Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a (t) = 8t i + sin t j + cos 2t k, v(0) = i, r(0) = jr(t) = ?(b) On your own using a computer, graph the path of the particle.

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.

{3 + i, 3 - i, 3, -1}

step-by-step explanation: