Two ships, A and 8, leave port at the same time. Ship A travels northwest at 23
ID: 1409692 • Letter: T
Question
Two ships, A and 8, leave port at the same time. Ship A travels northwest at 23 knots and ship 8 travels at 28 knots In a direction 34degree 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 8? (c) After how many hours will the ships be 140 nautical miles apart? (d) What will be the bearing of 8 (the direction of the position of 8) 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.) By accessing this Question Assistance, you will leam while you earn points based on the Point Potential Policy set by your instructor.Explanation / Answer
let's find "A" velocity but first let's set the following :
positive y axiss = north
negative y axiss = south
positive x axiss = west
negative x axiss = east
"A" velocity :
Northwest meanse : velocity making 45º with the horizontal and the vertical.
Va = 23*cos(45) i + 23*sin(45) j (knots)
"B" velocity :
Note 34 degrees west south means, 34 degrees with the horizontal
Vb = 28*cos(34) i - 28*sin(34) j (knots)
a) velocity of ship A relative to B :
23*cos(45) i + 23*sin(45) j - 28*cos(34) i + 28*sin(34) j
-6.95 i + 31.92 j (knots)
magnitude : sqrt(6.952 + 31.922) = 32.67 knots
direction = 77.72 north - east
c) to find the time when the ships are separate 160 nautical miles :
let's set a time : t
Da = 23t
Db = 28t
The angle that separates A from B = 45 + 34 = 79º
Using the law of cosines :
D^2 = (23t)^2 + (28t)^2 - 2*23t*28t*cos(79)
(D= 140 )
140^2 = 1313t^2 - 245.76 t^2
19600 = 1067.24 t^2
t = 4.28 hours
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