IIT JEE , AIEEE Forum - Physics , Mechanics - Four persons K, L , M, N are initially at the four corners of a square of side 'd'. Each

tut_einstein

Four persons K, L , M, N are initially at the four corners of a square of side 'd'. Each person now starts moving with a uniform speed 'v' IN SUCH A WAY THAT THAT K always moves directly towds. L, L towards M, M towards N AND N TOWARDS K. aFTER WHAT TIME WILL THEY MEET?

indr12365

U 'll get da path followed resembling an exhaust fan centred at da centre of square O.
Since each has same velocity all reach O at same time.
Distance moved by each is same as a quarter of circle i.e. pi r/2
where r=dis fron a vertex to O=d/root 2
Time=[pi *d/2root2 ] / v
Rate da ans if I'm correct..Please nudge if wrong

damasdelhi

i will give u a formula. time(t) is given by a/v+vcosk. here k=90
in equilateral triangle k=60, in rectangle k=90 and so on.

v.sabarish

I guess it's simply d/v...
relative vel of K wrt L is v... and d relative seperation just decreases linearly..
the answer must be d/v

sahilmadaan12

see is very easy ques i this type of ques

always take any two adjacent points of the figure

say P AND Q

NOW find their relatine velocity on the line joining them it come v + v cos $\theta$

since the distance b/w them is d

so time taken is d / v + v cos $\theta$

for square $\theta$ = 90

iit009

is answer d/v

cutepooja

ya answer is d/v
acc to the ques they move such that the form a square everytime wid a smaller side eventually they meet at a point
the speed of every one (along) the line joining initial position and the final point is v/root2
and they move distance d/root 2
thus time is d/v
rate me if correct

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