This is probably easy but I am completely stumped.

Given that the acceleration vector is a(t)=(-9cos(-3t))I+(-9sin(-3t))J+(-5t)K, the initial velocity is v(0)=I+K, and the initial position vector is X(0)=I+J+K, compute:

A. The velocity vector v(t)=_____I+_____J+_____K

B. The position vector X(t)=_____I+_____J+_____K

*note the coefficients in your answer must be entered in the form of the expressions in the variable t

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If the acceleration vector is the derivative of the velocity vector then would you just integrate to work backwards? I tried that but was wrong.