To find the acceleration during interval C, we will use the formula for acceleration:

$a = \frac{\Delta v}{\Delta t}$

where:

• $\Delta v$ is the change in velocity,
• $\Delta t$ is the change in time.

From the graph:

1. Interval C corresponds to the portion where the velocity is decreasing (from about 15 seconds to 30 seconds).
2. The velocity at 15 seconds is 20 m/s.
3. The velocity at 30 seconds is 0 m/s.

Now, let's calculate:

$\Delta v = 0 \, \text{m/s} - 20 \, \text{m/s} = -20 \, \text{m/s}$ $\Delta t = 30 \, \text{seconds} - 15 \, \text{seconds} = 15 \, \text{seconds}$

Thus, the acceleration is:

$a = \frac{-20 \, \text{m/s}}{15 \, \text{seconds}} = -1.33 \, \text{m/s}^2$

Conclusion:

The acceleration during interval C is -1.33 m/s² (negative because the velocity is decreasing).

Would you like more details or have any questions? Here are some related questions:

1. What does a negative acceleration indicate in a motion graph?
2. How would the acceleration be different if interval C represented an increase in velocity?
3. Can you explain the relationship between the slope of the graph and acceleration?
4. What units are typically used to measure acceleration, and why?
5. How would the graph change if the object were accelerating uniformly?

Tip: The steeper the slope of a velocity-time graph, the greater the acceleration or deceleration.