Answer the previous questions for the motion described by the graph of Fig.

The figure below shows x versus t graph of an object:

In the time interval OA, x increases non-linearly over time and tries to saturate at point A, therefore the velocity of the object reaches closer to zero about A. If one assumes that the curve OA is parabolic, one can assure that the slope of the graph in interval OA will be constant, and therefore the velocity time graph will be a straight line with a positive magnitude.
For a straight velocity time graph in interval OA, the acceleration would be a constant because the slope of the velocity time graph will be a constant values.
In the time interval AB , x remains constant over time t, and the slope of the graph will be zero accounting for no change in x . Therefore the velocity v_x in the interval AB will remain at zero, and so will be the acceleration.
In the time interval BC, x varies non-linearly with time again. It is important to note that the slope of x versus t curve increases with time, however if the change is at the constant rate one obtains a straight line in velocity time curve.
For a straight velocity time graph in interval BC , the acceleration would be a constant because the slope of the velocity time graph will be a constant values.
In the time interval CD, the x changes linearly with time (straight line). This ensures that slope of the graph remains constant and the velocity v_x has a positive constant value. For a constant velocity time graph, the acceleration tie would be zero accounting for the fact that there is no change in velocity.
The velocity time graph is shown below:

It is important to note that the slope of line in interval OAis not equal to the slope of line in interval BC. This highlights the difference in the curvature in x versus t graph.
The acceleration time graph is shown below:

It can be seen in the figure above that the acceleration is constant in the interval OA and BC, but is equal to zero in the interval AB and CD .