Question #241016
Solve the FF problems step by steps

1. A person has a resultant vector of 5.0 m in the Direction 37 degrees North of East. How far north and how far east had he walked?

2. A plane flying due north at 100 meters per second is blown by a strong wind at 500 meters per second due east. What is the plane's resultant velocity?
1
Expert's answer
2021-09-23T08:30:20-0400

(1)

dx=dcosθ=5.0 mcos37=4.0 m [E],d_x=dcos\theta=5.0\ m\cdot cos37^{\circ}=4.0\ m\ [E],dy=dsinθ=5.0 msin37=3.0 m [N].d_y=dsin\theta=5.0\ m\cdot sin37^{\circ}=3.0\ m\ [N].

(2) We can find the magnitude of the plane's resultant velocity from the Pythagorean theorem:


vplane,res=vwind2+vplane2,v_{plane,res}=\sqrt{v_{wind}^2+v_{plane}^2},vplane,res=(500 ms)2+(100 ms)2=510 ms.v_{plane,res}=\sqrt{(500\ \dfrac{m}{s})^2+(100\ \dfrac{m}{s})^2}=510\ \dfrac{m}{s}.

We can find the direction of the plane's resultant velocity from the geometry:


θ=tan1(vyvx)=tan1(100 ms500 ms)=11.3 N of E.\theta=tan^{-1}(\dfrac{v_y}{v_x})=tan^{-1}(\dfrac{100\ \dfrac{m}{s}}{500\ \dfrac{m}{s}})=11.3^{\circ}\ N\ of\ E.

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