1. The problem statement, all variables and given/known data A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical acceleration of the rocket is given by ay=(2.90m/s^3)t, where the +y-direction is upward. What is the height of the rocket above the surface of the earth at t = 10.0 s ? Express your answer with the appropriate units. 2. Relevant equations v=dy/dt a=dv/dt y =v*t + (1/2) a*t^2 3. The attempt at a solution I drew a diagram. A rocker going up, from 0 -10 a= 2.90 *t from 10 and higher, a = -9.802 I integrated the acceleration from 0-10, and got the velocity to be 2.90/2 t^2 I used the kinematics formula for position, height in this case and got h= 2.90/2 t^3 + .5*-9.802*t^2 = 959.9 its not correct, what am I doing wrong? What should I do instead and why?