Question

Four particle A,B,C,D are situated at the corners of a square ABCD of sides a at t=0. Each of the particles moves with constant speed V towards the next particle. At what time will these particles meet each other?

Answer 1

Answer:

ans should be t=a/v.....

Answer 2

The four particles will meet each other at  $\frac{a}{v}$  unit time.

Given:

The side of the square ABCD is 'a'.

Four particles A, B, C, and D are situated at the corners of a square ABCD.

Each of the particles moves with constant speed V towards the next particle.

To Find:

The time at which the particles meet each other.

Solution:

The particles will finally meet at centre 'O' after some time.

So the particle has to cover the displacement of AO.

According to the figure,

cos 45°=$\frac{AE}{AO}$

$AO= \frac{AE}{cos45} = \frac{\frac{a}{2} }{\frac{1}{\sqrt{2} } } =\frac{a}{\sqrt{2} }$

The time taken by particle at point A to go to point O.

$t= \frac{AO}{v cos 45}=\frac{\frac{a}{\sqrt{2} } }{\frac{v}{\sqrt{2} } }$$=\frac{a}{v}$

Hence, the four particles will meet each other at  $\frac{a}{v}$  unit time.

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