Given an acceleration vector, initial velocity (u0, v0), and initial position (x0, y0), find the velocity and position vectors for t>=0.
a(t)=(cos(t), 2sin(t))
(u0, v0)=(0, 4)
(x0, y0)=(2, 0)
We can integrate the a(t) to get v(t) up to a constant vector, v, to be determined.
v(t) = ∫a(t) dt = ( sin(t), -2cos(t) ) + v, and so v(0) = ( sin(0), -2cos(0) ) + v = (0,-2) + v
But v(0) = (u0, v0) = (0, 4). Thus, (0,4) = (0,-2) + v, or v = (0,6). We have v(t) = (sin(t), -2cos(t)) + (0,6).
Adding in this constant vector, we get v(t) = ( sin(t), -2cos(t) + 6 ).
Now we can repeat the same for v(t). Integrating v(t) will give the position vector, r(t), up to a constant, r.
r(t) = ∫v(t) dt = ( -cos(t), -2sin(t) + 6t ) + r, and so r(0) = ( -cos(0), -2sin(0) + 6(0)) + r = (-1,0) + r
But r(0) = (x0, y0) = (2, 0). Thus, (2,0) = (-1,0) + r, or r = (3,0). We have r(t) =( -cos(t), -2sin(t) + 6t ) + (3,0).
Adding in this constant vector, we get r(t) = ( -cos(t) + 3, -2sin(t) + 6t ).
I hope this helps. Let me know if I can clarify any points.
