Since the slope of the line in velocity-time graph is constant, the acceleration is constant.

The particle has negative velocity after t=10 s. This says that the particle is turning around at t=10 s.

The area under the graph gives the distance travelled by the particle. Since area under graph when velocity is positive is not equal to the area under graph when velocity is negative, distance covered when velocity is positive is not equal to the distance covered when velocity is negative. Thus displacement is not zero.

Since area under the graph from t=0 to t=10 s is equal to the area under the graph from t=10 to t=20 s, distance covered will be equal in both time intervals. Thus, average speed in both time intervals is equal. Therefore, only options A and D are correct.