A radar station, located at the origin of xz plane, as shown in the figure , detects an airplane coming straight at the station from the east.

Question

A radar station, located at the origin of xz plane, as shown in the figure , detects an airplane coming straight at the station from the east. At first observation (point A), the position of the airplane relative to the origin is $\vec{R_{A}}$ . The position vector $\vec{R_{A}}$ has a magnitude of $360^{m}$ and is located at exactly $40^{\circ}$ above the horizon. The airplane is tracked for another $123^{\circ}$ } in the vertical east-west plane for ${5.0}^{s}$ , until it has passed directly over the station and reached point B. The position of point B relative to the origin is $\vec{R_{B}}$ (the magnitude of $\vec{R_{B}}$ is $880^{m}$ ). The contact points are shown in the diagram, where the x axis represents the ground and the positive z direction is upward.

Define the displacement of the airplane while the radar was tracking it: ${\vec{R}}_{B A} = {\vec{R}}_{B} - {\vec{R}}_{A}$ . What are the components of ${\vec{R}}_{B A}$
Express ${\vec{R}}_{B A}$ in meters as an ordered pair, separating the x and z components with a comma, to two significant figures.

Answer 1

The position vector $\vec{R_{A}}$ is given by
${\vec{R}}_{A} = (r \cos θ) \hat{i} + (r \sin θ) \hat{k}$
Substitute (360m) for r and 40 deegres for $θ$ to find $\vec{R_{A}}$
${\vec{R}}_{A} = ((360 m) \cos 40^{\circ}) \hat{i} + ((360 m) \sin 40^{\circ}) \hat{k}$
$= (275.8 m) \hat{i} + (231.4 m) \hat{k}$
The position vector $\vec{R_{B}}$ is given by,
${\vec{R}}_{B} = (r \cos θ) \hat{i} + (r \sin θ) \hat{k}$
The angle for the position vector $o v e r \to {R}_{B}$ along vertical and horizontal plane is,
$θ = 123^{\circ} + 40^{\circ} = 163^{\circ}$
Substitute (880m) for r and 163 deegres for $θ$ to find $\vec{R_{B}}$
${\vec{R}}_{B} = ((880 m) \cos 163^{\circ}) \hat{i} + ((880 m) \sin 163^{\circ}) \hat{k}$
$= (- 841.5 m) \hat{i} + (257.3 m) \hat{k}$
The resultant displacement vectors components of the air plane are, ${\vec{R}}_{B A} = {\vec{R}}_{B} - {\vec{R}}_{A}$
Substitute $(- 841.5 m) \hat{i} + (257.3 m) \hat{k}$ for $\vec{R_{B}}$ and $(275.8 m) \hat{i} + (231.4 m) \hat{k}$ for $\vec{R_{A}}$ to find $\vec{R_{B A}}$ .
${\vec{R}}_{B A} = (- 841.5 m) \hat{i} + (257.3 m) \hat{k} - (275.8 m) \hat{i} + (231.4 m) \hat{k}$
$= - (1117.3 m) \hat{i} + (25.9 m) \hat{k}$ Round off the components of resultant displacement vectors in to two significant figures, ${\vec{R}}_{B A} = - (1100 m) \hat{i} + (26 m) \hat{k}$
Therefore, the displacement of the airplane as an ordered pair separated by the x and z components is (-1100m,26m).

Answer 2

Vector A:

$\cos 40 = \frac{x}{360}$ ; x = 276

$\sin 40^{\circ} = \frac{y}{360}$ ; y = 231

$R_{A_{x}}, R_{A_{z}} = (275.775, 231.40)$

Vector B:

$∠ (123 + 40 = 163; 180 - 163 = 17^{\circ})$

$\cos 17 = \frac{x}{880}; x = - 841.55$

$\sin 17 = \frac{y}{880}; y = 257.29$

$R_{B_{x}}, R_{B_{z}} = (- 841.55, 257.29)$

$R_{B A} = \vec{B} - \vec{A} = (- 841.55, 257.29) - (275.775, 231.40) = (- 116.84, 25.89)$

A radar station, located at the origin of xz plane, as shown in the figure , detects an airplane coming straight at the station from the east.

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