91Ó°ÊÓ

If $/$ is the length of minute hand of a watch representing the magnitude of a vector making an angle $\mathrm{θ}$anticlockwise with the positive direction of $x$-axis.

Vector $\stackrel{→}{l}$can be represented in terms of its $x$and $y$components as,

$\stackrel{→}{l}=l_x\widehat{i}+l_y\widehat{j}$

Where its $x$and $y$components are,

$\stackrel{→}{l}=l_x\widehat{i}+l_y\widehat{j}$

Where its $x$and $y$components are,

$\begin{array}{l}{l}_x=l\cos \mathrm{θ}\\ l_y=l\sin \mathrm{θ}\end{array}$

Substitute $l\cos \mathrm{θ}$for $l_x$and $l\sin \mathrm{θ}$for $l_y$in equation (1) position vector of tip of minute hand of watch can be written as,

$\stackrel{→}{l}=l\cos \mathrm{θ}\widehat{i}+l\sin \mathrm{θ}\widehat{j}……\left(2\right)$

If $\stackrel{→}{l}$is initial position vector of tip of minute hand and ${\stackrel{→}{l}}_f$is final position vector of tip of minute hand then displacement vector of $\mathrm{Δ}\stackrel{→}{l}$tip of minute hand of watch

$\stackrel{→}{\mathrm{Δ}l}=\stackrel{→}{l_f}−{\stackrel{→}{l}}_i……\left(3\right)$

When minute hand is at its position is drawn as below.

Angle made by minute hand anticlockwise with the positive direction of $x$-axis is ${90}^{°}$.

Substitute $2\text{cm}$for / and ${90}^{°}$for in $\mathrm{θ}$equation (2).

Initial position vector of tip of minute hand

$\begin{array}{l}{\stackrel{→}{l}}_i=(2\text{cm})\cos {90}^{°}\widehat{\text{i}}+(2\text{cm})\sin {90}^{°}\widehat{\text{j}}\\ {\stackrel{→}{l}}_i=(2\text{cm})\widehat{\text{j}}\end{array}$

When minute hand is $8.20\text{AM}$at its position is drawn as below.

If watch is divided into 12 parts, each part makes an angle ${30}^{°}$,

Then angle made anticlockwise with the positive direction of $x$-axis by 11 parts is $\left(11\right)\left({30}^{°}\right)={330}^{°}$

So, angle made by minute hand anticlockwise with the positive direction of x-axis is ${330}^{°}$.

Substitute$2\text{cm}$for $/$ and${330}^{°}$for$\mathrm{θ}$

Final position vector of tip of minute hand

$\begin{array}{l}{\stackrel{→}{l}}_f=(2\text{cm})\cos {330}^{°}\widehat{\text{i}}+(2\text{cm})\sin {330}^{°}\widehat{\text{j}}\\ {\stackrel{→}{l}}_f=(\sqrt{3}\text{cm})\widehat{i}−(1\text{cm})\widehat{j}\end{array}$

Substituting values of $\stackrel{→}{l_f}$and $\stackrel{→}{l}$in equation (3).

$\begin{array}{l}\stackrel{→}{\mathrm{Δ}l}={\stackrel{→}{l}}_f−{\stackrel{→}{l}}_i\\ \stackrel{→}{\mathrm{Δ}l}=(\sqrt{3}\widehat{i}−\widehat{j})−\left(2\widehat{j}\right)\\ \mathrm{Δ}\stackrel{→}{l}=(\sqrt{3}\text{cm})\widehat{\text{i}}−(3\text{cm})\widehat{\text{j}}\end{array}$

Therefore, the displacement vector of the tip of minute hand of watch is $\mathrm{Δ}\stackrel{→}{l}=\sqrt{3}\widehat{i}−3\widehat{j}\text{cm}$.

When minute hand is at $8.00\text{AM}$its position is drawn in blue color as below. (Red arrow represents hour hand)

Angle made by minute hand anticlockwise with the positive direction of x-axis is ${90}^{°}$.

Substitute $2\text{cm}$for / and ${90}^{°}$for $\mathrm{θ}$ initial position vector of tip of minute hand.

$\begin{array}{l}{\stackrel{→}{l}}_i=(2\text{cm})\cos {90}^{°}\widehat{\text{i}}+(2\text{cm})\sin {90}^{°}\widehat{\text{j}}\\ {\stackrel{→}{l}}_i=(2\text{cm})\widehat{\text{j}}\end{array}$

When minute hand is at $9.00\text{AM}$its position is drawn in blue color as below. . (Red arrow represents hour hand).

So, angle made by minute hand anticlockwise with the positive direction of x-axis is ${90}^{°}$.

Substitute $2\text{cm}$for / and ${90}^{°}$for $\mathrm{θ}$final position vector of tip of minute hand,

$\begin{array}{l}{\stackrel{→}{l}}_f=(2\text{cm})\cos {90}^{°}\widehat{\text{i}}+(2\text{cm})\sin {90}^{°}\widehat{\text{j}}\\ {\stackrel{→}{l}}_f=(2\text{cm})\widehat{\text{j}}\end{array}$

Substituting values of $\stackrel{→}{l_f}$and $\stackrel{→}{l}$in equation (3).

$\begin{array}{l}\stackrel{→}{\mathrm{Δ}l}={\stackrel{→}{l}}_f−{\stackrel{→}{l}}_i\\ \mathrm{Δ}\stackrel{→}{l}=(2\text{cm})\widehat{\text{j}}−(2\text{cm})\widehat{\text{j}}\end{array}$

Initial position of minute hand of a clock is in $+Y$direction and final position is also in $+Y$directic so displacement vector is a null vector because initial and final position vector are same.

Therefore, the displacement vector of the tip of minute hand of watch is$\mathrm{Δ}\stackrel{→}{l}=\stackrel{→}{0}\text{cm}$.