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:


vres=vx2+vy2=(5 kmh)2+(12 kmh)2=13 kmh.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:



θ=tan1(vyvx)=tan1(12 kmh5 kmh)=67.4 N of W\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.


OA=30cos30°,30sin30°\overrightarrow{OA}=\langle30\cos 30\degree, 30\sin 30\degree\rangle

OB=20cos120°,20sin120°\overrightarrow{OB}=\langle20\cos 120\degree, 20\sin 120\degree\rangle

OC=10cos225°,10sin225°\overrightarrow{OC}=\langle10\cos 225\degree, 10\sin 225\degree\rangle

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

=1531052,10+10352=\langle15\sqrt{3}-10-5\sqrt{2}, 10+10\sqrt{3}-5\sqrt{2}\rangle

8.90969,20.24944\approx\langle8.90969, 20.24944\rangle

R=(1531052)2+(10+10352)2|\overrightarrow{R}|=\sqrt{(15\sqrt{3}-10-5\sqrt{2})^2+(10+10\sqrt{3}-5\sqrt{2})^2}

22.122895\approx22.122895

Rx=8.90969>0,Ry=20.24944>0R_x=8.90969>0, R_y=20.24944>0

Then it is the first Quadrant and 0°θ90°0\degree\leq\theta\leq 90\degree


tanθ=RyRx\tan \theta=\dfrac{R_y}{R_x}


θ=tan110+10352153105266.25°\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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment