Question

34. 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 an acceleration of 1.2 m/s2. How long does it take the passenger to reach the door?

Answer 1

4.08 s

Step-by-step explanation:

Let the passenger took "t" time to catch the train

so in this case the total distance moved by the train + 5 m = total distance moved by the passenger

so we will have

distance moved by train is given as

`d_1 = (1)/(2)(0.6) t^2`

also the distance moved by passenger

`d_2 = (1)/(2)(1.2) t^2`

so we will have

`d_1 + 5 = d_2`

`0.3 t^2 + 5 = 0.6 t^2`

`0.3 t^2 = 5`

t = 4.08 s