Question

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Three particles A B and C are situated at vertices of an equilateral triangle ABC of side B at t = 2 0. each of the particle moves with a constant speed v. a always has its velocity along a b, b along b c and C along CA. at what time will the particle meet each other ​

Answer 1

The time at which the three particles meet each other is equal to the time at which any of the particle reaches O.

From the fig., we get the time of meeting is equal to the time during which the particle at B moves towards O. The particle has a velocity component $\sf{v\cos30^o}$ along OB.

Let $\sf{l}$ be the length of each side of the equilateral triangle. Then the time of meeting is,

$\longrightarrow\sf{t=\dfrac{OB}{v\cos30^o}\quad\quad\dots(1)}$

But from ΔOBD,

$\longrightarrow\sf{BD=OB\cos30^o}$

$\longrightarrow\sf{\dfrac{l}{2}=OB\cos30^o}$

$\longrightarrow\sf{OB=\dfrac{l}{2\cos30^o}}$

Then (1) becomes,

$\longrightarrow\sf{t=\dfrac{l}{2v\cos^230^o}}$

$\longrightarrow\sf{t=\dfrac{4l}{2v\cdot3}}$

$\longrightarrow\sf{\underline{\underline{t=\dfrac{2l}{3v}}}}$

Answer 2

Correct Question :-

Three particles A, B and C are situated at the varies of an equilateral of triangle ABX of side d at t=0. Each of the particle moves with constant speed v. A always has its velocity along AB, B along BC and C along CA.

At what time will the particles meet each other ?

Solution :-

▪ They will meet at the centre O of the ΔABC.

▪ At any time of motion, all three particles will form an ΔABC (equilateral) with the centre O.

▪ Let, concentrate the motion of particle A.

▪ At any instant, velocity of particle A makes angle 30°.(with AO)

▪ Velocity component of particle A along AO at any instant = v cos30°

▶ Calculation of AO

$\implies\sf\:AO=\dfrac{2}{3}\sqrt{d^2-(\dfrac{d}{2})^2}\\ \\ \implies\sf\:AO=\dfrac{2}{3}\sqrt{d^2-\dfrac{d^2}{4}}\\ \\ \implies\sf\:AO=\dfrac{2}{3}\sqrt{\dfrac{3d^2}{4}}\\ \\ \implies\sf\:AO=\dfrac{2\sqrt{3}d}{6}\\ \\ \implies\bf\:\red{AO=\dfrac{d}{\sqrt{3}}}$

▶ Calculation of time

$\twoheadrightarrow\bf\:time=\dfrac{displacement}{velocity}\\ \\ \twoheadrightarrow\sf\:t=\dfrac{d}{\sqrt{3}v\cos30\degree}\\ \\ \twoheadrightarrow\sf\:t=\dfrac{2d}{(\sqrt{3}\times \sqrt{3})v}\\ \\ \twoheadrightarrow\boxed{\bf{\purple{\large{time=\dfrac{2d}{3v}}}}}$

✴ Please, see the attachment for better understanding....