Two boys enter a running escalator at the ground floor in a shopping mall and they do some fun on it. The first boy repeatedly foolows `p_(1)=1` step up and then `q_(1)=2` steps down whereas the second body repeatedly follws `p_(2)=2` steps up and then `q_(2)=1` step down. Both of them move rlative to escalator with speed `v_(r) =50 cm s^(-1)`. If the first boy takes `t_(1)=250 s` and the second boy takes the first boy takes `t_(1)=50 s ` to reach the first floor, how fast is escalator running ?.

Explanation

Let the length of each step is `x`. Time taken by the first bod to complete `P_(1)=1` step up and `q_(1)=2` steps down is
`t=(3x)/(v_(r)) =(3x)/(50)`
Average speed w.r.t. Escalateo, `v_(b1//e)=(-x xx50)/(3x) =(-50)/(3) cm s^(-1)`
Velcity of by (1) w.r.t. ground `=v_(e)-(50)/(3)`
So finally `(v_(e)-(50)/(3))250 =(v_(e)+(50)/(3))50`
`rArr v_(e)=25 cm s^(-1)`.