 # 2 identical trains take 3 sec to pass 1 another when going in the opposite direction but only 2.5 sec if speed of 1 is increased by 50%.The time one would take to pass the other when going in the same direction at their original speed is? 10 years ago

let train A is moving along positie x axis with vel = +V

train B along negative x axis = -U

vel of train A with respect B is = V-(-U) = V+U

let S be distance

S=(V+U)3

S=(V + U + U/2)2.5

=> S= (V+3/2U)2.5

get v and u in terms of S

then apply S=(V-U)t    (now V along and u in same direction v=V-U)

and find t

10 years ago

working in ref frame attached to 2, velocity of 1 wrt 2= v1+v2

let length of train be x metres each.

therefore time taken to cross each other= 2x/v1+v2=3

=> 2x= 3v1+3v2----------------------------------------------(1)

Case 2:

velocity of 1= v1+50v1/100=3v1/2

now, time taken to cross each other=      2x                                                             -------------- = 2.5

3v1/2- v2

=> 2x= 2.5v2+ 7.5v1/2---------------(2)

equating (1) and (2)

3v1+3v2= 2.5v2+7.5v1/2

=> 1.5v1/2= v2--------------------------------(3)

now if they move in same direction: velocity of 1 wrt 2= v1-v2

time taken to cross each other= 2x/v1-v2

= 2x/v1-0.75v1 (from 3)

= 2x/0.25v1

= 3v1+3 v2/ 0.25v1 (from 1)

= 3v1+ 0.75v1/0.25v1 (from 3)

= 3.75v1/0.25v1

= 15 sec

all the best!