Question

Four particles A, B, C and D are kept at the vertices of a square of side T. Now all of them simultaneously start moving with same uniform speed 'V' such that 'A' is moving towards 'B', B is moving towards 'C' and so on. The time after the area of square formed by the particles becomes half is (va-1/vb)The value of (a+b) is:​

Answer 1

The time after which the area of the square formed by the particles becomes half is a/2v

Explanation:

The sid of the square = a

When the particles move towards each other with uniform speed v then in the end they all will meet at the center of the square

Thus, the displacement of each particle will be $a\sqrt{2}/2=a/\sqrt{2}$

The component of velocity towards the centre of the square will be

$v\cos 45^\circ=v/\sqrt{2}$

Therefore, the time taken by the particles when they meet

$=\frac{a/\sqrt{2}}{v/\sqrt{w}}$

$=\frac{a}{v}$

When the particles meet, the area of the square formed by them is zero

In time $a/v$ the area is reduced to zero

Thus, in time $a/2v$, the area will be reduced to half.

Hope this answer is helpful.

Know More:

Q: 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?

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