Question

Two trains enter at the opposite sides of a tunnel of length L with speeds 'V'. A particle enters the tunnel at the same time with a speed 'v' and it vibrates in the tunnel[i.e. if it reaches the end of the tunnel then it comes back]. What is the position of the particle by the time the 2 trains meet?

Answer 1

vL/2V

Answer 2

considering v<=v

Answer 3

can u explain it clearly?? actual answer is If v<=2V then the position is (v*L)/(2*V) from the starting point else it is 2*L -(v*L)/(2*V) from the starting point.. but i need some clear explanation....

Answer 4

let at time t they meets so 2Vt=L
t=L/(2V)
vt=vL/2V

Answer 5

if v<2V then vL/2V <L so the particle is inside the tunnle . if v>2V then vL/2V >L so then the particle willbe reflected back

Answer 6

so it will be then vL/2V - L from the 2nd end
and hence it will be L-(v/2V-L)=2L-vl/2V from the first end

Answer 7

thanks dipankarstudy............ get an medal

Answer 8

thank u..

Answer 9

can u answer that n queen problem question...

Answer 10

i think it is in wikipedia...

Answer 11

oh kkkkkk

Answer 12

http://en.wikipedia.org/wiki/Eight_queens_puzzle
check this...

Answer 13

thanks.........