16sec

answer is 13 sec

there are 2 cars moving opposite each other with 50cm/s for 100cm in forward direction and 50cms with 25cm/sec backward direction

therefore it is enough to calculate for 1 car ie 250cms

at 2 sec car moves 100cms forward and by next 2 sec it comes backward by 50cms so total 50cms for 4sec.

for 6th sec car moves 150cms forward total and by 8th sec 100cms total.

by 10th sec car moves 200cms forward and by 12th sec 150cms total.

by 13secs car moves 250cms and collides with other car.

so ans is 13sec

Upto 12th Sec 150 cm the above person wht he did is correct

From that in 2 sec it moves 100cms

so in 14 sec 250cms by single car another car also 250 cms total 500cms so both collide at 14th Sec

a------------------b

----> 500m <----

forward direction 100cm-----50cm/s

backward direction 50cm-----25cm/s

as we know the formulae: speed distance/time

i.e time distance/speed

now for forward direction t 100/50 2sec

for backward direction t 50/25 2sec

now since the 2 toys are travelling with equal speed so we can guess that the 2 toys collide exactly at half the way so 500cm/2 250cm

we can directly calculate for 1 toy

at 14sec the 2 toys collide.................

14 secs

at 2 secs each car will be moved to a dist of 100cms

at 4 secs each car will be moved to a dist of 50cms

at 6 secs each car will be moved to a dist of 150cms

at 8 secs each car will be moved to a dist of 100cms

at 10 secs each car will be moved to a dist of 200cms

at 12 secs each car will be moved to a dist of 150cms

at 14 secs each car will be moved to a dist of 250cms that is at the point they collide

Ya

That perfect answer is 14 sec..

kanagaraj

Each car moves a 100 cm forward and 50 cms backwards

the total forward and a backward moment is defined as a cycle.

each cycle takes 4 seconds as 2 seconds 100 cms forward at 50 cm/s and backwards at 25 cm/s

in each cycle of 4 seconds each car moves 50 cms forwards and distance betweeen car is reduced by 100 cms

cycle 1 4 seconds

Car A moves 50 cms and same the car B

now the distance is 400 cms

cycle 2 8 seconds

Car A moves 100 cms and same the car B

now the distance is 300 cms

cycle 3 12 seconds

Car A moves 150 cms and same the car B

now the distance is 200 cms

now in the last cycle the collision occurs

as each car moves 100 cms in 2 seconds

so the 200 cms is covered by both cars

so after an instant collision at the 12+2 14 seconds

the car moves backwards

hence the answer is 14 seconds

raghavan iyengar

hcl technologies