A ship,A moves due north at a speed of 30km/h while a second ship B moves at a speed of 20km/h on a bearing of 300' determine

Question

(a) the magnitude and direction of the velocity of B relative to A.
(B) the shortest distance of approach of the two ship,if ship B is 10km due east of A.?

Answer 1

You do not move on a "bearing" of 300 deg.
You mean a "heading" of 300 deg.
A bearing is a direction from you to a target.
Mathematicians drive me nuts.
Anyway:
a heading of 300 deg is 60 deg West of North
so
A is 30 km/h at 0 deg
and
B is 20 km /h at 60 deg left of that
so boat B is going 20 cos 60 North and 20 sin 60 west. That is 10 North and 17.3 km/h West.
(a) asks for B - A
north = 10 - 30 = -20 km/h or falling behind at 0 km/h
west = 17.3 - 0 = 17.3 km/h to the left (West, port side of A)
magnitude = sqrt (400 +17.3)^2
tangent of direction West of South = 17.3/ 20

Answer 2

typo
sorry sqrt (400 + 17.3^2)

Answer 3

B-A distance N = 0 - 20 t
B-A distance west = 17.3 t - 10
so
r = [400 t^2 + (17.3 t - 10)^2)]^.5
r = [400 t^2 + (299 t^ 2 - 346 t +100) ]^.5
r = [101 t^2 - 346 t + 100 ]^.5
dr/dt = 0 at desired time
dr/dt = 0 =.5 * (202 t -346) / [101 t^2 - 346 t + 100 ]^.5
so when t = 346/202
then go back and get r (after checking my arithmetic carefully)

Answer 4

obviously wrong
r = [400 t^2 + (299 t^ 2 - 346 t +100) ]^.5
r = [699 t^2 - 346 t + 100 ]^.5
dr/dt = 0 at desired time
dr/dt = 0 =.5 * (1400 t -346) / [101 t^2 - 346 t + 100 ]^.5
so when t = 1400/202 about 7 hours
then go back and get r (after checking my arithmetic carefully)