Question

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

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Answer 1

Answer:

Suppose, the man catches the train after that train moves a distance ‘x’.

So, distance moved by the man is, S = 9+x

Let, u be the speed of the man with which he runs to catch the train. If ‘t’ is the time after which he catches then,

S = ut

=> 9+x = ut ………………..(1)

Now, the train starts from rest with an acceleration of 2m/s^2

So, it travels a distance ‘x’ before the man boards it in time ‘t’.

x = 0 + ½ × 2 × t^2

so x = t^2 ............(2)

Since, the man is just able to get into the train, it means the man catches the train when the speed of the train becomes ‘u’. (for speed of the train more than ‘u’ the man will miss it)

Using first kinematics equation,

u = 0 + 2 × t

=> u = 2t ………………….(3)

(1) & (3) => 9+x = (2t)t

now

9 + x = 2t^2

9+x= 2x using (2)

x= 9m

(2) = x = t^2

=> t = 3 s

(3) => u = 2t

=> u = 6 m/s

So, the speed with which the man has to run to catch the train is 6 m/s.