Question

Ships A and B leave port together. For the next two hours, ship A travels at 15.0 mph in a direction 10.0 ∘ west of north while the ship B travels 45.0 ∘ east of north at 40.0 mph .

Part A What is the distance between the two ships two hours after they depart? Part B What is the speed of ship A as seen by ship B?

Answer 1

velocity of A is

vA= 15(-sin 10 degree x+cos 90 degree y) =-2.60472 x + 14.7721 y mph

velocity of B is

vB=40(cos 45 degree+sin 45 degree) = 28.2843 x + 28.2843 y mph

A) Distance vector between the two ships is

d=2(va-vb) = -61.778 x - 27.0243 y

magnitude is d = 67.43 miles

B) The velocity of ship A as seen by B is

v=d/2 = 33.72 mph