On the platform

relative speed 60-6=54km/hr=15m/s time=distance/speed=120/15=8sec

it is clear that velocity appear to the man is 60-6=54km/hr=15m/sec , and we have distance 120m so required time is 120/15=8

120m/15ms-1=8s

The man is moving with respect to ground so in order to make the man an inertial frame of reference we need to stop him by using vector reversal technique.Here we simply reverse the velocity vector of the man and use vector addition to add it with velocity of train.We get velocity of train with respect to man as (60-6=54)km/hr =15 m/s.

Note: Here we have subtracted 6 from 60 because in vector addition direction matters and since we have reversed the velocity vector of man and then we are adding it to train's velocity vector we get relative velocity with respect to man as 15 m/s.

Now taking length of train as distance with respect to man (i.e distance=120 meter) and using formula time=distance/speed we get Time=120/15 = 8 sec.

For simplicity we will assume the train moves first, then the man moves a distance according to the time.

For the train to travel 0.12km at a speed of 1/60 km/s, it will take 7.2s

In 7.2s, the man traveling at 1/600 km/s will move 0.012km

For the train to travel 0.012km at a speed of 1/60 km/s, it will take 0.72s

Repeating this cycle will give time values of the train: 7.2, 0.72, 0.072, 0.0072, etc.

This is identified as a geometric progression where the common ratio = 0.1

Using the equation to find the sum to infinity of a GP

S = a/(1-r)

S = (7.2)/(1-0.1)

S = 8s