A person standing on a platform finds that a train moving with velocity, 0.6c takes one second to pass by him. Find (a) the length of the train as seen by the person and (b) the rest length of the train .

Answer

`3.5 s`

Explanation

`(n-2)l=(1)/(2) at^(2)`
`(n-1)l=(1)/(2)a(t+3)^(2)`
`nl=(1)/(2)a (t+5)^(2)`
`l=(1)/(2)a [(1+3)^(2)-t^(2)]=(1)/(2) a[(t+5)^(2)-(t+3)^(2)]`
`t^(2)+9+6t-t^(2)=t^(2)+25+10 t-t^(2)-9-6t`
`9+6t =16 +4t`
`2t=7`
`t=3.5 sec`.