Let's analyze the graph and the given problem.

The graph shows the position $s(t)$ of an object as a function of time $t$, and the task is to determine the nature of acceleration (negative, zero, or positive) on each of the intervals [0,10], [10,20], [20,30], and [30,40].

1. Acceleration on [0,10]:

   • In this interval, the graph is increasing and concave up, meaning that the slope (velocity) is increasing. Therefore, the acceleration is positive, not negative.

2. Acceleration on [10,20]:

   • In this interval, the graph is linear with a constant slope, indicating that velocity is constant. Since the rate of change of velocity is zero, the acceleration is indeed zero. This statement is correct.

3. Acceleration on [20,30]:

   • In this interval, the graph continues to increase but now shows concave down behavior, which indicates that the velocity is still increasing but at a decreasing rate. This means that acceleration is negative, not positive.

4. Acceleration on [30,40]:

   • In this interval, the graph continues to increase and is concave up again, indicating that the velocity is increasing at an increasing rate. Hence, the acceleration is positive. This statement is correct.

Summary of corrections:

• [0,10]: Positive (was incorrectly marked negative).
• [10,20]: Zero (correct).
• [20,30]: Negative (was incorrectly marked positive).
• [30,40]: Positive (correct).

Do you have any questions, or would you like further details? Here are some related questions:

1. How is acceleration related to the concavity of a position-time graph?
2. What does it mean when a position-time graph is concave up or concave down?
3. How do you determine the sign of acceleration from a velocity-time graph?
4. Can acceleration be zero when velocity is non-zero?
5. What is the physical significance of negative acceleration in motion?

Tip: Always remember, when analyzing acceleration from position graphs, positive acceleration corresponds to concave up curves, and negative acceleration corresponds to concave down.