Question

can someone try to do this amazing question?

Four persons K, L, M, and N are initially at the four corners of a square of side 'd'. Each person now moves with a uniform speed of 'v' in such a way that K always moves directly towards L, L directly towards M, M directly towards N and N directly towards K. At what time will the four persons meet?

Answer 1

$\boxed{ \bf \orange {Hello \: Loco + Maniac } }$

Well thanks alot for asking such a great question over this haveli xD , well without wasting our time let's drink it xD

$\bf \pink {see \: the \: attached \: picture \Rightarrow }$

Suppose there be a centre O , since it's making a square

ON = d / √2

Given that N is moving towards K

so , KNO = 45°

obiviously Now the component of velocity of that will cover the distance ON ,

=> V cos45°= v /√2 { since cos45° = 1/√2 }

$Now \: time \: taken = \frac{d}{ \sqrt{2} } \div \frac{v}{ \sqrt{2} } \\ \\ \Rightarrow \: \frac{d}{ \sqrt{2} } \times \frac{ \sqrt{2} }{v} \\ \\ \Rightarrow \boxed { \bf \frac{d}{v} }$

$\bf Hope \: This \: helps \: you : )$