Question

A train, 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?

Answer 1

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

Answer: 2.3 m

The train started from rest. Initial velocity, $u_t=0 m/s$

Acceleration of train$a_t=0.6m/s^2$

The distance travelled by train in t=3 s is:

Using the equation of motion:

$s=ut+\frac{1}{2}at^2$

$s=\frac{1}{2}0.6\times3^2m=2.7m$

Hence, distance travelled by train in 3 s is 2.7 m.

Initial velocity of man, $u_m=0 m/s$

Acceleration of man, $a_m=1.2m/s^2$

Distance covered by man in 3s:

$s=0+\frac{1}{2}1.2\times3^2=5.4m$

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