Ships A and B leave port together. For the next two hours, ship Atravels at 20 mph in a direction 30° west of north while theship B travels 20° east of north at 25 mph.
a. What is the distance between the two ships two hours after theydepart?
b. What is the speed of ship A as seen by ship B?

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

i think it would be 5