Question

Four persons p,q,r, and s are initially at the four corners of a square of side d ' . each person moves with a constant speed "v" in such a way that p always moves directly towards q, q towards r r towards s and s towards p . after what time will the four persons meet ?

Answer 1

Explanation:

Hola!!!

Please refer attachment!!!

Trick:

If you calculate when p & q will meet will be time when four of them will meet each other!

First calculate

Relative velocity!

then,

speed = distance/time

Hope it helps uh!

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• P always moves directly towards Q, Q towards R , R towards S and S towards P
• They are moving with a velocity "v".
• Side of square "d".

$\huge{\bold{\underline{Explanation:-}}}$

#refer the attachment for figure,

CASE-1

As from the figure, they will come closer and they will meet at the centre of the square.

Therefore the net displacement will be,

On the diagonal ,

As it is a square.

$\large{\bold{(Hypotenuse)^2 = (side)^2 + (side)^2}}$

$\large{(H)^2 = (d)^2 + (d)^2}$

$\large{(H)^2 = 2d^2}$

$\large{H = \sqrt{2d^2}}$

$\large{H = d \sqrt{2}}$

Now, Finding the displacement,

$\large{\bold{S = \dfrac{H}{2}}}$

As the displacement is half of diagonal.

$\large{S = \dfrac{d \sqrt{2}}{2}}$

$\large{S = \dfrac{ d \sqrt{2}}{ \sqrt{2} . \sqrt{2}}}$

$\large{S = \dfrac{ d \cancel{\sqrt{2}}}{ \cancel{ \sqrt{2}} . \sqrt{2}}}$

It becomes,

$\large{\boxed{S = \dfrac{d}{ \sqrt{2}}}}$

CASE-2

Say only two persons are there (For easy convenience)

One is at first vertex of square, second at second vertex of square.

The first person moves towards person, hence has velocity towards him. But the second person moves towards third vertex.

( vetex ----> Persons)

So, velocity vectors of both are in two different directions.

Resultant would be along the diagonal.

Angle = 90°.

Therefore the resultant will be v√2.

But the velocity vector at center of the square will be.

$\large{u = \dfrac{v \sqrt{2}}{2}}$

As it will be half.

$\large{u = \dfrac{ v \sqrt{2}}{ \sqrt{2} . \sqrt{2}}}$

$\large{u = \dfrac{ v \cancel{ \sqrt{2}}}{ \cancel{ \sqrt{2}} . \sqrt{2}}}$

$\large{\boxed{u = \dfrac{v}{ \sqrt{2}}}}$

Now,

we know,

$\large{\bold{Time = \dfrac{Distance}{Speed}}}$

Now,

$\large{t = \dfrac{d}{u}}$

Substituting the values,

$\large{t = \dfrac{d \times \sqrt{2}}{ v \times \sqrt{2}}}$

$\large{t = \dfrac{d \times \cancel{ \sqrt{2}}}{ v \times \cancel{ \sqrt{2}}}}$

Now,

$\huge{\boxed{\boxed{t = \dfrac{d}{v} \: Seconds}}}$

So, the four persons meet at d/v.