A train begin to move with an acceleration of 2m/s0¡­5. When a man 9m away from the core of the train, the man begins to run and catches in 3 seconds. What is the acceleration of the man?

Answer 1

Not certain what you mean by "core"

I think this is what you mean. The man runs 9 meters more than the train goes in the same time.

distance man=9+2m/s*3 seconda
=15m

for the man:
vf^2=1/2 a t^2
now his vf must be at least the speed of the train at that point, so
vf=>2m/s^2*3=6m/s
or

6m/s<=1/2 a 3^2
a>=12/9 m/s^2