Two sea trawlers left a sea port simultaneously in two mutually perpendicular directions. Half an hour later, the shortest distance between them was 17 km and another 15 minutes later, one sea trawler was 10.5 km farther from the original than the other. Find the speed of each sea trawler

Assume the speed in km/hr be a and b.

Then, the distance travelled after half an hour = $$\ \frac{\ a}{2}$$ and $$\ \frac{\ b}{2}$$

The distance between them = $$\ \ \ \frac{\ 1}{2}\sqrt{\ a^2+b^2}$$  = 17

=> $$\sqrt{\ a^2+b^2}$$ = 34

=> $$a^2+b^2$$ = 1156  .....(1)

After 45 minutes, the difference of the distance travelled = $$\ \frac{\ 3}{4}\left(a-b\right)$$ = 10.5

=> a-b = 14   => $$a^2+b^2-2ab\ $$ = 196.......(2)

From 1 and 2, 2ab = 960 

$$a^2+b^2$$ +2ab = 1156+960= 2116

=> a+b=46

Hence a=30 and b=16