91影视

• The magnitude of the training flight isfor the vector A is,$\stackrel{鈫�}{A}=147\mathrm{km},85掳\mathrm{east}$

• The magnitude of the training flight is for the vector B is, $\stackrel{鈫�}{B}=106\mathrm{km},167掳\mathrm{south}鈥�\mathrm{east}$
• The magnitude of the training flight is for the vector C is,$\stackrel{鈫�}{C}=166\mathrm{km},235掳\mathrm{south}鈥�\mathrm{west}$

The figure has been provided below-

The displacement vector is just the original position vector minus the end position vector.

This is the quickest path from the beginning location of the moving body to the end position.

a)

The diagram has been provided below-

The magnitude of flight in x-direction is,

$D_x=-\left[\stackrel{鈫�}{A}\sin 85掳+\stackrel{鈫�}{B}\sin 13掳-\stackrel{鈫�}{C}\sin 55掳\right]$

Here,$D_x$ is the magnitude of the flight in the x-direction,

Substituting the values in the above equation,

$$\begin{gathered} D_x=-\left[\left(147\mathrm{km}\right)\sin 85掳+\left(106\mathrm{km}\right)\sin 13掳-\left(166\mathrm{km}\right)\sin 55掳\right] \\ {D}_x=-34.3\mathrm{km} \end{gathered}$$

The magnitude of flight in y-direction is,

$D_y=-\left[\stackrel{鈫�}{A}\cos 85掳-\stackrel{鈫�}{B}\cos 13掳-\stackrel{鈫�}{C}\cos 55掳\right]$

Here,$D_y$ is the magnitude of the flight in the y-direction,

Substituting the values in the above equation,

$$\begin{gathered} D_y=-\left[\left(147\mathrm{km}\right)\cos 85掳-\left(106\mathrm{km}\right)\cos 13掳-\left(166\mathrm{km}\right)\cos 55掳\right] \\ {D}_y=185.7\mathrm{km} \end{gathered}$$

The resultant magnitude of flight is,

$$\begin{gathered} D=\sqrt{D_x^2+D_y^2} \\ D=\sqrt{{\left(34.30\mathrm{km}\right)}^2+{\left(-185.7\mathrm{km}\right)}^2} \\ D=189\mathrm{km} \end{gathered}$$

Thus, The distance that the student flies from Manhattan to get back to Lincoln is,$D=189\mathrm{km}$

b)

The direction of flight can be evaluated by,

$蠁={\tan }^{-1}\left(\frac{D_y}{D_x}\right)$

Substituting the values in the above equation,

$$\begin{gathered} 蠁={\tan }^{-1}\left(\frac{185.7\mathrm{km}}{34.3\mathrm{km}}\right) \\ =79.53掳 \end{gathered}$$

Since, magnitude of direction in x-direction is less than zero and in y-direction it is more than zero.

Thus, the direction relative to north-west is $79.53掳$.