Answer:
remember that:
a*i + b*j + c*k can be written as a vector: (a, b, c)
We know the acceleration of the particle, and we want to find the velocity of the particle, so we just need to integrate two times.
a(t) = (8*t, sin(t), cos(2*t))
integrating that, we get:
V(t) = ( (1/2)*8*t^2, -cos(t), sin(2*t)/2) + v0
where v0 is the vector that defines the velocity at t = 0
in the question you wrote:
V(0) = i
so i suppose that this is:
V(0) = (1, 0, 0)
Then the velocity equation gives:
V(t) = ( (1/2)*8*t^2, -cos(t), sin(2*t)/2) + (1, 0, 0)
V(t) = (4*t^2 + 1, -cos(t), sin(2*t)/2)
Now to get the position equation, we integrate it again
r(t) = ((4/3)*t^3 + t, -sin(t), -cos(2*t)/4) + r0
where r0 is the initial position, in the question you wrote:
r(0) = j
so we get:
r(0) = (0, 1, 0)
replacing that we get:
r(t) = ((4/3)*t^3 + t, -sin(t), -cos(2*t)/4) + (0, 1, 0)
r(t) = ((4/3)*t^3 + t, -sin(t) + 1, -cos(2*t)/4)
Writing this in the same notation than in the question, we get:
r(t) = [(4/3)*t^3 + t}*i + [-sin(t) + 1]*j + [-cos(2*t)/4]*k
b) Now we want to graph this:
The image isn't really good but can be used to understand the motion of the particle.