Chapter 04, Problem 079 Two ships, A and B, leave port at the same time. Ship A

Question

Chapter 04, Problem 079 Two ships, A and B, leave port at the same time. Ship A travels northwest at 22 knots and ship B travels at 25 knots in a direction 40° 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 150 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.) (a) Number (b) Number (c) Number (d) Number Units Units Units Units

Explanation / Answer

Horizontal velocity of ship A, VAx = -22 x cos45 = -15.56 knots

Vertical velocity of ship A, VAy = 22 x sin45 = 15.56 knots

Velocity vector of A, VA = -15.56 i + 15.56 j

Horizontal velocity of ship B, VBx = -25 x cos40 = -19.15 knots

Vertical velocity of ship B, VBy = -25 x sin40 = -16.07 knots

Velocity vector of B, VB = -19.15 i - 16.07 j

A) VAB = sqrt (3.592 + 31.632) = 31.83 knots

B) Theta = arctan (3.59/31.63) = 6.48 degree

C) Time, t = 150/ 31.83 = 4.71 h

D) Since, VAB = - VBA

So, B stays at bearing of 6.48 degree west of south relative to A

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