A Quick Trip Part 2 of 6 - Position Function Since we have the velocity and time functions, we can reconstruct (a) What is the position function for this velocity? r(t) = ( + 3) E+ + 1 (b) What is the position of your destination B? r = IN (c) Rewrite this in physical form. Submit Forfeit Part 1 of 6 You have sailed from town A to town B over a time of 20 hours. Choose your origin to be the launch point of the boat at town A. v(t) = ( 0.012 mph/hr2 t2 - 0.5 mph/hr t + 4.62 mph ) E + 2.5 mph N The first step in a kinematics problem should be to find the initials conditions. (a) What is the initial position? F = 0 m Somi E + 0m Som N (b) What is the initial velocity? V1 = 4.62 mph| | 4.62 mph E + 2.5 mph 2.5 mph N (c) Write this in physical form. G, = [0 mph x 5.25 mph ( 3.98 x 0.879 E + (No Response) 0.476 NQues. Part(a) Here velocity function is given tes! V' = (0.012 + 2 - 0.5+ + 4.62 ) E + 2.5 1 We know that y = di dt di = V. di Now integrate both side : E ( t ) = / V.dt = 1 1 1 0 . 012 + 2 - 0.5+ + 4.62 ) E + 2.5m ) dt E (t ) = 10.01) +3 - 0.5+2 + 1 4-6 2 ) t ) E + 2.5 try 3 E / H = ( 4.62 + - 0. 25 +2 + 0. 00 4+3) E + 2-5 t m Part(b): Now wee have to find out position after t= 20 he Put t = zohr 2 ( 20 ) = | 4.62 X20 - 0.25 ( 20 ) 2 + 0. 104 ( 20 ) 3 ) E + ( 25) ( 20 ) m 2 ( 20 ) = 24. 4 E + 50 N Hence position of final destination B is 24.4 E tSON. Part ( c) : 121 = magnitude of vector = 1 ( 24.4) 2 +(50 ) 2 12'1 = 55.63. Now 2 ( 20 ) = 24.4 E + 50 M Take common 55-63 ( 20) = = ( B ) = 55. 63 ( 0.43 86 E+ 0.898 TH) Hence physical from of 2 is 55.63 (0.4386E + 0.8987)