Question

A passenger plane departs from an airport and takes the following route: it first travels from a point O to city A, located 175 km in a direction 30 north of the East. Then it goes to city B at 153 km. Heading 20 ° west of north. Finally fly 195 km. Directly west towards city C. The position of city C with respect to the position of the starting point is:

Answer 1

The locations where the passneger travels can be shown as,

The position of passenger at city A can be expressed as,

`\begin{gathered} \vec{A}=(175km)\cos 30^(\circ)\hat{i}+(175km)\sin 30^(\circ)\hat{j} \\ =(175\text{ km)(}0.866)\hat{i}+(175\text{ km)(}0.5)\hat{j} \\ =151.55\text{ km }\hat{\text{i}}+87.5\text{ km }\hat{\text{j}} \end{gathered}`

The position of passenger at city B can be expressed as,

`\begin{gathered} \vec{B}=((151.55km)-(153km)\sin 20^(\circ))\hat{i}+(87.5km+(153km)\cos 20^(\circ))\hat{j} \\ =((151.55km)-(153\text{ km)(}0.342))\hat{i}+(87.5km+(153\text{ km)(}0.94))\hat{j} \\ =(151.55km-52.33\text{ km) }\hat{\text{i}}+(87.5km+143.82\text{ km) }\hat{\text{j}} \\ =99.22\text{ km }\hat{\text{i}}+231.32\text{ km}\hat{\text{j}} \end{gathered}`

The position of passenger at city C can be expressed as,

Therefore, the magnitude of distance of city C can be calculated as,

`\begin{gathered} C=\sqrt[]{(-95.78km)^2+(231.32km)^2} \\ =\sqrt[]{9173.81km^2+53508.94km^2} \\ =252.3\text{ km} \end{gathered}`

Thus, the distance of city C from point O is 252.3 km.

The direction of location is calculated as,

`\begin{gathered} \cos \theta=\frac{-95.78^{}}{231.32} \\ \theta=-65.5^(\circ) \end{gathered}`

Thus, the direction of city C from point O is -65.5 degree.