A man is 48m behind a bus which is at rest. The bus starts accelerating at the rate of `1m//s^(2)`, at the same time the man starts running with uniform velocity of 10 m/s. What is the minimum time in which the man catches the bus ?

Explanation

Bus is at rest.
`thereforeu=0,a=1m//sec^(2)`
Let the bus covers the distance 's' in 'n' seconds
`therefores=ut+1/2at^(2)rArrs=0+1/2*1*n^(2)`
`therefores_((bus))=n^(2)/2`
A man running with uniform velocity, v = 10m/sec.
Distance covered by man in n seconds = 10 nm.
But after 'n' seconds the man catches the bus.
`therefore` After n seconds, `s_((man))=48+s_((bus))rArr10n=48+n^(2)/2rArr20n=96+n^(2)+96-20n=0`
`(n-12)(n-8)=0rArrn=8or12`
`therefore` The minimum time in which the man catches the bus is 18 sec.