To solve this problem, we need to analyze the given motion of the object, which is described as follows:

1. The object decelerates (negative acceleration) for 2 seconds.
2. Then, it moves at a constant velocity for 1 second.
3. Finally, it experiences a positive acceleration for 3 seconds.

Key points to analyze:

• Negative acceleration for 2 seconds: This implies that the velocity should be decreasing during the first 2 seconds.
• Constant velocity for 1 second: During this period (from 2 to 3 seconds), the velocity should remain flat.
• Positive acceleration for 3 seconds: After 3 seconds, the velocity should increase until the 6-second mark.

Now, let's examine the graphs:

1. Graph A:

   • The graph starts with a slight slope downward (indicating negative acceleration), followed by a flat section, and then an upward slope.
   • This matches the description: negative acceleration, constant velocity, and positive acceleration.

2. Graph B:

   • This graph shows an increase in velocity at the start, which contradicts the negative acceleration requirement.

3. Graph C:

   • This graph shows a steep decrease in velocity followed by an upward slope, but it lacks the constant velocity section.

4. Graph D:

   • This graph also shows a decrease, then a flat section, followed by a slight upward slope, but the shape of the acceleration phases is inconsistent.

Correct Answer:

• Graph A is the correct representation of the motion described.

Would you like further clarification or details on this problem?

Here are five related questions:

1. What is the physical meaning of acceleration being positive or negative?
2. How does constant velocity appear on a velocity-time graph?
3. How would the graph look if the object was at rest after deceleration?
4. What would change if the object accelerated for the entire 6 seconds?
5. How do we determine displacement from a velocity-time graph?

Tip: Always remember, the slope of a velocity-time graph represents acceleration.