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)

answer: group b.

step-by-step explanation:

many reason,

step-by-step explanation: