Initially, the two ships A and B are 5 km apart with A due west of B. Ship A is sailing due south at 5 km/h and ship B is sailing due west at 8 km/h.

The velocity of A relative to B has a magnitude of `sqrt(5^2 + 8^2)`...

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Initially, the two ships A and B are 5 km apart with A due west of B. Ship A is sailing due south at 5 km/h and ship B is sailing due west at 8 km/h.

The velocity of A relative to B has a magnitude of `sqrt(5^2 + 8^2)` = `sqrt(25 + 64) = sqrt 89` km/h. The direction of the velocity of A relative to B is `tan^-1(5/8) = 32` degrees in the south-east direction.

Let the time after which A has bearings of 225 degrees to B be t.

`sin 45 = (-5 + 8*t)/D` and `cos 45 = (5*t)/D`

=> `1/sqrt 2` = `(-5 + 8*t)/D`

=> `D = (-5 + 8t)*sqrt 2`

`cos 45*sqrt 2 = 5*t/(-5 + 8t)`

=> `5t = -5 + 8t`

=> 3t = 5

=> `t = 5/3`

D = `(-5 + 8*5/3)*sqrt 2`

=> 11.78 km

**The velocity of A relative to B is `sqrt 89` km/h at 32 degrees in the south-east direction and the distance between the two when A has bearings of 225 degrees to B is 11.78 km.**