91Ó°ÊÓ

• The displacement vector of ship is,$\stackrel{→}{A}=285\mathrm{km},62.0^{°}\mathrm{nth}-\mathrm{east}$.
• The resultant vector of ship is,$\stackrel{→}{R}=115\mathrm{km}$.

The term "displacement vector" refers to the change in an object's position vector.

Vector distance between object's starting position and its end point may also be used to measure its displacement.

The diagram has been provided below-

The direction of the ship can be evaluated by,

$$\begin{gathered} \stackrel{→}{A}+\stackrel{→}{B}=\stackrel{→}{R} \\ \stackrel{→}{B}=\stackrel{→}{R}-\stackrel{→}{A} \end{gathered}$$

Here, role="math" localid="1655809188002" $\stackrel{→}{B}$is the resultant displacement vector

The displacement vector$\stackrel{→}{B}$divided in x-component and y-component is,

$$\begin{gathered} B_x=R_x-A_x \\ {B}_y=R_y-A_y \end{gathered}$$ …1)

Here, role="math" localid="1655809377112" ${\stackrel{→}{B}}_x\mathrm{and}{\stackrel{→}{\mathrm{B}}}_{\mathrm{y}}$are the x and y component of the displacement vector ${\stackrel{→}{A}}_x$and${\stackrel{→}{A}}_x$are the x and y component of the resultant vector and are the x and y component of the displacement vector of the ship.

The equation of the displacement vector in the x direction can be expressed as-

Substituting the values in the above equation,

$A_x=-A\cos 62.0°=-133.8\mathrm{km}$

The equation of the displacement vector in the y direction can be expressed as-

$\stackrel{→}{A_y}=285\mathrm{km}×\sin 62°$

Substituting the values in the above equation,

$A_y=A\sin 62.0°=251.6\mathrm{km}$

The resultant displacement vector is,

$R_x=115\mathrm{km},{\mathrm{R}}_{\mathrm{y}}=0$

Here, role="math" localid="1655809935734" $\stackrel{→}{R_x}\mathrm{and}\stackrel{→}{{\mathrm{R}}_{\mathrm{v}}}$are the resultant vectors in the x and y direction

Substitute value in the displacement vector in the equation 1),

$$\begin{gathered} B_x=R_x-A_x \\ {B}_x=\left(115\mathrm{km}\right)-\left(-133.8\mathrm{km}\right) \\ {B}_x=248.8B_y \\ {B}_y=R_y-A_y \\ {B}_y=0-\left(251.6\mathrm{km}\right) \\ {B}_y=-251.6\mathrm{km} \end{gathered}$$

The equation of the resultant displacement vector is,

$B=\sqrt{B_x^2+B_y^2}$

Here, B is the resultant displacement vector

Substituting the values in the above equation,

$$\begin{gathered} B=\sqrt{{\left(248.8\mathrm{km}\right)}^2+{\left(-251.6\mathrm{km}\right)}^2} \\ =353.8\mathrm{km} \end{gathered}$$

Thus, the direction of the resultant displacement is $353.8\mathrm{km}$.

The equation of the resultant displacement vector is,

$\tan \mathrm{Î\pm }=\frac{B_y}{B_x}$

Substituting the values in the above equation,

$$\begin{gathered} \tan \mathrm{Î\pm }=\left(\frac{-251.6\mathrm{km}}{248.8\mathrm{km}}\right) \\ \mathrm{Î\pm }={\tan }^{-1}\left(\frac{-251.6\mathrm{km}}{248.8\mathrm{km}}\right) \\ =-45.3° \end{gathered}$$

Thus, the direction of the resultant displacement is, $45.3°$south of east.