Particles in triangular motion

Classical Mechanics Level pending

Any three particles $A$ , $B$ and $C$ are situated at the vertices of a triangular shaped wooden board $ABC$ of identical side length $l$ at time $t = 0$ (in seconds).

If each of the particles moves with constant speed $v$ (in $\text{ m/s}$ ) and the velocity of particle $A$ , $B$ and $C$ is always along $AB$ , $BC$ and $CA$ respectively.

Then at what time(in seconds) will these particles meet each other ?

Detail and Assumptions:

$v = 2\text{ m/s}$ .
$l = 6\text{ m}$ .

2.25 seconds 1 second 2 seconds Will never meet

1 solution

Naren Bhandari
Feb 28, 2017

Considering any wooden board as per the condition supplied. Since wooden board is of identical length $l$ which shows that it is in the equilateral triangular form with each interior $\angle60^\circ$ .

Now on resolving the component velocity $B$ along $BA = vcos60^\circ = \frac{v}{2}$

Thus the separation of particles at $A$ along $B$ decreases with the speed of $v + \frac{v}{2} = \frac{3v}{2}$ (relative velocity of $B$ with respect to $A$ in opposite direction)

Since the speed is constant, the time taken in reducing the speed is also constant, the time taken in reducing the separation in $AB$ from $l$ to zero is given

$t = \frac{l}{\frac{3v}{2}}$

$t = \frac{2l}{3v}$

Now plugging the value of $l$ and $v$

$t = \frac{2\times6}{3\times2} = 2seconds$

Please note a similar problem I have posted, https://brilliant.org/problems/four-boats-chasing/

Laszlo Mihaly - 3 years, 2 months ago

Particles in triangular motion

1 solution

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