Let ship A be the origin of the x-y plane. So we can give initial coords to A(0,0) and B(0,25). Let’s say that westward is the positive x direction, so the x coord increases with time. The picture shows westward as negative x values.
After time t, the new coords of A are (16t,0) and B are (0,25-20t). When t=0, we have the starting positions of A and B. The distance between them we can call r, and
r=√(x²+y²)=√((16t)²+(25-20t)²)=√(256t²+625-1000t+400t²)=
√(656t²-1000t+625).
The rate of change of r=dr/dt=½(1312t-1000)/√(656t²-1000t+625).
[The differentiation can also be done as follows:
r²=x²+y², 2rdr/dt=2xdx/dt+2ydy/dt, rdr/dt=xdx/dt+ydy/dy⇒
rdr/dt=16t.16-(25-20t).20=256t-500+400t=656t-500.
dr/dt=(656t-500)/r=(656t-500)/√(256t²+(25-20t)²)...]
When t=0 (1pm), dr/dt=½(-1000)/25=-20, that is, initially B is moving towards A at a speed of 20mph. t is measured in hours. Note that t=½, the distance AB=17 miles.
When t=½ (1.30pm), dr/dt=½(656-1000)/√(164-500+625)=-172/17mph=-10.12mph approx.
This is a negative value, which means that the distance between AB is shortening, that is, the velocity is negative. (At t=1000/1312 hrs or about 46 minutes, the instantaneous change of distance is 0mph. At t=1hr 15min, at 2.15pm, the velocity is 16mph, as expected, so the distance is continuing to widen because B has arrived at (0,0).)
Initial position of A at 1pm and position of B at 2.15pm are at the origin. A1 and B1 are the ships’ positions at 1.30pm; A2 and B2 their positions at minimal rate of change of length of the line joining them; A3 and B3 are the positions at 2.15pm. The lengths AB, A1B1, A2B2, A3B3 show the varying length (distance between the ships).