Question

A man is 9m behind the door of a train when it starts moving with acceleration of 2m/s2. The man runs at full speed . How far does he run and after what time does he get into the train? What is his full speed?

Answer 1

Distance run by the train at the moment t:


                 x = a * t^2 / 2 

             
Distance run by the man, D


                 D = 9 m + distance run by the train

                 D = 9m + a(t^2) / 2


Velocity of the man, V = D/t => D = V*t


Equal D:

V*t = 9m + a(t^2) / 2


Now, you have to make an additional assumption. It is that when the man reaches the traing their velocities are the same. This if because, if the train is faster, the man will not reach it, and if the train is slower then he will pass the train (which is an unnecessary waste of effort from the man)


So, the best contition is that the speed of the train equals the speed of the man.


The speed of the train follows the equation of the uniformly accelerated motion: V = a.t


So from substitute that V in the equation V*t = 9m + a(t^2)/2

    
 => (at) t = 9m + a(t^2) / 2


=> a(t^2) - a(t^2) / 2 = 9m


=> a(t^2) / 2 = 9m


=> (t^2) = 9m * 2 / a = 9m * 2 / ( 2 m/s^2) = 9 s^2


=> t = 3 s


=> D = 9m + a(t^2) / 2 = 9 m + 2(m/s^2) (3s)^2 / 2 = 9m + 9 m = 18 m


=> V = D / t = 18 m / 3s = 6m/s


Answer:   

The man ran 18 m.

He got into the train after 3 s

The full speed is 6 m/s