Question

In the figure, a radar station detects an airplane approaching directly from the east. At first observation, the airplane is at distance di = 370 m from the station and at angle 91 = 37° above the horizon. The airplane is tracked through an angular change 40 = 120° in the vertical east-west plane; its distance is then d2 = 810 m. Find the (a) magnitude and (b) direction of the airplane's displacement during this period. Give the direction as an angle relative to due west, with a positive angle being above the horizon and a negative angle being below the horizon. Airplane Me ---- Radar dish (a) Number Units (b) Number Units

Answer 1

Let $\vec{A}$ be the distance $d_{1}\ at\ \theta _{1}$

$\vec{A}=370m\ at\ 37^{\circ}\ above\ horizon$

$\vec{A}=A_{x}i+A_{y}j=Acos\theta i+Asin\theta j=370cos37i+370sin37j$

${\color{Red} \vec{A}=295.50i+222.67j}$

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Let $\vec{B}$ represent displacement of plane

${\color{Red} \vec{B}=B_{x}i+B_{y}j}$

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Let $\vec{R}$ be the distance $d_{2}\ at\ \theta _{2}$

$\vec{R}=810m\ at\ (120^{\circ}+37^{\circ})157^{\circ}\ above\ horizon$

$\vec{R}=R_{x}i+R_{y}j=Rcos\theta i+Rsin\theta j=810cos157i+810sin157j$

${\color{Red} \vec{R}=-745.61i+316.49j}$

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$\vec{R}=\vec{A}+\vec{B}$

$-745.61i+316.49j=(295.50i+222.67j)+(B_{x}i+B_{y}j)$

$-745.61i+316.49j=295.50i+222.67j+B_{x}i+B_{y}j$

$-745.61i+316.49j=(295.50+B_{x})i+(222.67+B_{y})j$

Equate i terms

$-745.61=295.50+B_{x}$

$-745.61-295.50=B_{x}$

${\color{Red} B_{x}=-1041.11m}$

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Equate j terms

$316.49=222.67+B_{y}$

$316.49-222.67=B_{y}$

${\color{Red} B_{y}=93.82m}$

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(a)Magnitude

$B=\sqrt{B_{x}^{2}+B_{y}^{2}}$

$B=\sqrt{(-1041.11)^{2}+93.82^{2}}$

ANSWER: ${\color{Red} B=1045.33m}$

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(b) Direction

$\theta =tan^{-1}(\frac{B_{y}}{B_{x}})$

$\theta =tan^{-1}(\frac{93.82}{-1041.11})$

$\theta =-5.15^{\circ}$

ANSWER: ${\color{Red} \theta =5.15^{\circ}\ due\ West\ above\ horizon}$

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