For the answer to the question above, let coordinates of boat B at 10 am at the buoy be (0,0)
then co-ordinates of boat A is (3,0)
let dist. between A & B be z, then at time t hrs after 10
z² = (3+3t)² +(5t)² = 34t² +18t +9,
value 112.5 at t = 1.5
2z.dz/dt = 2*34t +2*9
dz/dt = (34t +9)/z
= (34*1.5 + 9)/√112.5
= 5.6569 mph