Question

A particle is moving on hexagon of side 'a' with speed v. what will be it's avg speed and avg velocity for A to C journey​

Answer 1

Let the time taken to travel from A to C be $\displaystyle\sf {t.}$ Assume the hexagon is of regular.

The distance between A and C will be $\displaystyle\sf {2a.}$ Then, average speed,

$\displaystyle\longrightarrow\sf{\underline {\underline {v'=\dfrac {2a}{t}}}}$

There will be a triangle ABC where $\displaystyle\sf {AB=BC=a}$ and $\displaystyle\sf {\angle ABC=120^{\circ}.}$

The displacement between A and C will be, by cosine rule,

$\displaystyle\longrightarrow\sf{AC=\sqrt{AB^2+BC^2-2\cdot AB\cdot BC\ \cos120^{\circ}}}$

$\displaystyle\longrightarrow\sf{AC=\sqrt{a^2+a^2-2a^2\cos120^{\circ}}}$

$\displaystyle\longrightarrow\sf{AC=\sqrt{2a^2(1-\cos120^{\circ})}}$

$\displaystyle\longrightarrow\sf{AC=\sqrt{2a^2\cdot2\sin^260^{\circ}}}$

$\displaystyle\longrightarrow \sf{AC=2a\sin 60^{\circ}}$

$\displaystyle\longrightarrow\sf{AC=2a\cdot\dfrac {\sqrt3}{2}}$

$\displaystyle\longrightarrow\sf{AC=a\sqrt3}$

Hence the magnitude of the average velocity is,

$\displaystyle\longrightarrow\sf{|\vec v|=\dfrac {a\sqrt3}{t}}$

It's direction is along $\displaystyle\sf {\vec {AC}.}$

Let the direction of $\displaystyle\sf {\vec {AC}}$ be $\displaystyle\sf {\alpha.}$ Then,

$\displaystyle\longrightarrow\sf{\alpha=\tan^{-1}\left (\dfrac {a\sin 60^{\circ}}{a+a\cos 60^{\circ}}\right)}$

$\displaystyle\longrightarrow\sf{\alpha=\tan^{-1}\left (\dfrac {\frac {a\sqrt3}{2}}{a+\frac {a}{2}}\right)}$

$\displaystyle\longrightarrow\sf{\alpha=\tan^{-1}\left (\dfrac {\frac {a\sqrt3}{2}}{\frac {3a}{2}}\right)}$

$\displaystyle\longrightarrow\sf{\alpha=\tan^{-1}\left (\dfrac {1}{\sqrt3}\right)}$

since $\displaystyle\sf {\alpha}$ is an acute angle,

$\displaystyle\longrightarrow\sf{\alpha=30^{\circ}}$

Let $\displaystyle\sf {\vec {AB}=a\ \hat i}$ and $\displaystyle\sf {\vec {BC}=a\cos 60^{\circ}\ \hat i+a\sin 60^{\circ}\ \hat j.}$

Then the average velocity is,

$\displaystyle\longrightarrow\sf{\underline {\underline {\vec v=\dfrac {a\cos 30^{\circ}\sqrt3}{t}\ \hat i+\dfrac {a\sin 30^{\circ}\sqrt3}{t}\ \hat j}}}$