Can someone verify this calculus problem solution for me? From the origin point O, a particle moves with velocity 4m/min, in the direction $30$ degrees north-east. After one minute, another particle B moves from O on the line $x + \sqrt3y = 0$, with velocity 6m/min, in the direction (BOA), which is an acute angle. Evaluate the rate of change of the distance between A and B after 2 mins, from moving the particle B.
I think X= 12 because 
X=x○ + dx/dt ×t 
= 0+4×(2+1) = 4×3 =12 
because it says after moving B and B moves after A with 1 min . 
then Y = y○ + dy/dt ×t 
= 0+6×2 = 12 
So dS/ dt = 5(root)3
am I right ???
I think X= 12 because 
X=x○ + dx/dt ×t 
= 0+4×(2+1) = 4×3 =12 
because it says after moving B and B moves after A with 1 min . 
then Y = y○ + dy/dt ×t 
= 0+6×2 = 12 
So dS/ dt = 5(root)3
am I right ???