An automobile accelerates from rest at 2/m per second squared for 20 s. (continues on add details..)?

Question

The speed is then held constant for 20s , after which there is an acceleration of -3m per second squared until the automobile stops. What is the total distance traveled?

My T.A. explained but I dont understand!! I need help!

Answer 1

Break it into 3 pieces.

First piece

distance = 1/2 at^2

Gets you up to a velocity v=at

Second piece

distance = vt (using that velocity above)

Third piece

distance = v^2 / (2a) (using that same velocity, and a is now the absolute deceleration)

Add up the 3 pieces.

Answer 2

Second 1: 2m
Second 2: 4m (total 6m)
Second 3: 16m (total 22m)
etc...

When you reach the constant stage, multiply the last value by 20, and add it to the accumulated total.

For deceleration,
Second 1: -3m
Second 2: -9m
Second 3: -81m
etc...