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!