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 26 knots in a direction 34° 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 200 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.)

Explanation / Answer

here,

Vae is velocity of ship A with respect to the earth= 2145 deg

this takes north as 0 and counts CCW
then Vbe= 26146 deg (B with respect to earth)

we want Vab
so
Vae + Veb = Vab
and Veb is the opposite of Vbe which is either -26146
or better, 26326

Vab = 2145 deg + 26326

draw these two vectors as sides of a parallelogram and the diagonal from the origin will be Vab
use trig to solve for magnitude and direction

Vab = sqrt(21^2 + -26^2)
Vab = 16.32 knots

= arctan(y/x)
= arctan(326/45)

I get 16.32-7.24 deg

or

16 knots at 7 deg east of north

Now d= 200 nautical mile apart

S = D / T

T = D / S = 200 / 16.32

T = 12.25 hr

Now bearing of B,

D = Vba * T
= -16.32 / 12.25
= - 1.33 nautical mile, 360-7 = 353

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