Two particles, A and B, are moving along arbitrary paths and are at positions rA

Question

Two particles, A and B, are moving along arbitrary paths and are at positions rA and rB from a common origin. The relative position of point B with respect to A is designated 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 results in the following equation that relates the velocities of the particles, vA and vB, with the relative-velocity vector v{BA}: {v}B={v}A+{v}{BA} A cruise ship is traveling at a speed of v1 = 23.6 ft/s. A speedboat with a late passenger is heading toward the cruise ship at an angle of theta1 = 46.0 degree; its speed is v2 = 39.0 ft/s. uploaded image What is v, the magnitude of the speedboat's velocity relative to the cruise ship?

Explanation / Answer

we have that v_boat = v_ship + v_boattoship so v_boattoship = v_boat-v_ship=v2-v1. the angle between them is 90-46=44(deg). magnitude of v_boattoship. v_boattoship^2=(v2^2+v1^2-2v1v2)(vector) so v^2=v2^2+v1^2-2v1v2*cos44 (magnitude). so v=sqrt(v1^2+v2^2-2v1v2*cos44)=27.5(ft/s)

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