Answer:
Thus we find that velocity vector at time t is
(5t+15, 5t^2/2, 4t^2)
Step-by-step explanation:
given that acceleration vector is a funciton of time and at time t
[tex]a(t) = (5,5t, 8t)[/tex]
v(t) can be obtained by integrating a(t)
v(t) = [tex](5t, 5t^2/2, 4t^2)+(u_0,v_0,w_0)\\=(5t+15, 5t^2/2, 4t^2)[/tex]
Thus we use the fact that acceleration is derivative of velocity and velocity is antiderivative of acceleration.
The arbitary constant normally used for integration C is here C vector = initial velocity (u0,v0,w0)
Position vector can be obtained by integrating v(t)
Thus we find that velocity vector at time t is
(5t+15, 5t^2/2, 4t^2)