Stanton D. answered 11/01/21
Tutor to Pique Your Sciences Interest
So James H., you integrate the acceleration expression twice with respect to time. Each integration gives you one arbitrary constant; you may best insert these at each integration, using the "initial conditions".
So for the first integration, int[36t + 6]dt = (36/2)t^2 + 6t + C(1). This is the expression for velocity! So if v(0) = 11, we solve for C(1): 18*0^2 + 6*0 +C(1) = 11 ; I think you can handle that?
Then you integrate the complete expression for velocity again, which will require the constant C(2) in a similar fashion, to give you the position expression. Then you plug in to find your answer.
If this is all new material to you, it's time to find a regular tutor to walk you through integration and differentiation.
-- Cheers, --mr. d.