Find the position vector of a particle whose acceleration vector is a = (6t, 2) with an initial velocity vector (0, 0) and initial position vector (3, 0)

Question

Please I need your help with explanation. Thank you

Answer 1

this is just straightforward integration and evaluation at given points to determine the constant.
acceleration: a = (6t,2)

velocity: v = (3t^2,2t) + c
since v(0) = 0, c = (0,0), and thus v = (3t^2,2t)

position: s = (t^3,t^2) + c
since s(0) = (3,0), c = (3,0) and thus

s(t) = (t^3+3, t^2)

Answer 2

What else I need to do to get the answer, please?

Answer 3

huh? I gave you the position vector. Was there something else you wanted????