Two ships, A and B, leave port at the same time. Ship A travels northwest at 21

Question

Two ships, A and B, leave port at the same time. Ship A travels northwest at 21 knots and ship B travels at 28 knots in a direction 40° west of south. (1 knot 1 nautical mile per hour; see Appendix D.) (a) What is the magnitude the velocity of ship A relative to B? knots (b) What is the direction of the velocity of ship A relative to B? o east of north (c) After what time will the ships be 155 nautical miles apart? (d) What will be the bearing of B (the direction of B's position) relative to A at that time? o west of south

Explanation / Answer

here,

speed of ship A , vA = 21 kmots * ( - cos(45) i + sin(45) j)

vA = ( - 14.8 i + 14.8 j ) knots

speed of ship B , vB = 28 knots * ( - sin(40) i - cos(40) j)

vB = - 18 i knots - 21.45 j knots

a)

the speed of A relative to B , vAB = vA - vB

vAB = 3.2 i knots + 36.25 j knots

the magnitude of velocity of A relative to B , |vAB| = sqrt(3.2^2 + 36.25^2)

|vAB| = 36.4 knots

b)

the direction of velocity of ship A relative to B , theta = arctan(3.2/36.25) = 5.04 degree east of North

c)

the time taken , t = 155 /|vAB|

t = 155 /36.4 = 4.26 h

d)

the bearing of B relative to A is in opposite direction of theta

so, the bearing of B relative to A is 5.04 degree West of South

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