Answer:
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)
