Respuesta :
The passenger is 2.3 m away from the door after 3 seconds
Since both train and passenger is moving with a constant acceleration so we can use the second equation of motion to calculate the distance cover by train and the passenger,
Distance cover by the train in 3 seconds
S=ut-0.5at^2
S=0*3-0.5*0.6*3^2=2.7 m
Distance cover by passenger in 3 seconds
S=0*3-0.5*1.2*3^2=5.4 m
Now distance of passenger from the train
s=5+2.7-5.4=2.3 m
Therefore after 3 seconds passenger is 2.3 m away from the door.
Answer: 2.3 m
The train started from rest. Initial velocity, [tex]u_t=0 m/s[/tex]
Acceleration of train[tex]a_t=0.6m/s^2[/tex]
The distance travelled by train in t=3 s is:
Using the equation of motion:
[tex]s=ut+\frac{1}{2}at^2[/tex]
[tex]s=\frac{1}{2}0.6\times3^2m=2.7m[/tex]
Hence, distance travelled by train in 3 s is 2.7 m.
Initial velocity of man, [tex]u_m=0 m/s[/tex]
Acceleration of man, [tex]a_m=1.2m/s^2[/tex]
Distance covered by man in 3s:
[tex]s=0+\frac{1}{2}1.2\times3^2=5.4m[/tex]
train door was already ahead of man by 5 m. Hence, train door is now at (5+2.7)m=7.7 m from the initial point of man.
Therefore, the distance between train's door and man after 3 s is=(7.7-5.4) m =2.3 m
