91Ó°ÊÓ

1. The position of an airplane at ${40}^{°}$above the horizon(x-direction) is${\mathit{d}}_{\mathbf{1}}\mathbf{=}\mathbf{360}\mathbf{}\mathit{m}$

2. The position of an airplane at ${\mathbf{123}}^{\mathbf{°}}$from $d_1$is ${\mathit{d}}_{\mathbf{2}}\mathbf{=}\mathbf{790}\mathit{m}$

Using the relative motion concept, we can find the magnitude and direction of the vector.

Formulae:

The position vector in unit-vector notation,$\stackrel{\xcancel{}\xcancel{}}{d}\mathrm{θ}\hat{i}=dd\cos \mathrm{θ}\hat{j}+\cos$ …(¾±)

Magnitude of the resultant vector $\stackrel{\xcancel{}\xcancel{}}{V}$is given by,$V=\sqrt{V_x^2-V_y^2}$ …(¾±¾±)

The position of an airplane at${40}^{°}$above the horizon(x-direction) can be written in the vector notation using equation (i) as,

$$\begin{gathered} \stackrel{\xcancel{}\xcancel{}}{d_1}=\left(360m\right)\cos {40}^{°}\hat{i}+\left(360m\right)\sin {40}^{°}\hat{j} \\ =\left(276m\right)\hat{i}+\left(231m\right)\hat{j} \end{gathered}$$

The position of an airplane at ${123}^{°}$from $d_1$can be written in the vector notation using equation (i) as,

$$\begin{gathered} \stackrel{\xcancel{}\xcancel{}}{d_2}=\left(790m\right)\cos {163}^{°}\hat{i}+\left(790m\right)\sin {163}^{°}\hat{j} \\ =\left(-755m\right)\hat{i}+\left(231m\right)\hat{j} \end{gathered}$$

The displacement of an airplane is given as:

$$\begin{gathered} \text{'}d=d_2-d_1 \\ =\left[\left(-755m\right)\hat{i}+\left(231m\right)\hat{j}\right]-\left[\left(276m\right)\hat{i}+\left(231m\right)\hat{j}\right] \\ =\left(-1031m\right)\hat{i} \end{gathered}$$

Hence, the magnitude value of the displacement is $1031m$.

From the part (a) calculations, we can get that the airplane’s displacement is along west$\left(-\hat{i}\right)$direction.