Question

Find the position vector for a particle with acceleration, initial velocity, and initial position given below. a(t)=⟨6t,2sin(t),cos(2t)⟩

v(0)=⟨4,−2,−5⟩
r(0)=⟨−4,0,1⟩
r(t)=⟨

Answer 1

To get the velocity you integrate acceleration using the initial conditions. That is v(t) = v(0) + integral a(t) from 0 to t. This gives v(t) = (1, -2, 5) + (5/2 t^2, -2cos t + 2, 1/2 sin(2t)) = (5/2 t^2 + 1, -2cos t, 1/2 sin(2t) + 5). Similarly, for the position, you integrate velocity and use the initial conditions. That is r(t) = r(0) + integral v(t) from 0 to t. This gives r(t) = (5, 1, 2) + (5/6 t^3 + t, -2sin t, -1/4 cos(2t) + 5t + 1/4)= (5/6 t^3 + t + 5, -2sin t + 1, 1/4 cos(2t) + 5t + 9/4)