velocity and position vectors?

Question

Given that the acceleration vector is a (t ) =
(−9 cos (3t )) I + (−9 sin (3t )) J + (−2t ) K, the initial
velocity is v (0) = I + K, and the initial position vec-
tor is X (0) = I + J + K, compute:
A. The velocity vector v (t ) = ___I+ ___J+ ___K
B. The position vector X (t ) = ___+ ___J+ ___K

Answer 1

The velocity vector is the integral of the acceleration vector
so V=-3sin(3t) I +3cos(3t) J -t^2 K +C where C is a constant vector
At t=0 V(0)= 3J+ C= I+k so C = I-3J+K and
v(t) -3sin(3t)I+3cos(3t)J -t^2K +I-3J+K
The position vector is the integral of the velocity vector

x(t) = cos(3t) I +sin(3t) J -t^3/3K +[I-3J+K]t +C where C is a constante vector
x(0)= I+C = I+J+K so C= J+K
so x(t) = [cos(3t)+t]I+[sin(3t) -3t+1]J +[-t^3/3+t+1]K