Question

See the sketch below (which is not to scale). Ships A and B leave port P (stationary) at the same time. A sails at 17 degrees south of east at a speed of 20 km/hr. B sails at 10 degrees west of north at 30 km/hr. Take east x, and northy. (a) Set up proper vector expressions (which means with & and y components) for the velocities of A and B as seen by P: vP and vap. (10 points). (b) From your results in (a), get an expression for the velocity of B as seen by A, VBA (5 points) What is the speed of B as seen by A? (2 points) What is the direction of travel of B as seen by A, as measured in degrees from east? (2 (c) (d) points) (e) After 4 hours, how far apart will the ships be from each other? (1 point) North (y) 30 km/hr T > East (x) Port 17 20 km/hr

Answer 1

V_AP = V_A cos (17)i^- V_A sin (17) J^ V_BP = -V_B sin (10) i^+ V_B cos (10) j^ V_AP = 20 cos (17) i^- 20 sin (17) j^ V_AP = 19.12 i^- 5.85 j^ V_A = 20 km/hr V_B = 30 km/Hr similarly V_BP = -30 sin (10) i^+ 30 cos (10) j^ V_BP = -5.21 i^+ 29.5 j^ \r\n b) V_BA = V_B - V_A = -5.2 i^+ 29.5 j^- 19.5 j^- 19.12 i^+ 5.85 j^ V_BA = -24.4 i^+ 35.35 j^ c) (V_BA) = squareroot (24.4)^2 + (35.35)^2 = 42.95 km/hr d) tan theta = 35.35/- 24.4 theta = -55.38 degree e) As we know that d = \V_BA Vertical-bar times t = 42.95 times 4 = 171.8 km/hr