$$d = vi + 1/2at^2$$
where:
* d is the distance traveled (in meters)
* vi is the initial velocity (in meters per second)
* a is the acceleration (in meters per second squared)
* t is the time (in seconds)
In this case, the initial velocity is 0 m/s, the acceleration is a constant, and the time is 2 seconds. We can plug these values into the equation to find the distance traveled after 2 seconds:
$$d = 0 + 1/2(a)(2^2)$$
$$d = 2a$$
So after 2 seconds, the ball has traveled a distance of 2a meters.
To find the distance traveled in the next second, we can use the same equation, but this time we will use a time of 3 seconds:
$$d = 0 + 1/2(a)(3^2)$$
$$d = 4.5a$$
So in the next second, the ball will travel a distance of 2.5a meters.
