Asked by dillip
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?
Answers
Answered by
bobpursley
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
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
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