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?
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
To find the acceleration of the man, we first need to determine the final velocity of the train. We can use the equation of motion:
v^2 = u^2 + 2as
Where,
v = final velocity of the train
u = initial velocity of the train (Assuming the train starts from rest, u = 0)
a = acceleration of the train
s = displacement
Given:
a = 2m/s^2
s = 9m
Substituting the values into the equation, we have:
v^2 = 0^2 + 2(2)(9)
v^2 = 36
v = √36
v = 6m/s
Now, let's find the acceleration of the man. We can use the equation:
s = ut + 0.5at^2
Where,
s = displacement (9m)
u = initial velocity of the man (Assuming the man starts from rest, u = 0)
a = acceleration of the man
t = time taken (3s)
Substituting the given values into the equation, we have:
9 = 0 + 0.5a(3)^2
9 = 0.5a(9)
9 = 4.5a
a = 9/4.5
a = 2 m/s^2
Therefore, the acceleration of the man is 2 m/s^2.