Given that the acceleration vector is a (t) = (-4cos (2t),-4sin (2t), t)

the initial velocity is v (0) = (1,0,1),

and the initial position vector is r (0) = (1,1,1), compute:

A. The velocity vector v (t)

B. The position vector r (t)

a (t) = (-2sin2, - 2cos0, (1/2)

r (t) = (cos2, sin2, 1/6)

Step-by-step explanation:

By definition, given a position vector, r (t),

- The velocity vector v (t) is the derivative of the position vector.

v (t) = r' (t)

- The acceleration vector a (t) is the derivative of the velocity vector.

a (t) = v' (t) = r'' (t).

Knowing that integration is the reverse of differentiation, we can obtain the velocity and position vectors from the given acceleration vector.

a (t) = (-4cos (2t), - 4sin (2t), t)

Integrating with respect t, we have:

v (t) = (-4/2) sin (2t), (-4/2) (-cos (2t)), (1/2) t²)

at (1, 0, 1), a (t) = (-2sin2, - 2cos0, (1/2) (1))

= (-0.0698, - 2, 0.5)

Now, we integrate the velocity vector with respect to t to obtain the position vector.

r (t) = (cos (2t), sin (2t), (1/6) t³)

at (1, 1, 1)

r (t) = (cos2, sin2, 1/6)

= (0.999, 0.035, 0.167)