Answer:
The answers to the question are
a. The distance between the two ships two hours after they depart is 39.1 miles
b. The speed of ship A as seen by ship B is 12.14 m/h south and 15.32 m/h east or Ship B sees ship A has having a magnitude velocity of 19.5 m/h ∡38.4 ° South East
Explanation:
Firstly we list out the known variables as follows
Ship A velocity = 20 m/h
Ship A direction = 30 ° West of North
Ship B velocity = 25 m/h
Ship B direction = 20 ° East of North
Distance covered after 2 hours for ship A = 20×2 = 40 miles
Distance covered after 2 hours for ship B = 25×2 = 50 miles
Dx component of ship A = 40×sin (30) = -20 miles and Dy = 40×cos (30) = 34.64 miles
While for ship B, we have
Dx = 50×sin (20) = 17.1 miles and Dy = 50×cos (20) = 46.98 miles
Therefore, the distance between the two ships is
((Ship B Dx - Ship A Dx)^2 +Ship B Dy - Ship A Dy)^2)^0.5
=((17.1-(-20))^2 + (46.98-34.64)^2)^0.5 = 39.1 miles
The speed of ship A as seen by ship B can be found by
The vy of ship A relative B = 20 cos 50 = 12.86 m/h
Therefore relative to ship B ship A is moving at a speed of 12.86 m/h - 25 m/h = -12.14 m/h north and 20 sin 50 or 15.32 m/h east
or 12.14 m/h south and 15.32 m/h east having a magnitude of 19.5 m/h 38.4 ° South East
This is solved graphically by the attached diagram, the required vector is the line between the green and blue dots
