, \text{m}\).
   - Thus, the total displacement is:
     \[
     24 - 2 = 22 \, \text{m}.
     \]
     
Would you like more details on any of these answers or need further clarification?

### Related Questions:
1. How do you calculate average velocity from a velocity-time graph?
2. What is the significance of the area under a velocity-time graph?
3. How does acceleration relate to the slope of a velocity-time graph?
4. Can the displacement ever be greater than the total distance travelled?
5. How would the graph change if the particle had a constant positive acceleration?

### Tip:
Always remember that in velocity-time graphs, the slope gives the acceleration, and the area under the graph gives the displacement.

The velocity-time graph for a particle is shown below. What is the acceleration of the particle? What is the total distance travelled by the particle? The particle started at the origin. What is its displacement after 8 seconds?

Learn how to calculate the acceleration, total distance traveled, and displacement for a particle using a velocity-time graph. Understand the key kinematic principles involved, including acceleration as the slope and distance as the area under the graph.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation