Two ships, A and B, leave port at the same time. Ship A travels northwest at 21 knots and ship B travels at 25 knots in a direction 32° west of south. (1 knot = 1 nautical mile per hour; see Appendix D.) What are (a) the magnitude (in knots) and (b) direction (measured relative to east) of the velocity of ship A relative to B? (c) After how many hours will the ships be 180 nautical miles apart? (d) What will be the bearing of B (the direction of the position of B) relative to A at that time? (For your angles, takes east to be the positive x-direction, and north of east to be a positive angle. The angles are measured from -180 degrees to 180 degrees. Round your angles to the nearest degree.)

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lcmendozaf lcmendozaf · Answer 1 · 2019-10-06T13:45:11+02:00

Answer:

a) 38.8224 knots

b) 86° due north of east

c) 4.63 h

d) 86° due south of east

Explanation:

The velocity of B relative to A:

[tex]V_{A/B}=V_A-V_B =[-21*cos(45),21*sin(45)]-]-25*sin(45),-25*cos(45)][/tex]

[tex]V_{A/B}=[2.7539,38.7246]knots/h[/tex]

[tex]|V_{A/B}|=38.8224 knots/h[/tex]

For the angle: It's on the first quadrant, so:

[tex]\alpha =atan(\frac{38.7246}{2.7539} )= 86°[/tex]

For the amount of time, we will use the relative velocity calculated:

[tex]D_{A/B}=V_{A/B}*t[/tex]

[tex]t=\frac{D_{A/B}}{V_{A/B}}=\frac{180knots}{38.8224knots/h} =4.63h[/tex]

The bearing of B relative to A will have the opposite direction of A relative to B, so:

α = -86° This is 86° due south of east