Answer:
Kinda? Depends what the question is fully asking
Explanation:
Acceleration is a change in velocity. So I guess if the velocity of something is -2 m/s and its positively accelerating at a value of +1 m/s, then that means every second its velocity changes by +1m/s.
So that -2 m/s thing after one second will be going -1 m/s.
After another second it'll be going 0 m/s.
After another itll be going +1 m/s and so on.
So at one point for a brief moment, it can have an acceleration but be at 0 m/s velocity.
displacement is given by equation

now at t = 5 s the position is

similarly position at t = 9 s

so the displacement of object in given interval of time will be

time interval

now the average velocity will be given as


so its average speed is 252 m/s