Question

Four particles situated at the corners
of a square of side a, move at a constant
speed v. Each particle maintain a
direction towards the next particle in
succession. Calculate the time the
particles will take to meet each other

Answer 1

Given:

Four particles situated at the corners

of a square of side a, move at a constant

speed v. Each particle maintain a

direction towards the next particle in

succession.

To find:

Time taken by the particles to meet with one another.

Calculation:

First , we need to calculate the velocity of approach considering any two adjacent particles.

Let velocity of approach be v app :

$\therefore \: v \: app. = v - v \cos(90 \degree)$

$= > \: v \: app. = v - (v \times 0)$

$= > \: v \: app. = v$

Now , time taken by the particles to meet can be calculated by dividing the edge length of the square with the the velocity of approach.

$\therefore \: time = \dfrac{edge \: length}{v \: app.}$

$= > time = \dfrac{a}{v}$

So , final answer is :

$\boxed{ \bold{ time = \dfrac{a}{v} }}$

Answer 2

Answer:

t = a/v

Explanation:

Each particle move perpendicular with the neighbour particle so non component of V along the line of motion of neighbour velocity so vel. of

approach = V

=> t = a/v