Atrain, starting from rest, accelerates along the platform at a uniform rate of 0.6 m/s2. a passenger standing on the platform is 5 m away from the door when the train starts to pull away and heads toward the door at acceleration of 1.2 m/s2. how far away from the door is the passenger after 3 seconds?

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.

hèllo, sorry i really need some more points so if you need this qustion answered than retype it