Answer to Question #284645 in Trigonometry for Charles

Question

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.

Answer 1

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