Question

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.)

Answer 1

a) 38.8224 knots

b) 86° due north of east

c) 4.63 h

d) 86° due south of east

Step-by-step explanation:

The velocity of B relative to A:

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

`V_(A/B)=[2.7539,38.7246]knots/h`

`|V_(A/B)|=38.8224 knots/h`

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

`\alpha =atan((38.7246)/(2.7539) )= 86°`

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

`D_(A/B)=V_(A/B)*t`

`t=(D_(A/B))/(V_(A/B))=(180knots)/(38.8224knots/h) =4.63h`

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