Question

A radar station detects an airplane approaching directly from the east. At first observation, the range to the plane is 375.0 m at 43.0 ° above the horizon. The plane is tracked for another 123.0 ° in the vertical east-west plane, the range at final contact being 797.0 m. What is the magnitude of the displacement of the plane (in meters) during the period of observation?

Answer 1

819.78 m

Step-by-step explanation:

Given:

• OA = range of initial position of the airplane from the point of observation = 375 m
• OB = range of the final position of the airplane from the point of observation = 797 m
• 
  `\theta` = angle of the initial position vector from the observation point =
  `43^\circ`
• 
  `\alpha` = angle of the final position vector from the observation point =
  `123^\circ`
• 
  `\vec{AB}` = displacement vector from initial position to the final position

A diagram has been attached with the solution in order to clearly show the position of the plane.

`\vec{OA} = OA\cos \theta \hat{i}+OA \sin \theta \hat{j}\\\Rightarrow \vec{OA} = 375\ m\cos 43^\circ \hat{i}+375\ m\sin 43^\circ \hat{j}\\\Rightarrow \vec{OA} = (274.26\ \hat{i}+255.75\ \hat{j})\ m\\\vec{OB} = OB\cos \alpha \hat{i}+OB \sin \alpha \hat{j}\\\Rightarrow \vec{OB} = 797\ m\cos 123^\circ \hat{i}+797\ m\sin 123^\circ \hat{j}\\\Rightarrow \vec{OB} = (-434.08\ \hat{i}+668.42\ \hat{j})\ m`

Displacement vector of the airplane will be the shortest line joining the initial position of the airplane to the final position of the airplane which is given by:

`\vec{AB}=\vec{OB}-\vec{OA}\\\Rightarrow \vec{AB} =  (-434.08\ \hat{i}+668.42\ \hat{j})\ m-(274.26\ \hat{i}+255.75\ \hat{j})\ m\\\Rightarrow \vec{AB} =  (-708.34\ \hat{i}+412.67\ \hat{j})\ m`

The magnitude of the displacement vector =
`√((-708.34)^2+(412.67)^2)\ m = 819.78\ m`

Hence, the magnitude of the displacement of the plane is 819.67 m during the period of observation.