I totally forget how to do this at the moment.

acceleration is the derivative of velocity, and acceleration is given.
because there is acceleration, your velocity is not going to be constant.

so, you can integrate your acceleration " -.06t " to get -.03(t^2) +C
VELOCITY EQUATION , V(t)=-.03(t^2) +C , Set t=0 and solve initial condition .V=200

---
velocity is the derivative of position, so you can integrate your velocity equ to get the position equ

This post was edited by saber_x3 on Sep 8 2013 07:40pm

would the acceleration equation be x0 + v0t -.03t^2?

I think you mean the position equation, if so, still no
you gotta integrate the velocity equation
you'll have a polynomial with a 3rd power

your problem does not have a constant acceleration

This post was edited by saber_x3 on Sep 8 2013 07:52pm