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

Answer:

4.08 s

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

$d_1 = \frac{1}{2}(0.6) t^2$

also the distance moved by passenger

$d_2 = \frac{1}{2}(1.2) t^2$

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