Question

+ Relative-Motion Analysis of Two Particles Using Translating Axes < 5 of 5 n Review Learning Goal: Part A To analyze the motion of a particle using a translating frame of reference. Two particles, A and B, are moving along arbitrary paths and are at positions r A and rb from a common origin. The relative position of point B with respect to A is designated by a relative-position vector ТВА and is specified with the equation A cruise ship is traveling at a speed of vi = 23.4 ft/s . A speedboat with a late passenger is heading toward the cruise ship at an angle of 0 = 52.5°; its speed is v2 = 36.6 ft/s . (Figure 1) What is v, the magnitude of the speedboat's velocity relative to the cruise ship? Express your answer numerically to three significant figures in feet per second. rb = ra +rBA Taking the time derivative of the relative-position equation above results in the following equation that relates the velocities of the particles, VA and VB, with the relative-velocity vector VBA ► View Available Hint(s) v = 23.0 ft/s VB=VA + VBA Taking the time derivative of the above relative velocity equation results in the following equation that relates the accelerations of the particles, a A and ab, with the relative- acceleration vector aBA Submit Previous Answers ав = ад +aBA Correct

Answer 1

4 Son” ) V1 = 23.4 HIS N36-6fts 0 2 (2. พบ 23-13 *V2 = -V2 osoituasino's .. V1 - siๆ 2 ) (2) /4 - Y2- Vi . - 3 4 - 6 Lnt (2.5 i + 36.6 siๆ 82.53 - 13.43 - 22.28 ) + C 36 Vฯ - - 1 Val = (21- 29 +(64432 Vฯ 2 2 2 98 V1 2 3

by V3=7.1 FHS 38" #zl= 8z + V3 :-V2 cosoitua sino - 13 Los41-va sine = -36-6 os 525it 36.6 sin sus ? 7.1 WS 38.1-7.1 sinar vo! --27.871 +24.665 ار .: 224' - vz! -V =-27.87 17 24.665 23-4 = -27.87 +1.265 SU, ra ry, 110 - 27.87 1,265 ): Vse, vy = - 27.9 1. 27 this Ary