A cruise ship it traveling at a speed of v3 = 22.6 ft/s. A speedboat with a late

Question

A cruise ship it traveling at a speed of v3 = 22.6 ft/s. A speedboat with a late passenger is heading toward the cruise ship at an angle of theta = 46/5*; As speed is r2 = 39.2 ft/s (Part A figure) 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. v = 27.6 ft/s Correct As the speedboat approaches the cruise ship, an ocean current that is moving at v2 = 6.40 ft/s and that is oriented at an angle phi = 36.6 psi . (Part B Figure) What are vx and vy the scalar components of the speedboat's velocity relative to the cruise ship as the speedboat crosses this ocean current? Express your answer numerically to three significant figures In feet per second separated by a comma. vx, vy = ft/s. Only image found

Explanation / Answer

Because you are finding the velocity relative to the cruise ship you must subtract 22.6 ft/s from the y component of the speedboats velocity. Don't worry about rounding until the end.
V=39.2 ft/s
Vx=39.2cos(46.5)=31.8151979

Vy=39.2sin(46.5)-22.6=0.300506109

now to find the magnitude use the pathagorean theorem to find the hypotenuse and tan-1(31.8152/.3005) to find angle.

V=(31.81522+.30052) @ tan-1(31.8152/.3005) degrees.

Answer to part a:

V=31.8 ft/s @ 89.5 degrees

Now we have to find the components of the ocean current and add them to the components of the speedboat.

Vox=6.4sin(36.5)=5.96298808

V0y=6.4cos(36.5)=2.32438661

subtract from the components we found from the boat earlier.

Vx=31.8151979-5.96298808=25.8522098

Vy=0.300506109-2.32438661=-2.0238805

use the same method as earlier to find the vector

V=(25.85220982+2.02388052) @ tan-1(25.8522098/-2.0238805) degrees.

V=25.9313101@ 94.47635933 degrees

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