Answer to Question #138929 in Physics for hannah

Question

Q5) A plane, traveling with a velocity relative to the air of 320 km/h [28° S of W], passes over Winnipeg. The wind
velocity is 72 km/h [S]. Determine the displacement of the plane from Winnipeg 2.0 h later.

Answer 1

$$\bold{V=v_p+v_w}\\=(320\cos{28},320\sin{28})+(0,72)\\=(282.5,222.2)\frac{km}{h}$$

$$s_x=282.5(2)=565\ km\\s_y=222.2(2)=444.4\ km$$

The displacement is

$$s=\sqrt{565^2+444.4^2}=720\ km$$

$$\theta=\arctan{\frac{444.4}{565}}=38\degree\text{ S of W}$$