Please help with Part B. Will rate right after To analyze the motion a particle

Question

Please help with Part B. Will rate right after

To analyze the motion 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 r_B from a common origin. The relative position of point B with respect to A is designed by a relative-position vector r_BA and is specified with the equation r_B = r_A + r_BA Taking the time derivative of the relative-position equation above in the following equation that relates the velocities of the particles, v_A and v_B, with the relative-velocity vector v_BA. V_B = v_A + v_BA Taking the time derivative of the above relative velocity equation results in the following equation that relates the accelerations of the particles. a_A A cruise ship is traveling at a speed of v_1 = 24.6 ft/s. A speedboat with a late passenger is heading toward the cruise ship at an angle of theta = 47.5 degree ; its speed is v_2 = 35.8 ft/s. (Figure 1) What is v, the magnitude of the speedboat's velocity relative to the cruise ship? As the speedboat approaches the cruise ship, it encounters an ocean current that is moving at v_3 = 8.80 ft/s and that is oriented at an angle phi = 46.0 degree. (Figure 2) What are v_x and v_y, the scalar components of the speedboat's velocity relative to the cruise ship as the speedboat crosses this ocean current? This question will be shown after you complete previous question(s).

Explanation / Answer

Since you have asked only for second bit I presume that you have already solved first part.

Answers of first part

V21 = -21.18i -3.2j ft/s

The magnitude of this vector is

V21 = 24.3 ft/s

For b part, Velocity of sea current would also influence the boat velocity. Therefore,

V = V21 + V3

= (-24.18i - 3.2j) + (-8.8cos46 i - 8.8 sin46 j)

= -30.3i -9.53j

The magnitude of this vector is

V=(30.3^2 + 9.53^2)^0.5 =31.76 ft/s

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