Answer to Question #284645 in Trigonometry for Charles

Question #284645

1. A commercial ship is heading due North


at 12.0 km/hr but drifts westward at 5.0


km/hr. determine the magnitude and


direction of the resultant velocity of the


ship.




2. Find the resultant of the following


vectors.


A = 30 m at 30° with the x-axis


B = 20 m at 120° with the x-axis


C = 10 m at 225° with the x-axis.


1
Expert's answer
2022-01-05T17:14:59-0500

1.

Find the magnitude of the resultant velocity from the Pythagorean theorem:


"v_{res}=\\sqrt{v_x^2+v_y^2}=\\sqrt{(5\\ \\dfrac{km}{h})^2+(12\\ \\dfrac{km}{h})^2}=13\\ \\dfrac{km}{h}."

Find the direction of the resultant velocity from the geometry:



"\\theta=\\tan^{-1}(\\dfrac{v_y}{v_x})=\\tan^{-1}(\\dfrac{12\\ \\dfrac{km}{h}}{5\\ \\dfrac{km}{h}})=67.4^{\\circ}\\text{ N of W}"



2.


"\\overrightarrow{OA}=\\langle30\\cos 30\\degree, 30\\sin 30\\degree\\rangle"

"\\overrightarrow{OB}=\\langle20\\cos 120\\degree, 20\\sin 120\\degree\\rangle"

"\\overrightarrow{OC}=\\langle10\\cos 225\\degree, 10\\sin 225\\degree\\rangle"

"\\overrightarrow{R}=\\overrightarrow{OA}+\\overrightarrow{OB}+\\overrightarrow{OC}"

"=\\langle15\\sqrt{3}-10-5\\sqrt{2}, 10+10\\sqrt{3}-5\\sqrt{2}\\rangle"

"\\approx\\langle8.90969, 20.24944\\rangle"

"|\\overrightarrow{R}|=\\sqrt{(15\\sqrt{3}-10-5\\sqrt{2})^2+(10+10\\sqrt{3}-5\\sqrt{2})^2}"

"\\approx22.122895"

"R_x=8.90969>0, R_y=20.24944>0"

Then it is the first Quadrant and "0\\degree\\leq\\theta\\leq 90\\degree"


"\\tan \\theta=\\dfrac{R_y}{R_x}"


"\\theta=\\tan^{-1}\\dfrac{10+10\\sqrt{3}-5\\sqrt{2}}{15\\sqrt{3}-10-5\\sqrt{2}}\\approx66.25\\degree"

22.123 m at 66.25° with the x-axis


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