Step 1:

Let the ships A and B be at the points Q and P at 12 noon (when t=0).

$PQ=100 km$

At time t, let the positions of the ships be Q' and P'.

The distance travelled by the ships would be 35t due east (ship A) and 25t due to north (ship B)

Therefore $PQ'=35t-100$ and

$PP'=25\;t$

The distance between the two ships at time t, is $P'Q'=s$

Step 2:

$s^2=(35t-100)^2+(25t)^2$

$\quad= (35t-100)^2+625\;t^2$ -----(i)

differentiating with respect to t,

$2s \large\frac{ds}{dt}$$=2(35t -100) \times 35 +1250 \;t$

$2s \large\frac{ds}{dt}$$=70(35t -100) +1250 \;t$

Step 3:

The rate of seperation at time t is

$\large\frac{ds}{dt}=\large\frac{1}{2s}$$\bigg[70 (35t-100)+1250\bigg]$

$\qquad=\large\frac{1}{s}$$\bigg[35 (35t-100)+625\;t\bigg]$-----(ii)

Step 4:

When the time is $4.00\;pm, t=4$

and the distance between them (from(i))

$s=\sqrt {(35 \times 4 -100)^2+(25 \times 4)^2}$

$\quad=\sqrt {1600+10000}$

$\quad=\sqrt {11600}$

$\quad= 20 \sqrt {29} km$

Step 5:

Therefore their rate of separation at this time (from (ii))

$\large\frac{ds}{dt}=\frac{1}{20 \sqrt {29}}$$(35(140-100)+2500)$

$\large\frac{ds}{dt}=\frac{1}{20 \sqrt {29}}$$(1400+2500)$

$\large\frac{ds}{dt}=\frac{3900}{20 \sqrt {29}}$

$\qquad=\large\frac{195}{\sqrt {29}}$$\;km/h$